UNIT ONE
PARTIAL DIFFERENTIATION
1.0
INTRODUCTION
So far, the derivatives that you have studied in basic calculus are concentrated
on functions of one variable. In partial differentiation, we will extend our
knowledge of calculus into functions of two or more variables. As we will see,
while there are differences with derivatives of functions of one variable, if you
can do derivatives of functions of one variable you shouldn’t have any problems
differentiating functions of more than one variable. This unit shows how partial
derivatives arise and how to calculate partial derivatives by applying the rules for
differentiating functions of a single variable.
1.1
Partial Derivatives of a Functions of Two variables
Let z be called a function of two variables x and y if z has a definite value for
every pair of values x and y . Thus z  f  x, y  . Here x and y are called independent variables of the dependent variable z . z could similarly be defined for
more than two variables. z could similarly be defined for more than two variables.
We define the partial derivative of z with respect to x at the point  x0 , y0  as
the ordinary derivative of f  x, y0  with respect to x at the point x  x0 .
1
Partial Derivative with Respect to x
Definition 1.1
The partial derivative of f  x, y  with respect to x at the point  x0 , y0  is:
z
f

x  x0 , y0  x

 x0 , y0 
f  x0  h, y0   f  x0 , y0 
d
f  x, y0   lim 
h0
dx
h
provided the limits exists.
In general, if x and y are any points being differentiated to and held constant
respectively, then we denote the partial derivatives as follows:
z
f

x  x , y  x
 f x  x, y   lim 
h0
 x, y 
f  x  h, y   f  x, y 
.
h
provided the limits exists.
Definition 1.2
Partial Derivative with Respect to y
The partial derivative of f  x, y  with respect to y at the point  x0 , y0  is:
z
f

y  x0 , y0  y

 x0 , y0 
f  x0 , y0  h   f  x0 , y0 
d
f  x0 , y   lim 
h0
dy
h
provided the limits exists.
In general, if x and y are any points being differentiated to and held constant
respectively, then we denote the partial derivatives as follows:
z
f

y  x , y  y
 f y  x, y   lim 
h0
 x, y 
provided the limits exists.
2
f  x, y  h   f  x , y 
.
h
The following notations are used for the derivatives of function with respect to a
particular variable;
-
with respect to x of z  f  x, y  , we have:
zx 
-
z f

 f x  x, y 
x x
with respect to y of z  f  x, y  , we have:
zy 
z f

 f y  x, y 
y y
Note: We will be using the notations interchangeable, depending on the occasion.
Example 1.1
Finding Partial Derivatives at a Point
Find the values of f / x and f / y at the point  4, 5  if
f  x, y   x 2  3xy  y  1
Solution: To find f / x , we treat y as a constant and differentiate with
respect to x :
f

  x 2  3xy  y  1  2 x  3  1  y  0  0  2 x  3 y
x x
The value of
f / x at  4, 5 is 2  4   3 5  7 .
To find f / y , we treat x as a constant and differentiate with respect to y
:
f

  x 2  3xy  y  1  0  3  x  1  1  0  3x  1
y y
The value of
f / y at  4, 5 is 3 4   1  13 .
3
Example 1.2
Finding a Partial Derivative as a Function
Find f / y if f  x, y   y sin xy .
Solution:
We treat x as a constant and f as a product of y and sin xy :
f



  y sin xy   y  sin xy   sin xy  y 
y y
y
y

1.2
f

  y cos xy   xy   sin xy  xy cos xy  sin xy
y
y
Functions of More Than Two Variables
Although we have introduced the idea of a partial derivative for a function of two
variables, we can also find partial derivatives of a function of three or more variables, and the rule is the same: treat every variable – expect the one you are differentiating with respect to – as a constant and apply the usual differentiation
rules.
Example 1.3
A Function of Three Variables
If x , y and z are independent variables and
f  x, y, z   x sin  y  3z  ,
then
f


  x sin  y  3z    x sin  y  3z 
z z
z
 x cos  y  3z 
Example 1.4

 y  3z   3x cos  y  3z 
z
Other examples with functions of More Than Two variables
1. Find the f x and f y of f  x, y   2 x y  x sin y
3
Solution:
f x  6 x 2 y  sin y and f y  2 x3  x cos y
4
1
2. If f  x, y   x y  xy , find 
3
3
 f x

1

f y 
x 1, y  2
Solution:
f x  x, y   3 x 2 y  y 3
f y  x, y   x3  3xy 2
f x 1,2   6  8  2
f y 1,2   1  12  11
1 1
 1
1   11   2  13



  

22
22
 f x f y  x1, y 2   2   11 
1
v
v
 y
1  y 
  tan   , find the value of x x  y x .
x
 x
3. If v  sin 
Solution:
v 1
y
x 2  y  v
x
x
 
 2
 2 ,

 2

2
x y y 2  x 2 x  y  x  y y y 2  x 2 x  y 2
x
x
4. If z  e
v

x
x
y 2  x2
v
v
y 
x
y
ax by

xy
,
x  y2
y
2
x
y 2  x2

v

y
xy

x2  y 2
 f  ax  by  , prove that b
x
y 2  x2
x
y 2  x2


x
x  y2
2
xy
0
x2  y 2
z
z
 a  2abz .
x
y
Solution:
z
 aeaxby  f  ax  by   eaxby  af   ax  by 
x
z
 beaxby  f  ax  by   eaxby   b  f   ax  by 
y
5

b

a
Hence b
z
 abeaxby  f  ax  by   abeax by  f   ax  by 
x
z
 abe axby  f  ax  by   abe ax by  f   ax  by 
y
z
z
 a  2abeaxby  f  ax  by  .
x
y
5. Find all the first partial derivatives of:
g  w, x, y, z   x 2e 2 w  wyz 3  8 x  1
Solution:
gw  2 x2e2 w  yz 3 , g x  2 xe2 w  g , g y  wz 3 , and g z  3wyz 2
Implicit Partial Differentiation
Example 1.5
Find z / x if the equation
yz  ln z  x  y
Defines z as a function of the two independent variables x and y and the partial derivative exists.
Solution:
We differentiate both sides of the equation with respect to x , hold-
ing y constant and treating z as a differentiable function of x :


x y
 yz   ln z  
x
x
x x
y
z 1 z

1 0
x z x
with y constant,
1  z

  y   1
z  x


z
z

x yz  1
6

z
 yz   y
x
x
Exercise 1.1
I.
Find f / x and f / y of the following (1-10)
1.
f  x, y   2 x 2  3 y  4
7.
f  x, y   e x y1
2.
f  x, y    xy  1
8.
f  x, y   ln  x  y 
3.
f  x, y    g  t dt
9.
f  x, y   sin 2  x  3 y 
4.
f  x, y   x2  y 2
10.
f  x, y   x y
5.
f  x, y   x /  x2  y 2 
11.
f  x, y   tan 1  y / x 
6.
f  x, y   log y x
12.
f  x, y   e  x sin  x  y 
II.
III.
2
y
x
Find f x , f y and f z of the following


1.
f  x, y, z   1  xy 2  2 z 2
6.
f  x, y , z   e
2.
f  x, y, z   xy  yz  xz
7.
f  x, y, z   sin 1  xyz 
3.
f  x, y, z   x  y 2  z 2
8.
f  x, y, z   sec1  x  yz 
4.
f  x, y, z    x 2  y 2  z 2 
5.
f  x, y, z   ln  x  2 y  3z 
1/2

 x2  y 2  z 2
9. f  x, y, z   sinh xy  z
10.
2

f  x, y, z   e  xyz
Differentiating Implicitly
1. Find the value of z / x at the point 1,1,1 if the equation
xy  z 3 x  2 yz  0
defines z as a function of the two independent variables x and y
and the partial derivatives exists
2. Find the value of z / x at the point 1, 1, 3 if the equation
7
xz  y ln x  x 2  4  0
defines x as a function of the two independent variables y and
and the partial derivatives exists
8
z
1.3
Geometric Interpretation of f x and f y
For a function f  x, y  of two variables, what o the partial derivatives f x and
f y represent geometrically? Remember that for a function of one variable, the
derivative is the slope of the line tangent to the curve. When we want to take a
partial derivative of f  x, y  , we treat one of the variables as a constant. in effect, we consider f as a function of one variable, so the value of the derivative
should again be the slope of the line tangent to a curve. Let us look at this a little more closely. To form f x , we hold y as a constant (let us call it y0 ), so the
graph of the function z  f  x, y0  is a curve: It is the intersection of the surface z  f  x, y  with the plane y  y0 . The partial derivative of f with respect
to x gives the slope of the tangent line to this curve. Similarly, the partial derivative with respect to y gives the slope of the tangent line to the curve
z  f  x0 , y  , which is the intersection of the surface z  f  x, y  with the plane
x  x0 .
tangent line
to curve
EY z  f x, y at
 0
P   x0 , y0 
9
EY
tangent line
to curve
EY z  f x , y at
 0 
P   x0 , y0 
EY z
surface
z  f  x, y 
EY
curve
EY z  f x, y
 0
EY y0
EY x0
EY x
EY z
EY
EY
EY y
surface
z  f  x, y 
curve
EY z  f x , y
 0 
EY y0
EY P
EY x0
 f 

 x  P x0 , y0 
EY slope  
EY y
EY
EY P
 f 
EY slope  

 y  P x0 , y0 
EY x
Example 1.6
x2
Determine if f  x, y   3 is increasing or decreasing at  2,5 ;
y
a) If we allow x to vary and hold y fixed
b) If we allow y to vary and hold x fixed.
Solution:
a) We need f x  x, y  and its value at the point .
f x  x, y  
2x
y3

f x  2,5  
4
0
125
Hence the function is increasing at  2,5 as we vary x and hold y fixed.
b) We need f y  x, y  and its value at the point
3x 2
f y  x, y    4
y

f y  2,5   
12
0
625
Hence the function is decreasing at  2,5 as we vary y and hold x fixed.
10
Remark:
For a function of one variable, f  x  , f   a  represents the slope of the tangent
line to y  f  x  at x  a . The Partial Derivatives f x  a, b  and f y  a, b  also represent the slopes of the traces of f  x, y  for the plane y  b and x  a
at the point  a, b  respectively.
Example 1.7
Find the slopes of the traces to z  10  4 x  y at the point 1,2 
2
2
Solution:
The two partial derivatives for the slopes are:
f x  x, y   8x ,
f y  x, y   2 y
and the slopes are:
f x 1,2   8,
f y 1,2   4
So the tangent line at 1,2  for the trace to z  10  4 x  y
2
2
for the plane
y  2 has a slope of 8. Also the tangent line at 1,2  for the trace to
z  10  4 x 2  y 2 for the plane x  1 has a slope of 4.
1.4
Vector Function Equation
Let z  f  x, y  be a surface. If  a, b  are the x  y coordinates of a point on
the surface, then the equation of the surface as a vector function is:
  x, y   x, y, z  x, y, f  x, y 
If we differentiate the above with respect to x we will get a tangent vector to
traces for the plane y  b (i.e. for y ) and if we differentiate with respect to y
we will get a tangent vector to traces for the plane x  a (or fixed x ).
11
Hence the tangent vector for traces with fixed y is:
 x  x, y   1,0, f x  x, y 
and for traces with fixed x , the tangent vector is:
 y  x, y   0,1, f y  x, y 
The equation for the tangent line to traces with fixed y is then:
  t   a, b, f  a, b   t 1,0, f x  a, b 
and the tangent line to traces with fixed x :
  t   a, b, f  a, b   t 0,1, f y  a, b 
Example 1.8
Find the vector equation of the tangent lines to the traces to
z  10  4 x 2  y 2 at the point 1,2 .
Solution:
f x  x, y   8x  f x 1,2   8
f y  x, y   2 y
and
1,2, f 1,2   1,2,2
Hence the equation of the tangent line to the trace for the plane y  2 is:
  t   1,2,2  t 1,0, 8  1  t,2,2  8t
and the equation of the tangent line to the trace for the plane x  1 is:
  t   1,2,2  t 0,1, 2  1,2  t ,2  2t
Example 1.9
What is the equation of the line tangent to the intersection of the surface
 1 
z  arctan xy with the plane x  2 , at the point  2, ,  ?
 2 4
12
Solution:
Since x  2 , it means x is held constant, the slope of the tangent line we are
looking for is equal to the value of the partial derivative   / y  arctan xy 
 1
 2
at  x0 , y0    2,  :



x
x
 slope  
 arctan xy  
2
2
y
1   xy 
1   xy  


2

 1
1 2
1

2





2
 2, 12  
Since the tangent line is in the plane x  2 , this calculation shows that the
line is parallel to the vector v   0,1,1 , so we can write the equation of the
equation of the tangent line in parametric form, as follows:
x  2, y 
1

 t, z   t
2
4
Exercise 1.2
1. The plane y  1 slices the surfaces
z  arctan
x y
1  xy
in a curve C . Find the slope of the tangent line to C at the point where
x  2.
2. The equation x z  y z  3xy  1 defines an implicit function z  f  x, y  .
3 5
What is the value of
2 3
f
at the point  x, y    1,1 ?
y
3. If the variables P , V , and T are related by the equation PV  nRT , where
n and R are constants, simplify the expression
V T P


T P V
13
1.5
Partial Derivatives of Higher - Order
Just as we had higher order derivatives with functions of one variable we will also have higher order derivatives of functions of more than one variable.
However, this time we will have more options since we do have more than one
variable.
A partial derivative of a function of two or more variables is itself a function, so it
may be differentiated with respect to any of the variables. For example, after
finding f x and f y for a function f  x, y  , we may differentiate the derivatives
with respect to either x or y . The derivative of f x with respect to x is denoted
2 f
or f xx
x 2
and the derivative of f y with respect to y is denoted:
2 f
or f yy .
y 2
Let z  f  x, y  , then
z
z
and
being functions of x and y can further be
x
y
differentiated partially w.r.t. x and y . Thus,
  z   2 z  2 f

 f xx ;
 
x  x  x 2 x 2
  z   2 z  2 f


 f yy ;
y  y  y 2 y 2
The second – ordered mixed partials – which are the derivative of f x with
respect to y and of f y with respect to x - are denoted, respectively
2 f
 f xy
yx
2 f
 f yx
xy
For almost all commonly encountered functions, these mixed partials are
continuous, which ensures that they are identical; that is f xy  f yx , so the order
14
in which the derivatives are taken does not matter. That defines a theorem.
Theorem 1.1(Clairaut’s Theorem)
If f  x, y  and its partial derivatives f x , f y , f xy , and f yx are defined throughout
an open region containing a point  a, b  and are all continuous at  a, b  , then
f xy  a, b   f yx  a, b  .
Example 1.10
Let f  x, y   y ln  4 x  y 2 
4y
fx 
4x  y2
2 y2
 ln  4 x  y 2 
and f y 
2
4x  y
2
  4 y  4 4x  y 
f xy   f x  y  

y  4 x  y 2   4 x  y 2 2
  2 y2
8 y2
4
2 
f yx   f y   
 ln  4 x  y    

2
2
2
x
x  4 x  y

 4x  y2  4x  y
 f yx 
4 4x  y2 
 4x  y 
2 2
Hence f xy  f yx
Example 1.11
2
2 y
2  y
a
Prove that y  f  x  at   g  x  at  satisfies
t 2
x 2
Solution:
y
 f   x  at   g   x  at 
x
2 y
 f   x  at   g   x  at 
x 2
15
y
 af   x  at   ag   x  at 
t
2 y
 a 2 f   x  at   a 2 g   x  at 
2
t
2
2 y
2
2  y
 a  f   x  at   g  x  at    a
t 2
x 2
Exercise 1.3
1. Find all the second – order partial derivatives of the functions in a to f:
a) f  x, y   x  y  xy
d)
s  x, y   tan 1  y / x 
b) g  x, y   x y  cos y  y sin x
e)
f  x, y   sin xy
c) h  x, y   xe  y  1
f)
r  x, y   ln  x  y 
c)
z  e x  x ln y  y ln x
d)
z  x sin y  y sin x  xy
2
y
2. In a to d, verify that z xy  z yx
a) z  ln  2 x  3 y 
b) z  xy  x y  x y
2
2
3
3
4
3. Which order of differentiation will calculate f xy faster: x first or y first? Try
to answer without writing anything down:
a) f  x, y   x sin y  e
y
b) f  x, y   1/ x
c)
f  x, y   y   x / y 

4. z  ln e  e
x
2 z
r 2,
x
y
d)
f  x, y   y  x2 y  4 y3  ln  y 2  1
e)
f  x, y   x 2  5 xy  sin x  7e x
f)
f  x, y   x ln xy
 show that rt  s  0
2
2 z
t 2,
y
and
2 z
s
,
xy
16

5. If v  1  2 xy  y
2
 2



 , prove that x 1  x  xv   y  y y   0
1
2
2
1  x  4 y
6. Prove that f  x, y  
e
 x , show that f xx  x, y   f yx  x, y 
y
 3u
 3u

7. If u  x , show that
x 2y xyx
2
2
  
9
   u
8. If u  ln  x  y  z  3xyz  , show that, 
2
 x  y  z
 x y z 
3
3

3

9. If z  yf x 2  y 2 , show that y
z
z xz
x 
x
y y
2
2
2 z
 y
 y
2  z
2  z
 2 xy
y
0.
10. If z  x    y   , show that x
x 2
xy
y 2
x
x
 2u
 y
2
1  x 
11. If u  x tan    y tan   , find the value of
xy
x
 y
2
12. If u 
1
 2u  2u  2u
1
 2  2.
find
the
value
of
,
2
2
2
2

x
y
z
x

y

z


13. If x  e
r cos 
cos  r sin   and y  er cos sin  r sin   , prove that
y
1 x

.
r
r 
 2 x 1 x 1  2 x


0
Hence deduce that
r 2 r r r 2  2
1.6
Partial Derivatives of Still Higher – Order
17
x 1 y

,
r r 
So far we have only looked at first and second order derivatives, which appear
frequently in applications, there is no theoretical limit to how many times we can
differentiate a function as long as the derivatives involved exist. We can alsoform
partial derivatives of orders higher than two: f xxy or f yxy , for example
3 f
or f xxy .
xyx
Notice that by invoking the equality of mixed partials,
f xxy is equal to f xyx or
f yxx . This is an extension to Clairaut’s Theorem that says if all three of these are
continuous then they should all be equal,
f xxy  f yxx  f xyx .
Example 1.12
Find the indicated derivative for each of the following functions.
1.
Find f xxyzz for f  x, y, z   z y ln x
2.
3 f
xy
Find
for f  x, y   e
2
yx
3
2
Solution:
1.
2z3 y
6z2 y
z3 y2
z3 y2
 f xx   2  f xxy   2  f xxyz   2
fx 
x
x
x
x
 f xxyzz  
2.
1.7
12 zy
.
x2
2 f
f
3 f
2 xy
xy
 ye  2  y e 
 y 2e xy  2 ye xy  xy 2e xy
2
x
x
yx
Differentiability
18
With the functions of a single variable that if y  f  x  is differentiable at
x  x0 , then the change in the value of f that result from changing x from x0
to x0  x is given by an equation of the form:
y  f   x0   x
in which   0 as x  0 . For functions of two variables, the analogous
property becomes the definition of differentiability.
Definition 1.4 (Increment of a function of two variables)
 x0 , y0  ,
If f is a function of two variables z  f  x, y  and P is the point
then the increment of f at P is given by:
f  x0 , y0   f  x0  x, y0  y   f  x0 , y0 
Definition 1.5 (Increment of a function of n variables)
If f is a function of n variables x1 , x2 ,..., xn and P is the point
 x , x ,..., x  ,
1
2
then the increment of f at P is given by:
f  P   f  P  P   f  P 
Definition 1.6 Differentiability of a Function of Two variables
A function z  f  x, y  is differentiable at  x0 , y0  if f x  x0 , y0  and
f y  x0 , y0  exist and z satisfies an equation of the form:
z  f x  x0 , y0  x  f y  x0 , y0  y  1x   2 y ,
in which 1 ,  2  0 as x, y  0 . We call f differentiable if it is
differentiable at every point in its domain.
Definition 1.7 Differentiability of a Function of n variables
19
n
If f is a function of n variables x1 , x2 ,..., xn and the increment of f at the
point P can be written as:
n
n
i 1
i 1
f  P    Di f  P  xi   i xi
where 1  0 ,  2  0 ,…,  n  0 as  x1 , x2 ,..., xn    0,0,...,0  , then f
is said to be differential at P .
Total Differentiation
In partial differentiation of a function of two independent parameters, only one
parameter varies at a time. In the case of total differentiation, all parameters of
the independent function vary at the same time.
Given the function z  f  x, y  the total differential dz or df is given by,
dz  f x dx  f y dy or df  f x dx  f y dy
There is a natural extension to functions of there more variables. For instance,
given the function w  g  x, y, z  the differential is given by,
dw  g x dx  g y dy  g z dz
Definition 1.8
Total differential of n variables
If f is a function of n variables x1 , x2 ,..., xn and f is differentiable at P , then
the total differential of f is the function df having function values given by
n
df  P, x1, x2 ,..., xn    Di f  P  xi
i 1
Letting w  f  x1 , x2 ,..., xn  , defining dx  xi , for i  1,..., n and using the notation
w
instead of Di f  P  , we can write the above equation as
xi
20
w
dxi

x
i 1
i
n
dw  
Example 1.13
Compute the differential for each of the following functions
a) z  e
x2  y2
tan 2 x
t 3r 6
u 6
s
b)
Solution:

a) dz  2 xe
x2  y 2



tan 2 x  2e x  y sec2 2 x dx  2 ye x  y tan 2 x dy
2
2
2
2
3t 2 r 6
6t 3r 5
2t 3r 6
dt  2 dr  3 ds
b) du 
s2
s
s
NB: Differentials are sometimes called total differentials, so don’t be confused in
these words are used interchangeably.
1.8
The Chain Rule for Partial Derivatives
Previously, we used the chain rule to differentiate composite functions: If y is
functions of x (that is y  y  x  ), and x is a function of t (that is x  x  t  ),
then we find the derivative of y with respect to t from the equations:
dy dy dx
 
dt dx dt
Diagrammatically: y
x
t
For functions of a single variable, this is the only form of the chain rule we need.
But in the case of functions of more than one variable, there many different
forms of the chain rule; which one to apply depends on how many varibles there
are and how they are interrelated.
Differentiation of composite function
21
I.
If f  f  u  and u  u  x  then:
df df du


dx du dx
Diagrammatically: f
II.
u
x
If z  f  x, y  , where x  x  t  and y  y  t  , then
dz f dx f dy
   
dt x dt y dt
x
t
z
y
Illustrative Explanation (mnemonic) : Moving from z (source variable), there are two options to get to t (destination variable), that is to
pass through x or y (hence the addition sign). On the decision to pass
through x or y : that is why we have partial derivatives. But from either
x or y to t is a single decision, that is why we have ordinary derivative
for x and y  with respect to t  . dz / dt is an ordinary derivative because
t is the only destination variable.
Remark: If there are more than one destination variable from source variable we will have a partial derivative of source variable with respect with
destination variable.
III.
If z  f  x, y  where x    u, v  and y    u, v  , then
z f x f y
    ;
u x u y u
x
22
z f x f y
   
v x v y v
u
z
v
y
Here we have two destination variables u and v , so we have two
partial derivatives with respect to the two.
If f  f  u, v, y  where u  u  v, y  and v  v  x, y  , then
IV.
 f   f 
 f   u   f   v   u 
 y    y    u    y    v    u    y 
  x  u ,v   v , y   v  u , y   x   v
The subscript shows the respective variables kept constant for the
differentiation.
u
x
f
y
v
Theorem 1.2(Chain Rule)
Suppose that u is a differentiable function of the n variables x1 , x2 ,..., xn , and
each of these variables is in turn a function of the m variables y1 , y2 ,..., ym .
Suppose further that each of the partial derivatives
xi
 i  1,2,..., n; j  1,2,...m  exists. Then u is a function of y1, y2 ,..., ym and
y j
n
u
u xi


for j  1,..., m
y j i 1 xi y j
Example 1.14
1. If v  x  y where x  a cos t , y  b sin t ; a, b are constants.
3
Find
3
dv
and verify our results by a straight conversion v  v  t 
dt
23
Solution:
v  x3  y3 where x  a cos t , y  b sin t
dx
dy
dv
dv
 a sin t;
 b cos t;
 3x 2 ;
 3y2
dt
dt
dx
dy
dv
 3x 2  a sin t   3 y 2  b cos t 
dt
Then
 3a3 cos2 t sin t  3b3 sin 2 t cos t
 cos t sin t  3a3 cos t  3b3 sin t 
Verification:
v  x3  y3 ,  v  a3 cos3 t  b3 sin 3 t

dv
 3a 3 cos 2 t sin t  3b3 sin t cos t
dt
dv
 cos t sin t  3a 3 cos t  3b3 sin t 
dt
2. Let f  f  u, w, x  , where u  u  x, y  and w  w  x, y  .
a) Find a general expression for
f
x
b) Verify your answer to a) if f  u , w, x   w  2ux ,
3
u  x, y   4 xy  x  1 and w  x, y   y  xy 2
Solution:
a)
f f f u f w

  

x x u x w x
24
b) First we use the formulas given and write f explicitly in terms of
the independent variable x and y
f  f  u, w, x   w3  2ux
  y  xy 2   2  4 xy  x  1 x   y  xy 2   8x 2 y  2 x 2  2 x
3
3
2
f
 3 x  xy 2   y 2   16 xy  4 x  2
x
Now we can check our formula from part (a) gives the same result:
 f   f 
 f   u   f   w 


   
        
 x  y  x u ,w  u  w, x  x  y  w u , x  x  y
  2u    2 x  4 y  1   3w2  y 2
 2  4 xy  x  1  2 x  4 y  1  3 y 2  y  xy 2 
 16 xy  4 x  2  3 y 2  y  xy 2 
2
2
3. Let f  f  x, y, z   yz  x, where x, y, and z are the following functions
2
of t ; x  t   t , y  t   2t  3, and z  t   1  t. Find
2
df
at t  1 .
dt
Solution:
df f dx f dy y dz
    
dt x dt y dt t dt
df
  1 2t    z 2  2   2 yz  1
dt
  2t   2 1  t 2   2  2t  31  t 
Therefore,
df
2
  2t   2 1  t   2  2t  3 t  1   2
 t 1
dt t 1 
Exercise 1.4
25
1. If w  f  x, y  ; x  r cos , y  r sin  , show that
1  w   f   f 
 w 

  2
    
 r  r     x   y 
2
2
2
2. If v  f  y  z, z  x, x  y  , prove that
2
v v v
  0
x y z
 yx zx
u
u
u
,
, show that x 2
 y2
 z2
0

xy
xz

x

y

z


3. If u  u 
4. If v is a potential function, show that v  v  r  and r  x  y  z . Show
2
2
2
2
 2v  2v  2v  2v 2 v




that
x 2 y 2 z 2 r 2 r r
5. If z is a function of x and y and y and v are the two other variables such
that: u  Ix  my, v  Iy  mx, I , m are constants.
2
2 z 2 z
 2v
2
2  u

 I  m  2  2
Show that
x 2 y 2
y x
6. A function f  x, y  is rewritten in terms of new variables
u  e x cos y, v  e x sin y
Show that:
i)
f
f
f
u v
x
u
v
ii)
f
f
f
 v  u
y
v
u
2
2 f 2 f
2 f 
2
2  f

 v  x  2  2 
So deduce that
x 2 y 2
x 
 v
7. If by substitution, u  x  y , v  2 xy, f  x, y     u , v 
2
2
2
2 f 2 f
 2 
2
2  

 4 x  y  2  2 
Show that
x 2 y 2
v 
 u
Given the transformation x  cosh  cos, y  sinh  sin
Establish the following:
26
  2v  2v 
 2v  2v
2
2

  sinh   sin    2  2 
x 2 y 2
 x y 
Definition1.9 (Derivatives f :
Let f : D 
p
, where D 
n
n

p
)
, and let P0  int D . Then f is differentiable
at P0 if there is a linear function L such that
1
 f  P0  h   f  P0   L  h    0 .
h0 h 
lim
The linear function L is called the Derivative of f at P0 and L is
represented by an p  n matrix.
Note : When case n  p  1 , the matrix L is simply the 1 1 matrix whose entry
is the derivative of f .
1
 f  P0  h   f  P0   L  h    0 now translates inh0 h 
The requirement that lim
to mij 
fi
which are the components of the matrix, that is:
x j
 f1
 x
 1
 f 2

L   x1


 f p
 x1
f1
x2
f 2
x2
f p
x2
f1 
xn 
  m11 m12
f 2  
m
m22
xn    21
 
 
mp2
 m
f p   p1
xn 
Example 1.15
27
m1n 
m2 n 



m pn 
Let f :
2

2
3x1 sin x2 
. Assume f is differentia2
 x1  x1 x2 
be given by f  x1 , x2    3
ble, find the derivative (more precisely, the matrix of the derivatives) of f .
Solution:
 m11
 m21
m12 
.
m22 
The matrix of the derivatives is the 2  2 : L  
Now f1  x1 , x2   3x1 sin x2 and f 2  x1 , x2   x1  x1 x2 .
3
2
Computing the partial derivatives we have:
f1
 3sin x2
x1
f1
 3x1 cos x2
x2
f 2
 x12  x22
x1
f 2
 2 x1 x2
x2
The derivative is thus:
3sin x 3 x1 cos x2 
L   2 22

 x1  x2 2 x1 x2

Definition: 1.10
Suppose xi  xi  t  for i  1,..., n are the parametric equations of xi . Let
f :D
p
, where D 
n
and f is differentiable at P , then the total
differential of f is given by the p 1 matrix
28
df  P 
, i.e.,
dt
 m11 m12

df  P   m21 m22


dt

 m p1 m p 2
m1n   dxdt1 
m2 n   dxdt2 

or
 
 
m pn   dxdtn 
 m11 m12
m
m22
21
df  P   


 m p1 m p 2
m1n   dx1 
m2 n   dx2 
 
 

m pn   dxn 
When p  1 , we have df 
n
f
 x dx
i 1
i
i
Exercise 1.5
 xz  e y 


z log  x  y 2  

1. Find the derivative of f  x, y, z  


y


 xz 2  5 
2. Find the derivative of f  x, y   e

1.9
 x y 
, e x  y  
Implicit Differentiation
Suppose that F  x, y  is differentiable and that the equation F  x, y   c defines y as a differentiable function of x . Then at any point where Fy  0 ,
dy
F
 x
dx
Fy
29
That is: dF 
F
F
dx 
dy  0 the chain rule in differentiating F  x, y   c .
x
y
Simplifying, we get  Fx  Fy
Then 
dy
0
dx
dy
F
 x
dx
Fy
Example 1.16 Speedy Implicit Differentiation
Find the dy / dx if y  x  sin xy  0 .
2
2
Solution:
Let F  x, y   y  x sin xy
2
2
Then Fx  2 x  y cos xy and Fy  2 y  x cos xy
dy
F
2 x  y cos xy
 x 
dx
Fy
2 y  x cos xy

dy 2 x  y cos xy

dx 2 y  x cos xy
This calculation is faster than differentiating implicitly using single variable as you
have done in calculus past.
Suppose F  x, y, z   0 and assume that z  f  x, y  and we want to find
and/or
z
x
z
. In the first case we will differentiate both sides with respect to x ,
y
treating y as a constant and using the chain rule. i.e.,
F x F y F z


0
x x y x z x
Noting that
x
y
z
F
 1 and
 0 , we have
 x .
x
x
x
Fz
30
Similarly,
F
z
 y
y
Fz
Example 1.17
Find
z
z
2
and
for x sin  2 y  5 z   1  y cos6 zx
x
y
Solution:
Let F  x, y, z   x sin  2 y  5 z   1  y cos6 zx  0
2
2 x sin  2 y  5 z   6 yz sin 6 xz
z

x
5 x 2 cos  2 y  5 z   6 xy sin 6 xz
2 x 2 sin  2 y  5 z   cos6 xz
z

x
5 x 2 cos  2 y  5 z   6 xy sin 6 xz
Exercise 1.6
Assuming that the equations in 1 – 4 define y as a differentiable function of x ,
find the value of dy / dx at the given point.
1. x  2 y  xy  0
3
2
1,1
2. xy  y  3x  3  0
 1,1
3. x  xy  y  7  0
1,2 
2
2
2
4. xe  sin xy  y  ln 2  0
y
 0,ln 2 
1.10 Directional Derivatives and The Gradient
Definitions 1.11
I)
Field: If a function is defined in any region of space for every point of the
region, then this region is known as a field.
II)
Scalar point function: A function f  x, y  is called a scalar point function
if it associates a scalar with any point in space. Example: The temperature
31
distribution in a heated body, Density of a body and potential due to gravity.
III)
Vector point field: If a function F  x, y, z  defines a vector at every point
of a given region, then F  x, y, z  is called a vector point function. Examples: the velocity of a moving body, gravitational force.
Consider the surface z  f  x, y  and let P   x0 , y0  be a point in the domain
of f . What if we were asked to find the value of the derivative of f at P ? The
derivative should tell us how z changes as we move away from P , but a moment’s reflection tells us that the rate at which z changes depends on the direction in which we want to move.
z
surface
z  f  x, y 
32
y0
x0
y
P
x
For example, if we move away from P in the positive x direction, then we know
the rate of change of z will be the value of f x at P , and if we move from P in
the positive y direction, then the rate of change of z will be the value of f y at
P . But what if we wanted to know how z changes if we move from P in some
other direction? We need a way to figure out a general directional derivative.
The easiest way to determine a directional derivative is to first form the following
vector, whose components are the first – order partial derivatives of f :
 f  ˆ  f  ˆ  f f 
 i    j   , 
 x   y   x y 
This vector is called the gradient of f , and is denoted either by grad f , or
more commonly by f (the symbol  is pronounced del).
 f  ˆ  f  ˆ  f f 
 i    j   ,    fx , f y 
 x   y   x y 
Therefore f  
The Vector differential operator, del  is defined by (for three variables):
33

ˆ  ˆ  ˆ
i  j k
x y
z
The gradient of a scalar point function f  f  x, y, z  is defined by:
 

 
f
f
f
grad f  f   ˆi  ˆj  kˆ  f  ˆi  ˆj  kˆ   f x , f y , f z 
z 
x
y
z
 x y
grad f , f is a vector quantity.
Note that the unit vectors ˆi , ˆj, and k̂ will simply be written as i, j, and k
Let v  2,1 , then in this way we will know that x is increasing twice as fast as
y is. There is still a small problem with this however. There are many vectors
that point in the same direction (that x is increasing twice as fast as y is). For
instance all of the following vectors point in the same direction v  2,1 .
v
1 1
,
5 10
v  6,3
v
2 1
,
5 5
We need a way to consistently find the rate of change of a function in a given direction. We will do this by insisting that the vector that defines the direction of
change be a unit vector. Recall that a unit vector is a vector with length, or magnitude, of 1 . This means that for the example that we started off thinking about
we would want to use
v
2 1
,
5 5
since this is the unit vector that points in the direction of change
34
Sometimes we will give the direction of changing x and y as an angle. For instance, we may say that we want the rate of change of f in the direction of
   / 3 . The unit vector that points in this direction is given by,
u   cos ,sin  
The rate of change of f at the point P in the direction of u (where u is a vector whose initial point is P ) is denoted Du f P and is called the directional derivative of f at P in the direction of u . It is found by evaluating the dot product
of the gradient of f at P and the unit vector, û :
Du f P  f P  uˆ
,
which can be derived by finding the total differential:
df 
 f
f
f
f 
dx  dy   ˆi  ˆj   dxˆi  dyˆj
x
y
y 
 x


 df  f  dr
This equation verifies that f x is just the derivative of f in the  x direction
(that is in the direction î ), and f y is the derivative of f in the  y direction
(that is, in the direction ĵ ), since
Di f P  f P  ˆi   f x , f y   1,0  f x P and
Dj f P  f P  ˆj   f x , f y    0,1  f y P
but the real power of the equation is that it tells us how f will change as we
move in any direction.
Similarly, the scalar function f  x, y, z  , the total differentiation is:
df 
 f
f
f
f
f
f 
dx  dy  dz   ˆi  ˆj  kˆ   dxˆi  dyˆj  dzkˆ
x
y
z
y
z 
 x

35

 df  f  dr
Example 1.18
1. Let f  x, y   81  x  y . Find the rate of change of f as we move from
2
2
the point P  1,4  in the direction toward the point Q   4,8
Solution:
We want to move in the direction of the vector u  PQ   4  1,8  4    3,4  .
Because this vector has magnitude 5 , the unit vector in this direction is
uˆ   53 , 54  . Since the gradient of f is:


x
y
f   f x , f y   
,

 81  x 2  y 2 81  x 2  y 2 


our equation for the directional derivative tells us that:


x
y
ˆ
Du f P  f P  u  
,
   53 , 54 
2
2
2
2
 81  x  y
81  x  y  1,4

 
   18 ,  12    53 , 54 
  19
40
Since Du f P   19
40 , the line that is tangent to the surface at the point P  1,4 
and parallel to the vector PQ will descend by 19
40 unit as we move from P one
unit in the direction û .
2. Find the derivative of f  x, y   xe  cos xy at the point  2,0  in the direcy
tion of v  3i  4j
Solution:
The direction of v is obtained by dividing v by its length:
36
u
v v 3 4
  i j
v 5 5 5
The partial derivatives of f at  2,0  are:
f x  2,0    e x  y sin xy 
 2,0
 e0  0  1
f y  2,0    xe x  x sin xy 
 2,0
 2e0  2  0  2
The gradient of f at  2,0  is:
f  2,0  f x  2,0 i  f y  2,0 j  i  2 j
The derivative of f at  2,0  in the direction of v is therefore:
 Du f   2,0  f  2,0  uˆ
  i  2 j   34 i  54 j  53  85  1
Properties of Directional Derivatives
Evaluating the dot product in the formula
Du f P  f P  uˆ  f P uˆ cos  f cos (since uˆ  1 )
where  is the angle between the vectors û and f , reveals the following
properties:
1. The function f increases most rapidly when cos  1 or when û is
the direction of f . That is, at each point P in its domain, f increases most rapidly in the direction of the gradient vector f at P .
The derivative in this direction is:
Du f  f cos  0   f
2. Similarly, f decreases most rapidly in the direction of f . The derivative in this direction is Du f  f cos     f .
37
3. Any direction û orthogonal to the gradient is a direction of zero
change in f because  then equals  / 2 and
Du f  f cos  / 2   f  0  0
The properties can be summarized as the maximum value of cos  is 1 , which
occurs when   0 , so the maximum directional derivative is f P , which
occurs when û points in the same directions as f P . This property of the
gradient is important enough to repeat:
The vector f points in the direction in which f increases most rapidly,
and the magnitude of f gives the maximum rate of increase.
It is also easy to see that the vector f points in the direction in which f de-
creases most rapidly, and the negative of the magnitude of f gives the maximum rate of decrease.
Example 1.19 Finding Directions of Maximal, Minimal, and Zero Change

 
Find the directions in which f  x, y   x / 2  y / 2
2
2

(a) Increases most rapidly at the point 1,1
(b) Decreases most rapidly at 1,1
(c) What are the directions of zero change in f at 1,1 ?
Solution:
a) The function increases most rapidly in the direction of f at 1,1 . The
gradient there is:
 f 1,1   xi  yj1,1  i  j
Its direction is:
38
uˆ 
ij

ij
ij
1  1
2
2

1
1
i
j
2
2
b) The function decreases most rapidly in the direction of f at 1,1 ,
which is:
uˆ  
1
1
i
j
2
2
c) The direction of zero change at 1,1 are the directions orthogonal to f
:
n
1
1
1
1
i
i
j and n 
j
2
2
2
2
Algebra Rules for Gradients
1. Constant Multiple Rule:   kf   kf
2. Sum Rule:
  f  g   f  g
3. Difference:
  f  g   f  g
4. Product Rule:
  fg   f g  gf
5. Quotient Rule:
 f  gf  f g
  
g2
g
Example
We illustrate the rules with
f  x, y   x  y
g  x, y   3 y
f  i  j
g  3j
We have:
39
1.   2 f     2 x  2 y   2i  2 j  2  i  j  2f
2.   f  g     x  2 y   i  2 j  i  j  3j   i  j  3j  f  g
3.   f  g     x  4 y   i  4 j  i  j  3j   i  j  3jf  g


4.   fg    3xy  3 y 2  3 yi   3x  6 y  j
 3 y  i  j  3 yj   3x  6 y  j  3 y  i  j   3x  3 y  j
 3 y  i  j   x  y  3j  gf  f g
f 
x y
 x 1
 
   


g
 3y 
 3y 3 
5.  

1
x
3 yi  3xj 3 y  i  j   3x  3 y  j
i 2 j

3y 3y
9 y2
9 y2

3 y  i  j   x  y  3j gf  f g

9 y2
g2
Exercise 1.7
1. Find the gradient of the function at the given point
a) f  x, y   y  x

 2,1
 1,1
b) f  x, y   ln x  y
2
c) g  x, y   y  x
 1,0 
d) g  x, y  
2
2
 2,1
x2 y 2

2
2
2. Find f at the given point
a) f  x, y, z   x  y  2 z  z ln x
2
2
2

b) f  x, y, z   2 z  3 x  y
3
c)
2
2
1,1,1
 z  tan xz 1,1,1
1
f  x, y, z   e x y cos z   y  1 sin 1 x
40
 0,0,  / 6 
3. Find the derivative of the function at P0 in the direction of A .
a) f  x, y   2 xy  3 y , P0  5,5  ,
A  4i  3 j
b) f  x, y   2 x  y , P0  1,1 ,
A  3i  4 j
2
2
2


c) g  x, y   x  y 2 / x  3sec1  2 xy  , P0 1,1 ,
d) f  x, y, z   xy  yz  zx P0 1, 1,2  ,
e) f  x, y, z   3e cos yz P0  0,0,0  ,
x
A  12i  5j
A  3i  6 j  2k
A  i  jk
4. Find the directions in which the functions increase and decrease most rapidly
at P0 . Then find the derivatives of the functions in these directions.
a) f  x, y   x  xy  y , P0  1,1
2
2
b) f  x, y, z   ln xy  ln yz  ln zx P0 1,1,1 ,
UNIT 2
APPLICATIONS TO PARTIAL DERIVATIVES
2.0
INTRODUCTION
In this unit we will take a look at a couple of applications of partial derivatives.
Most of the applications will be extensions to applications to ordinary derivatives
41
that we have done in basic calculus. For instance, we will be looking at finding
the absolute and relative extrema of a function and we will also be looking at optimization.
2.1
Tangent Planes and Linear Approximations
Just as the derivative of a function of one variable can be used to find the equation of a tangent line to a curve, the partial derivatives of a function of two variables can be used to find the equation of a tangent plane to a surface.
For a one variable function y  f  x  , we talk about a tangent line, but for a
function of two variables we talk about tangent plane. In single variable,
y  f  x  , the equation of the tangent line is given as y  y0 
were
dy
 x  x0  ,
dx P
dy
is an ordinary derivative at P or gradient of the line at P .
dx
Let z  f  x, y  be the equation of a surface in 3  space, and let f x and f y denote, as usual, the first partial derivatives of f . Then the equation of the tangent plane to the surface at P   x0 , y0 , z0  is:
z  z0  f x P   x  x0   f y P   y  y0 
assuming that f x and f y are not both equal to zero at P . The equation of a
surface can also be given in the form f  x, y, z   c , for some constant c . The
difference is that z is now written explicitly in terms of x and y , as is the case
if the surface is described by an equation of the form z  f  x, y  . To take care
of this situation, simply use the equation f  x, y, z   c to to find z / x and
z / y by implicit differentiation; the equation of the tangent plane to the surface at the point P   x0 , y0 , z0  is:
42
z  z0 
z
z
  x  x0  
  y  y0 
x P
y P
assuming once again that z / x and z / y do not both equal zero at P .
Example 2.1
1. Find the plane tangent to the surface z  x cos y  ye at  0,0,0  .
x
Solution:
We calculate the partial derivatives of f  x, y   x cos y  ye at  0,0  :
x
f x  0,0    cos y  ye x 
 0,0
f y  0,0     x sin y  e x 
 1  0 1  1
 0,0
 0  1  1
For the tangent plane formula which is given as:
z  z0  f x P   x  x0   f y P   y  y0  ,
we have:
z  0  1  x  0  1  y  0
z x y
2. Find the equation of the tangent plane to the surface
x 2 z  y 3 z 5  xy  9
at the point P  1,2, 1 .
Solution:
Since the equation of this surface is given in the form f  x, y, z   c , we
use implicit differentiation to find the partial derivatives. First, we hold y
fixed and differentiate implicitly with respect to x to find z / x :
x 2 z  y 3 z 5  xy  9
43
z
 2 z

 2 xz   5 y3 z 4  y  0
x
x
 x


z 2 xz  y

x x 2  5 y 3 z 4
Next, we hold x fixed and differentiate implicitly with respect to y to find
z / y :
x 2 z  y 3 z 5  xy  9
x2

z  3 4 z
 5y z
 3 y2 z5   x  0
y 
y

z 3 y 2 z 5  x


y x 2  5 y 3 z 4
Evaluating each of these at the point P  1,2, 1 gives:
2 1 1  2
z
2 xz  y
 2
 2
0
3 4
x P x  5 y z 1,2,1 1  5  23   14
3 22   1  1 1
z
3 y2 z5  x



y P x 2  5 y 3 z 4 1,2,1 12  5  23   14 3
5
so the equation of the tangent plane at P is z  1 
equivalently;
2.1.1
1
 y  2  , or,
3
y  3z  5
Linearization and Linear Approximations
Definition 2.1
The linearization of a function f  x, y  at a point  x0 , y0  where f is differentiable is the function:
44
L  x, y   f  x0 , y0   f x  x0 , y0  x  x0   f y  x0 , y0  y  y0 
The approximation
f  x, y   L  x, y 
is the standard linear approximation of f at  x0 , y0  .
Example 2.2
Find the linearization of f  x, y   x  xy  12 y  3 at the point  3,2  .
2
2
Solution:
We first evaluate f , f x , and f y at the point  x0 , y0    3,2  :
f  3,2    x 2  xy  12 y 2  3
3,2
8
f x  3,2    2 x  y  3,2  4 f y  3,2     x  y  3,2  1
giving
L  x, y   f  x0 , y0   f x  x0 , y0  x  x0   f y  x0 , y0  y  y0 
 8   4 x  3   1 y  2   4 x  y  2
The linearization of f at  3,2  is L  x, y   4 x  y  2
Linear Approximation
The equation of the tangent plane to a surface z  f  x, y  at a point
P0   x0 , y0  provides a first – order (linear) approximation of the value of z at
45
points near P0 . To be more specific, let us say we wanted to evaluate f at a
point P1   x1 , y1  that is close to P0 . Since P1 is close to P0 , the value of z at
P1 on the tangent plane is close to the value of z at P1 on the surface. So, if
z1  f  x1 , y1  , we can use the equation of the tangent plane at P0 to say that:
z  z1  z0 f x P   x  x0   f y P   y  y0 
Example 2.2
What is an approximate value for this expression?
1.1 log 4  0.05  1.1 
2
Solution:


If we let f  x, y   y log 4 x  y 2 , then we are being asked to
approximate the value of f  0.05,1.1 is very close to the point
P0   0,1 , at which the exact value of f is easy to calculate:
z0  f  x0 , y0   1log  0  12   0
So if we figure out the equation of the plane tangent to the surface
z  f  x, y  at the point P0 ,
then, the value of f at P1 will be approximately equal to the value of z at
P1 on the tangent plane.
The first – order partial derivatives of f  x, y  are given as(already done):
4x
2 y2
fx 
and f y 
 log  4 x  y 2 
2
2
4x  y
4x  y
Evaluating each of these at the point P0 gives:
4y
2 y2
fx P 
 4 and f y 
 log  4 x  y 2 
2
2
2
P
0
0
4 x  y  0,1
4x  y
 0,1
46
so the equation of the tangent plane to the surface at P0 is:
z  0  4  x  0   2  y  1 , or,
more simply:
z  4x  2 y  2
Since P1   0.05,1.1 is close to P0   0,1 , the value of z at P1 on the
tangent plane,
z  4  0.05  2 1.1  2  0.4
Should give a close approximation to the value of z at P1 on the surface.
[Note: A calculator would tell you that 1.1 log  4  0.05   1.1   0.378 , so
2


our approximation of 0.4 is pretty good.]
Exercise 2.1
1. Find the equation of the tangent plane to z  ln  2 x  y  at  1,3 .
2. Find the linearization L  x, y, z  of f  x, y, z   x  xy  3sin z at the point
2
 x0 , y0 , z0    2,1,0 .
2.2
Gradient Vector, Tangent Planes and Normal Lines
In this unit we want to revisit tangent planes only this time we will look at
them in light of the gradient vector. In the process we will also take a look
at a normal line to a surface.
47
Before then, let us look at some vector calculus:
Formulae of Differentiation: Let F, G be vectors and  a scalar functions of the variable t .
I)
d
dF dG
(Linearity Rule)
F  G   
dt
dt
dt
II)
d
d
dF
 F   F   (Scalar Multiple,  )
dt
dt
dt
III)
d
dG dF

G (Dot Product Rule)
F G   F
dt
dt
dt
IV)
d
dF
dG
(Cross Product Rule):
F  G    G  F 
dt
dt
dt
Note: The order is important.
dh  t  dF  h  t  
d
F  h  t    h  t  F  h  t  

(Chain Rule)
dt
dt
dt

V)

Example 2.3
1. Find a tangent vector at the point where t  3 on the vector function
F  t   e2t i   t 2  t  j  ln t k
Solution:
1
F  t   2e2t i   2t  1 j  k
t
The required tangent vector at t  3
1
F  t  t 3  2e6i  5 j  k
3
2
2. A particle moves along the curve x  t  1, y  t , z  2t  5, where t is
3
the time. Find the component of its velocity and acceleration at time t  1
in the direction 2i  3j  6k .
48
Solution:
r  xi  yj  zk   t 3  1 i  t 2 j   2t  5 k
Velocity vector
v  3t 2i  2tj  2k

v t1  3i  2 j  2k
However the unit vector along:
2i  3j  6k :
2i  3j  6k 1
  2i  3j  6k  .
7
49
Therefore, component of velocity along the given vector:
  3i  2 j  2k  
Acceleration 
1
1
24
 2i  3j  6k    6  6  12   units
7
7
7
dv
 6t i  2 j. 
dt
dv
 6i  2 j
dt t 1
Hence acceleration along the given vector is:
  6i  2 j  0k  
1
1
18
 2i  3j  6k   12  6  
7
7
7
3. The position vector of a particle at time t is given by
r  cos  t  1 i  sinh  t  1 j   t 3k . Find the condition imposed on 
by requiring that at time t  1, the acceleration normal to the position
vector.
Solution:
r  cos  t  1 i  sinh  t  1 j   t 3k
dr
  sin  t  1 i  cosh  t  1 j  3 t 2k
dt
d 2r
   cos  t  1  i  sinh  t  1 j  6 t k

dt 2 
49
d 2r
 i  6 k
At t  1, r  i   k ,
dt 2
 1  6 2  0   2 
  
1
6
1
6
4. At any point of the curve x  3cos t , y  3sin t , z  4t , find
i)
Tangent vector
ii)
Unit tangent vector
iii)
Normal vector
iv)
Unit normal vector
Solution:
x  3cos t , y  3sin t , z  4t
r  xi  yj  zk  3cos t i  3sin t j  4t k
r
i)
Tangent vector, T 
ii)
Unit tangent vector
dr
 3sin t i  3cos t j  4k
dt
ˆ  T  3sin t i  3cos t j  4k  1  3sin t i  3cos t j  4k 
T
T
9sin 2 t  9cos 2 t  16 5
iii)
By definition, the normal vector:
However,

ds dr

 5,
dt dt

dT dT dt


ds dt ds
dt 1

ds 5
dT 1 d
1
   3sin t i  3cos t j  4k    3cos t i  3sin t j
ds 5 dt
5
50
iv)
Unit
Normal:
1
1
 3cos t i  3sin t j  3cos t i  3sin t j
 5
5
1
1
 3cos t i  3sin t j 5 9cos2 t  9sin 2 t
5
1
 3cos t i  3sin t j
5
  cos t i  sin t j
 Unit normal 
3
5
Let F  t   i  tj  t k and G  t   t i  e j  3k
t
2
5.
Verify that  F  G   t    F  G  t    F  G  t 
Solution:
i j
 F  G  t   1 t
t et
k
t 2   3t 2  t 2et  i   3  t 3  j   et  t 2  k
3
 F  G   t   3  2tet   t 2et  i  3t 2 j   et  2t  k
Next find  F  G  t  ,  F  G  t  and add.
i j
 F  G  t   0 1
t et
k
2t   3  2tet  i  2t 2 j  t k
3
i j
 F  G t   1 t
1 et
k
t 2  t 2et i  t 2 j   et  t  k
0
 F  G t    F  Gt   3  2tet  t 2et  i  3t 2 j   et  2t  k
Exercise 2.2
51
1. The coordinates of a moving particle are given by x  4t 
t2
and
2
t3
y  3  6t  . Find the velocity and acceleration of the particle when
6
t  2secs
2. A particle moves along the x  2t , y  t  4t and z  3t  5 where t is
2
2
time. Find the components of its velocity and acceleration at time t  1 in the
direction i  3j  2k.
3. Find the unit tangent vector and unit tangent vector at t  2 on the curve
x  t 2  1, y  4t  3, z  2t 3  6t where t is any variable.
4. Find the angle between the tangents to the curve r  t i  2tj  t k at
2
the point t  1 .
Tangent Plane Formula using the Gradient
52
3
Re – call from unit one that
f  x, y, z  be scalar point function, then the total
differential is:
df 
f
f
f
dx  dy  dz
x
y
z
 f
f
f 
  i  j  k    idx  jdy  kdz 
y
z 
 x
 f  dr where r  xi  yj  zk and dr  dxi  dyj  dzk
 f dr cos ,  is an angle between f and dr
Let f  x, y, z  be a given function, and let P   x0 , y0 , z0  be a point on the
level surface
f  x, y, z   c, c constant. Since the value of f does not change
on the surface, the directional derivative of f at P should equal zero no matter
in what direction we move. That is, if û is any vector tangent to the level surface at P , then   0 . So, according to our formula for the directional derivative, the dot product of the vector f with any vector û tangent to the surface
is equal to zero. This tells us that f is perpendicular to every tangent vector at
P , which means f is perpendicular to the tangent plane at P . In other
words:
For a function f  x, y, z  , the vector f p is perpendicular (normal) to the
curve of f that contains P.
By similar reasoning, we can also conclude that:
For a function f  x, y  , the vector f P is perpendicular (normal) to the
level curve of f that contains P .
53

Since the vector f p  f x p , f y , f y
p
p
 is normal to the level surface - and ,
therefore, to the tangent plane – of f  x, y, z  that contains P , the equation of
the tangent is given as:
f x p   x  x0   f y   y  y0   f z p   z  z0   0
p
Notice that every surface whose equation is given in the form z  f  x, y  can
be rewritten as
f  x, y   z  0 , which is a level surface of the function
g  x, y, z   f  x, y   z , whose gradient is g   f x , f y , 1 . So, the equation
of the tangent plane to the level surface g  x, y, z   0 is:
f x p   x  x0   f y   y  y0     z  z0   0
p
which is equivalent to the equation of the tangent plane to the surface
z  f  x, y  that we gave earlier.
Example 2.4
1. If   3x y  y z , find  at the point 1, 2, 1.
2
3 2
Solution:
 

 
   i
 j  k   6 xyi   3x 2  3 y 2 z 2  j   2 y 3 z  k
y
z 
 x
 1,2,1  12i  9 j  16k
2. Find the unit normal to the surface
x 2  y 2  z  1 at P 1,1,1.
Solution:
  x, y , z   x 2  y 2  z  1
 

 
   i
 j  k   2 xi  2 yj  k
y
z 
 x
54

 1,2,1  2i  2 j  k
 
 2i  2 j  k 1

  2i  2 j  k 

4  4 1 3
3. Let f  x, y   81  x  y . Find the rate of change of
2
2
f as we move
from the point P  1,4  in the direction toward the point Q   4,8
Solution:
Let u  PQ   4  1,8  4    3,4 

uˆ 
u 1
3 4
  3,4    , 
u 5
5 5


 
The Grad is: f  f x , f y  
x
2
2
 81  x  y
,


2
2 
81  x  y 
y
Our equation for the directional derivative tells that:


x
y
3 4
ˆ
Du f p  f p  u  f  
,
  , 
 81  x 2  y 2 81  x 2  y 2 

 1,4  5 5 
19
 1 1 3 4
  , g ,   
40
 8 2 5 5
Since Du f p  
19
, the line that is tangent to the surface at the point
40
P  1,4  and parallel to the vector PQ will descend by
move from P one unit in the direction û .
55
19
unit as we
40
4. Use the gradient to find the equation of the tangent plane to the surface
x 2 z  y 3 z 5  x  9 at the point P 1,2, 1.
Solution:
If f  x, y, z   x z  y z  xy, then the given equation represents the
2
3 5
level surface of f that contains the point P. Since the gradient of f is:
f   f x , f y , f z    2 xz  y, 3 y 2 z 5  x, x 2  5 y3 z 4 
the vector f p  f 1,2,1   0,13, 39 is normal to the tangent plane


at P .
Therefore, the equation of the tangent plane is:
0  x  1  13 y  2   39  z  1

y  3z  5
5. The temperature at any point in space is given by T  xy  yz  zx . De-
ˆ  3i  4k at
termine the derivative of T in the direction of the vector u
the point P 1,1,1
Solution:
T  xy  yz  zx but T  i
T
T
T
j
k
x
y
z
T  i  y  z   j x  z   k  y  x 
56
6. The temperature at a point in a space is T  x, y, z   x  y  z. A mos2
2
quito located at 1,1,2  desires to fly in such a direction that it will get
warm as soon as possible. In what direction should it move?
Solution:
T  x 2  y 2  z but T  i
T
T
T
j
k
x
y
z
T  2xi  2 yj  k T 1,1,2  2i  2 j  k
 T is the direction be the warmth it felt most
T 
2i  2 j  k 1
  2i  2 j  k 
4  4 1 3
Exercise 2.3

1. Evaluate grad   log x2  y 2  z 2

1
1
   3 r where R  r
R
R
2. If r  x i  yj  zk , prove that  
2
3. Find the directional derivative of the scalar function f  x, y, z  
1
at
x  y2  z2
2
the point 1, 2,2  in the direction u  i  j  k
4. For the function f  x, y  
x
, find the magnitude of the directional derivax  y2
2
tive along a line inclined at 30 with the axis at  0,1 .
5. Find the derivative of the scalar field function f  x, y, z   xyz in the directional
of the normal to the surface z  xy at the point  3,1,3.
57
2.3
Maximum and Minimum Problem
Relative Extrema: The function f  x, y  is said to have a relative maximum
at the point P0  x0 , y0  in the domain of f if f  x0 , y0   f  x, y  for all points
 x, y . Similarly, if f  x0 , y0   f  x, y  , then f  x, y  has a relative minimum at P0  x0 , y0  .
Therefore we say that P0 is a critical point of f  x, y  if the first partial
derivatives of f are both equal to zero there:
P0 is critical point of f  x, y  if:
f
f
0
 0 and
y P0
x P0
z
saddle
point
hj
y0
x0
y
P0
x
If at P0  x0 , y0  , the curve z  f  x, y0  has a maximum at x  x0 , but the
curve z  f  x, y0  has a minimum at y  y0 (vice versa), then we say that
f  x, y  has a saddle point at P0 . Although P0 is a critical point, a saddle
point is neither a local maximum nor a local minimum for the function.
58
The extreme values of f  x, y  can occur only at:
i)
Boundary points of the domain of f
ii)
Critical Points (interior points where f x  f y  0 or points where f x
or f y fail to exist).
Let  be the determinant of the following 2 by 2 matrix, which contains the
second partial derivatives of f , evaluated at P0 :
 f xx
  det 
 f yx
f xy 
2
 f xx  P0  f yy  P0    f xy  P0  

f yy 
(The matrix is called the Hessian Matrix of f  x, y  , and its determinant,  , is
called the Hessian of f . We also assume f xy  f yx if they are continuous at
P0 
Definition 2.2 Saddle Point
A differentiable function f  x, y  has a saddle point at a critical point P0 if in
every open disked centred at P0 there are domain points  x, y  where
f  x, y   f  x0 , y0  and domain points  x, y  where f  x, y   f  x0 , y0  . The
corresponding point
 x , y , f  x , y  on the surface z  f  x, y  is called a
0
0
0
0
saddle point of the surface.
Then the following statements hold:

If   0 and f xx  P0   0 , then f attains a local maximum at P0 .

If   0 and f xx  P0   0, then f attains a local minimum at P0 .

If   0, then f has a saddle point at P0 .

If   0, then no conclusion can be drawn and f may have either a relative extremum or a saddle point (that is any of these behaviours is possible).
59
Example 2.5
1. Locate and classify the critical points of the function:
f  x, y   x 2  y 3  6 xy
Solution:
The first step is to find f x and f y , and set both equal to zero:
f x  2x  6 y  0
f y  3 y2  6x  0
From the first equation, we see that x  3 y, and substituting this into the
second equation gives us:
3 y2  63 y   0

y2  6 y  y  y  6  0

y  0,6
Therefore the function has two critical points, P1   0,0  and P2  18,6  .
To classify them, we figure out the Hessian of
 f xx
  det 
 f yx
f xy 
 2 6 
 det 

  12 y  36
f yy 

6
6
y


The value of  at P1   0,0  is 12  0   36  36;
Since   0, P1 is a saddle point.
The value of  at P2  18,6  is 12  6   36  36;
Since   0, and f xx  2  0, the point P2 is a minimum.
60
2.4
Absolute Maximum and Minimum
In this section we are want to optimize a function, that identify the absolute
minimum and/or the absolute maximum of the function, on a given region in
2
We organize the search for the absolute extrema of a continuous function
f  x, y  on a closed and bounded region R into three steps.
Step 1: List the interior points of R where f may have local maxima and minima
and evaluate f at these points. These are the points where f x  f y  0 or
where one or both of f x and f y fail to exist (the critical points of f ).
Step 2: List the boundary points of R where f has local maxima and minima
and evaluate f at these points. We show how to do this shortly.
Step 3: Look through the lists for the maximum and minimum values of f .
These will be the absolute maximum and minimum values of f on R . Since absolute maxima and minima are also local maxima and minima, the absolute maximum and minimum values of f already appear somewhere in the lists made in
steps 1 and 2. We have only to glance at the lists to see what they are.
Example 2.6
Find the absolute maximum and minimum values of
f  x, y   2  2 x  2 y  x 2  y 2
on the triangular plate in the first quadrant bounded by the lines x  0 , y  0 ,
and y  9  x .
61
Solution:
Since f is differentiable, the only places where f can assume these values are
points inside the triangle where f x  f y  0 points on the boundary.
Interior Points
For these we have:
f x  2  2x  0 , f y  2  2 y  0 ,
yielding the single point 1,1 . The value of there is:
f 1,1  4 .
Boundary Points
We take the triangle one side at a time:
1. On the segment OA , y  0 . The function
f  x, y   f  x,0   2  2 x  x 2
may now be regarded as function of x defined on the closed interval
0  x  9 . Its extreme values may occur at the endpoints:
x  0 where f  0,0   2
x  9 where f  9,0   2  18  81  61
and at the interior points where f   x,0   2  2 x  0 . The only interior point
where f   x,0   0 is x  1 , where
f  x,0   f 1,0   3 .
2. On the segment OB , x  0 and
f  x, y   f  0, y  2  2 y  y 2
y is on the closed interval 0  y  9 .
Its extreme value may at the endpoints and interior points as we did for line
segment OA :
62
y  0 where f  0,0   2
so
y  9 where f  9,0   2  2  9   92  61
f   0, y   2  2 y  0  y  1, where
and
f  0,1  2  2 1  1  3 .
2
3. We have already accounted for the values of f at the endpoints of AB , so we need only look at the interior pins of
AB .
y  9 x,
With
we
have
f  x, y   2  2 x  2  9  x   x 2   9  x   61  18 x  2 x 2
2
Setting f   x,9  x   184 x  0 gives:
x  184  92 .
At this value of x ,
y  9  92  92 and f  x, y   f  92 , 92    412 .
Summary
We list all the candidates: 4,2, 61,3,   41/ 2  . The maximum is 4 , which
ulj y
f assumes at 1,1 . The minimum is 61 , which f assumes at  0,9  and
 9,0 .

ulj
ulj B 0,9

ulj y  9  x
ulj x  0
ulj
ulj
 92 , 92 
 
ulj 1,1
ulj ulj
ulj ulj
ulj 0
ulj y  0

ulj
ulj A 9,0
63

2.


Let S be closed square region S   x, y  : 0  x  2 and 0  y  2
Determine the absolute minimum and maximum values on S of the function
f  x, y   x 2  4 xy  y 2  5 x.
Solution:
Since S is closed (because it contains its boundaries) and bounded (because it’s a
subset of a circular region of finite radius), any continuous function defined on S
is guaranteed to attain an absolute maximum and absolute minimum value,
and these values may occur either at critical points or at points on the boundary of
S.
f x  2x  4 y  5  0
f y  4x  2 y  0

1
1 
x  , y  1, which is the critical point, i.e. P  ,1 , which is an
2
2 
interior of the square S .
On the boundaries, we have
f  0,0   0
f  0,2   4
f  2,0   6
f  2,2   6
 f xx
  det 
 f yx
f xy 
2 4 

det
 4 2  4  16  20  0
f yy 


1 
,1 is a saddle point.
2


The critical point, P 
The absolute minimum, f  2,0   6
The absolute maximum, f  2,2   6
64
Exercise 2.4
1. Find all the critical points for the following functions and classify each as a relative maximum, a relative minimum, or a saddle point.
i)
f  x, y   x 2  y 2
f  x, y   x 3  y 3  6 xy
iii)
f  x, y   12 x  x3  4 y 2
ii)
iv)
f  x, y   5 x 2  10 xy  20 x  7 y 2  5 y
f  x, y    x  4  ln xy
v)
f  x, y   xy 
vi)
8 8

x y
2. Find the absolute maxima and minima of the functions on the given domains.
a) f  x, y   2 x  4 x  y  4 y  1 on the closed triangular plate bounded by
2
2
the lines x  0 , y  2 , y  2 x in the first quadrant.
b) f  x, y   x  y on the closed triangular plate bounded by the lines x  0 ,
2
2
y  0 , y  2 x  2 in the first quadrant.
c) T  x, y   x  xy  y  6 x  2
2
2
on
the
rectangular
plate
0 x 5,
3  y  0 .
d) f  x, y   48 xy  32 x  24 y on the rectangular plate 0  x  1 , 0  y  1 .
3

e) f  x, y   4 x  x
f)
2
2
 cos y on the rectangular plate 1  x  3 ,  / 4  y / 4
f  x, y   4 x  8xy  2 y  1 on the triangular plate bounded by the lines
x  0 , y  0 , x  y  1 in the first quadrant.
3. Find two numbers a and b with a  b such that:
i)
  6  x  x dx
ii)
2
  24  2x  x  dx
b
2
a
b
1/3
a
has its largest value.
65
2.5
Maximum and Minimum Problems with a Constraint
Solving extreme value problems with algebraic constraints on the variables usually require the method of Lagrange multipliers in the next section. But sometimes
we can solve such problems directly as we want to do now.
Often, a question will ask us to maximize or minimize a function subject to a
constraint. For example, “What is the value of the function f  x, y   xy subject
to the constraint x  y  1 ?” The constraint equation limits the possible choic2
2
es of x and y , so the question really asks, “For all pairs  x, y  such that
x 2  y 2  1 , which pair(s) maximize the value of f  x, y   xy ?” One way to
solve this problem is to use the constraint equation to rewrite the function to be
maximized or minimized in terms of just one variable, and then proceed as we
did in the one – variable case. For example, solving this constraint equation for
y gives us:
y   1  x2
Let us ignore the minus sign for the moment, and substitute y  1  x
into f  x, y   xy :


g  x   f x, 1  x 2  x 1  x 2 , for 1  x  1
To maximize g , we first set its derivative equal to zero and solve for x :
g x  
 x2
1  x2
 1  x2  0
  x2  1  x2   0
x
1
2
66
2
(Had we used y   1  x , we would have found the same critical val2
ues of x ). Since g  x   0 at both endpoints of the closed interval
1  x  1 , we can say that g  x  attains a maximum value of 12 at
x  12 and a minimum value of  12 at x   12 . Since y   1  x 2 , we
see that there are four critical points:
P1 
 , , P   ,  , P   ,  , P   , 
1
2
1
2
2
1
2
1
2
3
1
2
1
2
1
2
4
1
2
We can now conclude that, subject to the given constraint, the function
f attains a maximum value of 12 at P1 and P4 and minimum value of  12
at P2 and P3 .
Example 2.7
1. Find
the
P  x, y, z 
point
closest
to
the
origin
on
the
plane
2x  y  z  5  0 .
Solution:
The problem asks us to find the minimum value of the function
OP 
 x  0   y  0   z  0
2
2
2
 x2  y 2  z 2
subject to the constraint that
2x  y  z  5  0
Since OP has a minimum value wherever the function
f  x, y , z   x 2  y 2  z 2
has a minimum value, we may solve the problem by finding the minimum value
of f  x, y, z  subject to the constraint 2 x  y  z  5  0 (thus avoiding square
67
roots). If we regard x and y as the independent variables in this equation and
write z as
z  2x  y  5
our problem reduces to one of finding the points  x, y  at which the function
h  x, y   f  x, y,2 x  y  5  x 2  y 2   2 x  y  5 
2
has its minimum value or values. Since the domain of h is the entire xy  plane,
the first derivative test tells us that any minima that h might have must occur at
points where
hx  2 x  2  2 x  y  5 2   0 hy  2 y  2  2 x  y  5   0
This leads to
10x  4 y  20
4x  4 y  10
and the solution:
x  53 ,
y  56
We may apply second derivative test to show that these values minimize h .
The z  coordinate of the corresponding point on the plane z  2 x  y  5 is
z  2  53   56  5   56
Therefore, the point we seek is
Closest point: P  53 , 65 ,  65 
The distance from P to the origin is
OP  x 2  y 2  z 2 
2.
 53    65     65   5 66  56  2.04
2
2
2
Find the points closest to the origin on the hyperbolic cylinder
x2  z 2  1  0
Solution:
Let
f  x, y , z   x 2  y 2  z 2
68
[Square of the distance]
if we regard x and y as independent variables in the constraint
equation, then
z 2  x2  1
and substituting into the square of distance function, we have:
h  x, y   x2  y 2   x2  1  2 x 2  y 2  1
hx  4 x  0  x  0, hy  2 y  0  y  0 ,
that is, at the point  0,0  .
But now we are trouble: Geometrically, there are no points on the
cylinder where both x and y are zero. Theoretically, finding
z
would gives us a complex number from z  x  1.
2
2
We can avoid this problem if we treat y and z as independent
variable such that:
k  y, z    z 2  1  y 2  z 2  2 z 2  y 2  1
Then k y  2 y  0
k z  4 z  0 or where y  z  0 . This leads to
x 2  z 2  1  1, x  1
The corresponding points on the cylinder are  1,0,0  . We can
see from the inequality
k  y, z   1  y 2  2 z 2  1
that the points  1,0,0  give a minimum value for k .
69
2.6
The Lagrange Multiplier Method
Another method of maximizing or minimizing a function subject to a constraint is
the Lagrange method. The Lagrange method is use if the constraint equation (s)
cannot be easily solved for one of the variable.
Here, if the function
f  x, y  or f  x, y, z  is subject to the constraint
g  x, y   c or g  x, y, z   c , then:
f  g
for some scalar  (assuming g  x   0 ). The scalar  is called the Lagrange
multiplier. Its value is unimportant; it only serves to help us figure out the critical point P .
Example 2.8
1. Use the Lagrange multiplier method to determine the extreme values of the
function f  x, y   xy , subject to the constraint x  y  1 .
2
2
Solution:
f  x, y   xy and g  x, y   x 2  y 2  1
f  f x , f y  y, x , g  g x , g y  2 x,2 y , g   2 x,2 y
f  g
 y, x     2 x,2 y 
The system of equations to solve is:
  y / x2 in the first equation,
  x / 2 y in the second equation and
x2  y 2  1
 x 2  y 2 , and substituting into the constraint, x 2  y 2  1 :
 x2  x2  1
70
x  1 / 2 and y  1 / 2 and we have four critical points:
 ,  , P   ,  , P   ,  , P   , 
f  x, y   xy , f  ,     , f   ,       
P1 
1
2
1
2
1
2
2
1
2
1
2

1
2
1
2
1
2
3
1
2
1
2
1
2
1
2
1
2

f  x, y   xy , f  12 , 12   12  12   12 , f
1
2
4
1
2
 ,  
1
2
1
2
1
2
1
2
1
2
  12   12
1
2
max  12 at P1 and P4 , min  12 at P2 and P3
2. Find the points closest to the origin on the hyperbolic cylinder x  z  1  0
2
2
Solution:
f  x, y, z   x 2  y 2  z 2 and g  x, y, z   x 2  z 2  1
f  2 x,2 y,2 z and g  2 x,0, 2 z
f  g
2x  2 x, 2 y  0, 2 z  2 z
The system of equations is given as:
2 x  2 x ...........(I)
2 y  0 ...............(II)
2 z  2 z ........(III)
x2  z 2  1........(IV)
From (I) x  0 , hence 2 x  2 x only if 2  2 or   1
For   1 , the equation 2 z  2 z becomes 2 z  2 z . If this equation
is to be satisfied as well, z must be zero. Since y  0 , from (II).
From (IV) x  z  1  x  0  1  x  1
2
2
2
2
The points on the cylinder closest to the origin are the points  1,0,0  ,
with a unit distance(i.e.
 1  02  02  1 )
2
71
3. Find the maximum and minimum values of f  x, y   4 x  10 y
2
2
on the disk
x2  y 2  4 .
Solution:
Note that the constraint here is the inequality for the disk.
The first step is to find all the critical points that are in the disk (i.e. satisfy
the constraint).
f  x, y   4 x 2  10 y 2
f x  x, y   8x  0  x  0 ,
f y  x, y   20 y  0  y  0 ,
So, the only critical point is  0,0  and it does satisfy the inequality.
At this point we proceed with Lagrange Multipliers and we treat the constraint as an equality instead of the inequality. We only need to do deal
with the inequality when finding the critical points.
So, we find the system of equation to solve:
f  x, y   4 x 2  10 y 2 , g  x, y   x 2  y 2  4
g x  x, y   2 x , g y  x, y   2 y
By Lagrange multiplier equation , f  g
f x , f y   g x , g y  8x,20 y   2 x,2 y
So the system of equations to solve is as follows:
8 x  2 x
20 y  2 y
x2  y 2  4
From the first equation:
2 x  4     0 ,  x  0 or   4
72
If we have x  0 then the constraint gives us y  2
If we have   4 the second equation gives us,
20 y  8 y  y  0
The constraint then tells us that x  2
If we’d performed a similar analysis on the second equation we would arrive at
the same points.
So, Lagrange Multipliers gives us four points to check:  0,2  ,  0, 2  ,  2,0  and
 2,0  .
f  0,0   4  0   10  0   0
2
2
f  2,0   4  2   10  0   16 , f  2,0   4  2   10  0   16
2
2
2
2
f  0,2   4  0   10  2   40 , f  0, 2   4  0   10  2   40
2
2
2
2
So min  0 and max  40 .
Lagrange Multipliers with Two Constraints
Many problems require us to find the extreme values of a differentiable function
f  x, y, z  whose variable are subject two constraints. If the constraints are
g  x, y, z   c and h  x, y, z   k
and g and h are differentiable, with g not parallel to h , we find the constrained local maxima and minima of f , which formula is given as:
f  g  h
Example 2.8
Find the maximum and minimum of f  x, y, z   4 y  2 z subject the constraints
2 x  y  z  2 and x 2  y 2  1 .
Solution:
Let g  x, y, z   2 x  y  z  2 and h  x, y, z   x  y  1
2
73
2
From the equations, we have:
f  0,4, 2 , g  2, 1, 1 , and h  2 x,2 y,0
Lagrange formula is:
f  g  h
Then the system of equations is given as:
f x   g x   hx
0  2  2 x ………………….(I)
f y   g y   hy
4    2 y ………………...(II)
f z   g z   hz
2   ………………………..(III)
2 x  y  z  2 ............................(IV)
x 2  y 2  1 ................................(V)
From the system of equations, equations (III) shows that   2
Plugging this into equation (I) and equation (III) and solving for x and y respectively gives,
0  2  2   2 x
x
4  2  2 y
y
2

3

Now, plug these into equation (V), we have:
4

2

9

2

13
2
1
    13
So, we have two cases to look at here. First, let us see what we get when
  13 . In this case we know that,
x
2
13
y
Plugging these into equation (IV) gives,
74
3
13
4
3
7

2 z  z 
2
13
13
13



So we have got one solution:  x  
2
3
7

,y
,z  
 2
13
13
13

Let’s now see what we get if we take    13 . Here we have,
x
2
13
y
3
13
Plugging these into equation (IV) gives,
4
3
7

2 z  z 
2
13
13
13


and the second solution:  x 
2
3
7

,y
,z 
 2
13
13
13

Now all that we need is to check the two solutions in the function to see
which is the maximum and which is the minimum.
3
7
26
 2

f 
,
,
 2  4 
 11.2111
13
13
13
13


3
7
26
 2

f
,
,
 2  4 
 3.2111
13 13
13
 13



So, we have a max  11.2111 at  
2
3
7

,
,
 2  and
13 13
13

3
7
 2

,
,
 2
13 13
 13

a min  3.2111 at 
Exercise 2.5
1. Find the points on the ellipse x  2 y  1 where f  x, y   xy has
2
extreme values.
2



Ans :   12 , 12 ,  12 ,  12 


75
2. Find the maximum value of f  x, y   49  x  y
2
2
on the line
Ans : 39
x  3 y  10
3. Find the points on the curve x y  54 nearest the origin.
2
Ans : 3, 3 2 
4. Find the maximum and minimum values of f  x, y, z   x  2 y  5z
on the sphere x  y  z  30 .
2
2
2
Ans :  f 1, 2,5  30max., f 1, 2,5  30min 
5. Find the point on the plane x  2 y  3z  13 closest to the point
1,1,1 . Hint : dist   x  1   y  1   z  1  Ans :  32 ,2, 52 
2
2
2
6. Find three real numbers whose sum is 9 and the sum of whose
squares is as small as possible. Ans :  3,3,3
7. Find the extreme values of the function f  x, y, z   xy  z
2
on the
circle in which the plane y  x  0 intersects the sphere
x 2  y 2  z 2  4 . Ans : max  4 at  0,0, 2.,min  2 at  2, 2,0 
8. Find the point closest to the origin on the curve of intersection of the
plane 2 y  4 z  5 and the cone z  4 x  4 y .
2
2
2
9. Minimize f  x, y, z   xy  yz subject to the constraints
x 2  y 2  2  0 and x2  z 2  2  0 .
10. Maximize f  x, y, z   x  y  z subject to the constraints
2
2
2
2 y  4 z  5  0 and 4 x2  4 y 2  z 2  0
76
UNIT THREE
TWO/THREE DIMENSIONAL SPACE, THE JACOBIAN, DIVERGENCE AND
CURL
3.0
INTRODUCTION
In this unit, we try to look at some transformations to be dealt with next units of
the book.
3.1
Two/Three Dimensional Space
Not only do we have the standard  x, y  or  x, y, z  coordinate system called
the Cartesian Coordinate System in the two or three dimensional spaces but
there
also
exist
alternate
systems
such
Polar
Coordinates
in
the
2  dimensional and Cylindrical and Spherical Coordinate Systems in the
3  dimensional .
3.1.1 Polar Coordinates
x P  x, y 
x Polar Coordinate
xr
x y
x Cartesian Coordinate
x x x
A point P with Cartesian Coordinate P  x, y  can equivalently be defined by the
polar coordinates P  r ,  . The Transformation between the two systems is given by:
P  x, y   P  r , 
That is: x  r cos , y  r sin 
With the inverse transformation: r 
77
x2  y 2 ,   tan 1  xy 
Example 3.1
1. Determine the equation r  a cos  in Cartesian system
2
2
2
Solution:
Put cos 
y
x
2
2
, sin   , r  x  y .
r
r
r 2  a2 cos2 
 x2 
x  y a  2
2 
x y 
2
2
2
  x2  y 2   a2 x2
2
2. Find the equation in polar coordinates:
a) x  y  a
2
2
2
b) y  x
2
Solution:
a) x  y  a
2
2
2
b)
y  x2
but x  r cos , y  r sin 
r sin  r 2 cos2 
 r 2 cos2   r 2 sin 2   a2
r  tan  sec
 r 2  cos2   sin 2    a2
 r 2  a2
3. Find the Cartesian coordinate
4.
 r,    2, / 3
Find the Polar Coordinate
 x, y    3,4
Solution:
x  r cos  2cos  / 3  1
r  x2  y 2  9  16  5
y  r sin  2sin  / 3  3
  tan 1  4 / 3  53.13

  x, y   1, 3

  r ,    5,53.13
78
3.1.2 Cylindrical Coordinates
We obtain cylindrical coordinates for space by combining polar coordinates
in the xy  plane with the usual z  axis. This assigns to every point in
space one or more coordinate triples of the form  r , , z  as shown in the
diagram below:
zz
cylindrical
coordinate
z P   r , , z 
zz
zo
zr
z x z
zy
zy
zx
Definition 3.1 Cylindrical Coordinate
Cylindrical coordinates represents a point P in space by ordered triples
 r , , z  in which
1. r and  are polar coordinates for the vertical projection of P on the xy 
plane
2.
z is the rectangular vertical coordinate.
Equations Relating Rectangular  x, y, z  and Cylindrical Coordinates  r , , z  :
Transformation equation:
x  r cos , y  r sin , z  z
With the inverse transforms:
r  x2  y 2 ,  tan 1  y / x  , z  z
79
where r  0 , 0    2 ,   z   .
3.1.3 Spherical Coordinate
To find the spherical coordinates of a point P in 3  space, we drop a perpendicular from P to the xy  plane, and let Q be the foot of this segment.
Then draw line segments form the origin, O , to both P and Q . The spherical
coordinates of the point P are   , ,  , where  is the distance OP ,  is the
angle that the segment OP makes the positive z  axis, and  is the angle that
the segment OQ makes with the positive x  axis (the same as the cylindrical
coordinate  )
J
z
J
spherical
coordinate
P    , , 
J
J 
J
J
J
J
J
xJ 
J
J
z
O
y
y
r
J
Q
x
Spherical coordinates locate points in space with angles and a distance, as
shown above.
80
Definition 3.2: Spherical Coordinates
Spherical coordinates represent a point
P in space by ordered triples
  , ,  in which:
1.
 is the distance from P to the origin
2.  is the angle OP makes with positive z  axis  0     
3.  is the angle from cylindrical coordinates.
Equations Relating Rectangular  x, y, z  and Spherical Coordinates   , ,  :
Since OPQ   , we have z   cos  and OQ  r   sin  ; this equation
tells us that the transformation equation is given by:
r   sin  , x  r cos   sin  cos , y  r sin   sin  sin
z   cos 
with the inverse transformation:
  r 2  z 2  x2  y 2  z 2 ,   cos1  z  ,   tan 1  xy 
Example 3.2
1. Find the corresponding Cartesian (Rectangular) equation to the cylindrical
equation.
a) r  6sin 
b)
r4
Solution:
Using the transformation equation:
x  r cos , y  r sin , z  z , r  x2  y 2 , sin 
a)
x2  y 2  6  y / x 
 x2  x2  y 2   36 y 2
x2  y 2  4
b)
 x 2  y 2  16
81
2. Find the Cartesian coordinates of  3,  / 2,5 in cylindrical coordinates.
Solution:
r  3,   / 2, z  5
x  r cos  3cos  / 2  0 , y  r sin   3sin  / 2   3 , z  z  5
The Cartesian coordinates is given as:  0,3,5 
3. Find the Cartesian coordinates of the points in spherical coordinates
 4, / 2, / 3 .
Solution:
  , ,    4, / 2, / 3
x   sin  cos  4sin  / 2 cos  / 3  2
y   sin  sin  4sin  / 2 sin  / 3  2 3
z   cos  4cos  / 2  0
 4, / 2,  / 3   2,2 3,0 
4. Find a spherical coordinate equation for the sphere
x 2  y 2   z  1  1
2
Solution:
x   sin  cos , y   sin  sin , z   cos 
 2 sin 2  cos 2    2 sin 2  sin 2     cos  1  1
2
  2 sin 2   cos2   sin 2     2 cos2   2 cos  1  1
  2  sin 2   cos2    2 cos
  2  2  cos 
   2cos
82
5. Find a spherical coordinate equation for the cone z 
x2  y 2 .
Solution:
z  x2  y 2
 cos   2 sin 2  by necessary substitution,
  cos   sin   cos  sin 
 1  tan 
Geometrically, the cone is symmetric with respect to the z  axis and cuts
the first quadrant of the yz  plane along the line z  y .
Hence,  

4
, is the spherical equation of the cone, and 0     .
Exercise 3.1
1. Plot the points with the given polar coordinates and find their Cartesian coordinates:
  2, / 4 
5, tan  4 / 3 g)  1,7  h)  2 3,2 / 3
 2, / 4  b)
1,0 c)
e)  3,5 / 6  f)
1
a)
 0, / 2 
d)
2. Plot the points with the given Cartesian coordinates and find two sets of polar
coordinates for each:
a)
 1,1
e)
  3, 1 f)
b)
1,  3 
c)
 0,3 d)
 1,0
 3,4
g)
 0, 2 
h)
 2,0 
3. Replace the polar equation by an equivalent Cartesian (Rectangular) equation. Then identify or describe the graph:
a) r sin   0
b)
d) r  4r sin e)
2
r  4csc
c)
r 2 sin 2  2
83
r cos  r sin   1
f)
r   csc  e r cos
g) r sin   ln r  ln cos
h) r  4r cos
2
i) r  2cos  2sin 
4. Replace the Cartesian (Rectangular) equation by an equivalent polar equation:
a) x  7 b)
x  y c)
e) y  4 x
f)
2
x  y  4 d)
2
2
x2 y 2

1
9
4
x 2   y  2   4 g) x 2  xy  y 2  1
2
5. Find the corresponding Cartesian equation to the cylindrical equations:
a) r  3cos  2sin    0
b)
d) z sin   r
r 2 cos2
2
3
e)
3
  / 4
c)
3  2cos  0
c)
  / 4
6. Change from spherical to Cartesian equations:
a) r  6sin  sin 
b)
  / 6
d) r  9 e)
r  2  tan 
f)
r  9sec
7. Find the Cartesian coordinates of the point in:
a) Cylindrical coordinates;
(i)
1,1,1
(ii)
 7,2 / 3,4
b) Spherical coordinates;
(i)
(iii)
 6, / 3,3 / 4  (ii)  4,2 / 3,5 / 6
 2,3 / 4,  (iv)  2 3, / 3, / 4 
8. Find the equations in cylindrical and spherical coordinates
a) x  y  4 z  16 b)
2
2
2
x 2  y 2  z 2 c)
d) x  y  9 e) x  y  z  8x  0
2
g) x  y  3z
2
2
2
2
2
2
2
84
f)
x2  y 2  2z
x2  y 2  z 2  9 z  0
3.2
Jacobian
If u and v are functions of two independent variables x and y , that is
u  u  x, y  and v  v  x, y  , the determinant:
u
x
v
x
u
y
v
y
is called the Jacobian of u , v with respect to x, y and
is written as
  u, v 
 u, v 
, or J 

  x, y 
 x, y 
Similarly, the Jacobian of u , v , w with respect to x, y , z is
u
x
  u , v, w  v

  x, y, z  x
w
x
u
y
v
y
w
y
u
z
v
z
w
z
Example 3.3
1. If x  r cos , y  r sin , evaluate
  x, y 
  r , 
.
and
  x, y 
  r , 
Solution:
x
  x, y  r

  r ,  y
r
x
 cos

y
sin 

r sin 
 r cos 2   r sin 2   r
r cos
85
r
  r ,  x

  x, y  
x
Note that
r
x
y
 r

y
y
r2
y
2
2
2
2
2
r  x  y  x y r 1
x
r3 r3
r3
r3 r
r2
  x, y    r , 
1

 r  1
  r ,    x, y 
r
 x, y 
Let x  u 2  v2 , y  2uv Calculate J 

 u, v 
Solution:
x
 x, y  u
J

 u, v  y
u
x
v 2u 2v

 4u 2  4v 2
y 2u 2v
v
Properties of Jacobian
I)
If u, v are functions of x, y then:
  u , v    x, y 

1
  x, y    u , v 
II)
If u, v are functions of r , s where r , s are functions of x, y, then:
  u, v    u, v    r , s 


  x, y    r , s    x, y 
e  e
e  e
, sinh  
.
Note that cosh  
2
2
Exercise 3.2
1. If x  a cosh  cos, y  a sinh  sin show that
  x, y  a 2
 cosh 2  cos 2 
  ,  2
86
yz
zx
xy
 u , v, w 
u

,
v

,
w

if

x
y
z
 x, y , z 
2. Find J 
Ans:  4
3. If x  r sin cos ; y  sin sin  , z  r cos , show that
  x, y, z 
 r 2 sin 
  r , , 
4. If u  x , v  y , find
2
2
  u, v 
  x, y 
Ans:  4xy 
5. If u  x  2 y, v  x  y, prove that
2
  u, v 
 2x  2
  x, y 
 x, y 

 u, v 
6. If x  u(1  v), y  v(1  u), find J 
Ans: 1  u  v 
  u, v 
y2
x2  y 2
, v
, find
7. If x 
2x
2x
  x, y 
Ans:  
 y
 2 x 
8. In a cylindrical coordinate, if x  r cos , y  r sin and z  r , find
 x, y , z 
J

 r , , z 
3.3
Ans:  r 
Divergence and Curl
In this section we are going to introduce a couple of new concepts, the curl and
the divergence of a vector.
Definition 3.3: Divergence
The divergence of a vector field F  M  x, y, z  i  N  x, y, z  j  Q  x, y, z  k
is denoted by divF where,
 


div F   i  j  k   M  x, y, z  i  N  x, y, z  j  Q  x, y, z  k 
y
z 
 x
87
Clearly, divF is a scalar. If div F  0 , F is called solenoidal.
Example 3.4


1. Evaluate div 3x i  5xy j  xyzk at the point 1,2,3 .
2
2
Solution:
 


div  3 x 2i  5 xy 2 j  xyzk    i  j  k   3x 2i  5 xy 2 j  xyzk 
y
z 
 x
 6 x  10 xy  3xyz 2
div  3x2i  5xy 2 j  xyzk 
1,2,3
 6  20  54  80
2. Let v   x  3 y  i   x  2 z  j   x   z  k . Find  if v is solenoidal.
Solution:
 


div v   i  j  k    x  3 y  i   x  2 z  j   x   z  k   0
y
z 
 x
 1  1    0    2
Note:  i  i  0, j  j  0, k  k  0, i  j  k , j  k  i, k  i  j
 j  i  k, k  j  i, i  k   j
Definition 3.4: Curl
The curl of a vector point function F is defined as: curl F    F . If
F  M  x, y, z  i  N  x, y, z  j  Q  x, y, z  k ,
then :
 


curl F    F   i  j  k    Mi  Nj  Qk 
y
z 
 x
 i  i
M
N
Q
M
N
  i  j
 i  k 
  j i
  j  j
x
x
x
y
y
88
 j k 
0k
Q
M
N
Q
 k  i 
  k  j
 k  k 
y
z
z
z
N
Q
M
Q
M
N
j
k
0i
j
i
0
x
x
y
y
z
z
 Q N   M Q 
 N M 
curl F    F  i 


j


k



 x  y 
x 
 y z   z


Alternatively,
i
curl F    F 

x
M
j

y
N
k

z
Q
 Q N   M Q 
 N M 
curl F    F  i 

 j


  k


y

z

z

x

x
y 





The curlF is a vector quantity. If curl F  0 , then field F is called irrotational.
Note: 0  0i  0 j  0k
Example 3.4


1. Calculate the curl to the vector F  xyzi  3x yj  xz  y z k .
2
2
2
Solution:
i
j


curl F    F 
x
y
xyz 3x 2 y
k

z
2
xz  y 2 z
   xz 2  y 2 z    3x 2 y  

curl F    F  i 




y

z


   xyz    xz 2  y 2 z  
   3x 2 y    xyz  
  k

 j


 z

 x


x

y




89
curl F   F  2 yzi   z 2  xy  j   6 xy  xz  k
2. Find the values of the constants a, b, c for which the vector
v   x  y  az  i   bx  3 y  z  j   3x  cy  z  k is irrotational.
Solution:
0  0i  0 j  0k   v
i

curl F    v 
x
 x  y  az 
j

y
 bx  3 y  z 
k

 0 , since it is
z
 3x  cy  z 
irrotational.
   3x  cy  z    bx  3 y  z      x  y  az    3x  cy  z  
 i


  j

y
z
z
x

 

   bx  3 y  z    x  y  az  
k 

0
x
y


  c  1 i   3  a  j   b  1 k  0
 c  1  0  c  1,3  a  0  a  3, b  1  0  b  1

 

3. The vector field defined by v  e sin y i  e cos y j  0k is rotational or
x
irrotational.
Solution:
i
j


curl F    v 
x
y
x
x
e sin y e cos y
90
k

z
0
x
   0    e x cos y     e x sin y    0  
  j

 i


 y
 


z

z

x

 

   e x cos y    e x sin y  
0
k 




x

y


 0i  0 j  e x cos y  e x cos y  k
 0i  0j  0k  0 , v is irrotational.
Exercise
1. Prove that for every field v , divcurl v  0 .
2. If   2x y  xz , find  and  
2
2
2
3. If   x yz and F  xzi  y j  2 x yk , find a) 
2
3
c)   F d) div  F  e)
2
2
curl  F 
91
b)   F
UNIT FOUR
MULTIPLE INTEGRALS
4.0
INTRODUCTION
Now that we have finished our discussion of derivatives of functions of more
than one variable we want to move on to integrals of functions of two or three
variables.
Most of the integration topics we know in single variable would be extended
here. However, because we are now involving functions of two or three variables
there will be some differences as well. There will be new notation and some new
issues that simply do not arise when dealing with functions of a single variable.
Multiple Integrals
Any integral of the form
    f ( x , x ,, x )dX
1
2
n
G
n times
where X  R n a G is an n  -dimensional solid is a multiple integral.
92
4.1
Elemental Areas and Volumes of Coordinates Systems
A one – dimensional linear coordinate is a directed line with an origin.
Ff
x1
x2
Ff
For any two arbitrary points x1 and x2 we write the length as x  x2  x1 .
If x1 and x2 are infinitesimally close then the elemental length is:
dl  dx  x2  x1
Form this we can calculate the length of any interval by integration, that is,
 dx   x  x  x
x2
x2
x1
x1
2
1
Two Dimensional Length
A point P in a plane is defined by the coordinates P   x, y  , and two points
P1  x1, y1  and P2  x2 , y2  are the opposite vertices of a rectangle formed form
the intersections of vertical and horizontal lines through the points. The side of
the rectangle are x  x2  x1 and y  y2  y1 . If the points are very close to
each other we have dx  x2  x1 and dy  y2  y1 .
The elemental area is then dA  dxdy
P2
D
P1
93
The integral evaluation of an area using the sum of the elemental area is:
P2
x2
y2
y2
x2
P1
x1
y1
y1
x1
A   dA   dA   dxdy   dx  dy   dy  dx
D
The last two expressions are called the iterated form of the integral and they
evaluated as:
A   dxdy   dx  dy   dy  dx   x x2   y  y2   x2  x1  y2  y1 
P2
x2
y2
y2
x2
x
y
P1
x1
y1
y1
x1
1
1
Example 4.1
1. The length of segments between (a) x1  6 and x2  12 (b) x1  2 and
x2  4 .
Solution:
a) L 
 dx   x  12  6  6
b) L 
 dx   x  4  2  6
x2
12
x1
6
x2
12
x1
6
The negative sign denotes movement in the negative sense( i.e. opposite direction)
2. Find the area of the rectangle with opposite vertices being 1, 2  and
 5,6  .
Solution:
This implies x1  1, x2  5, y1  2, y2  6 .
A
 5,6 
1,2
dxdy   dx  dy   x 1   y 2   5  1  6   2    48
5
6
1
2
5
6
The negative implies the coordinate’s direction produce left handed system
which is evident from the negative direction of the x  axis.
94
Three Dimensional Volumes
z
z
y
x
y
x
The diagram above is in Three dimensional Cartesian coordinate with mutually
perpendicular axes and fixed directions.
The elemental volume, dV  dxdydz .
P2
z2
P1
z1
V   dV  
y2
x2
y1
x1
 
dxdydz   dz  dy  dx   z z2  y  y2  x x2
z2
y2
x2
z
y
x
z1
y1
x1
1
1
1
Example 4.2
Find the volume of rectangular block bounded by 0  x  3,2  y  6,1  z  3 ,
we have:
V     dxdydz   dz  dy  dx
3
6
3
3
6
3
1
2
0
1
2
0
  z 1  y 2  x0   3  1 6  2 3  0  2  4  3  24
3
4.2
6
3
Double Integrals
Any integral of the form  f  x, y  dA is a double integral. The double
G
integral  f  x, y  dA where f  x, y   0 defines a surface S and G a
G
95
region in
2
represents the volume of the region under S (and above
the xy  plane).
Fubini’s Theorem
If f  x, y  is continuous on G  R   a, b    c, d   ( x, y ) 
2
: a  x  b, c  y  d 
then the integral  f  x, y  dA can be evaluated via iterated integrals as folG
lows:
d
 b f ( x, y )dx  dy  b  d f ( x, y )dy  dx
f
(
x
,
y
)
dA

R
c  a
a  c


R in this case defines a rectangle.
Example 4.3
Compute each of the following double integrals over the indicated
rectangles.
(a)  6 xy 2 dA, R   2, 4  1, 2
R
(b)  2 x  4 y 3dA, R   5, 4   0,3
R
(c)  x 2 y 2  cos  x   sin  y  dA, R   2, 1   0,1
R
(d) 
1
R  2x  3 y 
2
dA, R  0,1  1, 2
(e)  xe xy dA, R   1, 2   0,1
R
Solution
(a)
96
Solution 1
 6 xy dA     6 xy dy  dx    2 xy  dx   14 xdx  7 x
4
2
2
1
2
4
2
3
2
R
2
4
2 4
1
2
2
 84
Solution 2
 6 xy dA     6 xy dx  dy   3x y  dy   36 y dy  12 y
2
4
1
2
2
2
2
2
1
R
(b)
2
4
4
2
2
2
3 2
1
 84
  2 x  4 y dA     2 x  4 y  dydx
4
3
3
3
5 0
R
   6 x  81 dx   3x 2  81x 
4
4
5
5
 756
(c)
 x y  cos  x   sin  y  dA     x y  cos  x   sin  y   dxdy
2
1
2
1
2
2
0 2
R
   13 x3 y 2  1 sin  x   x sin  y  
1
1
0
2
dy
  73 y 2  sin  y  dy  79 y 3  1 sin  y   79  2
1
1
0
0
(d)
  2 x  3 y  dA  1 0  2 x  3 y  dxdy  1   12  2x  3 y   dy   12 1 213 y  31y dy
2
2
1
2
2
1
1
2
0
R
  12  13 ln 2  3 y  13 ln y 1   16  ln 8  ln 2  ln 5
2
(e) In this case it will be significantly easier to integrate with respect to y first as
we will see.
97
 xe dA    xe dydx
2
xy
1
xy
1 0
R
The y integration can be done with the quick substitution,
u  xy
du  xdy
which gives
 xe dA   e
xy
2
xy 1
1
0
R
dx   e x  1dx  e x  x
2
2
1
1
 e 2  2   e 1  1
 e2  e1  3
If f  x, y   g  x  h  y  and we are integrating over the rectangle
R   a, b    c, d  then,
 f  x, y  dA   g  x  h  y  dA    g  x  dx    h  y  dy 
R
b
d
a
c
R
Example 4.4
Evaluate  x cos 2  y  dA , R   2,3  0, 2 
R
Solution:


3
2
 2 cos 2 y dy   1 x 2 3  1 2 1  cos 2 y dy 
x
cos
y
dA

xdx


  
R
2  0     2  2  2 0 




  52  12  y  12 sin  2 y   
2
0
4.3
5
8
Evaluation of Double Integrals Over General Regions
In this section we will look at double integrals over non-rectangular
regions. i.e.,
 f  x, y  dA
D
where D is any region.
98
There are two types of regions for D that will be considered.
Case 1
D   x, y  | a  x  b, g1  x   y  g 2  x 
Case 2.
D   x, y  | h1  y   x  h2  y  , c  y  d 
The double integral for both of these cases are defined in terms of iterated integrals as follows.
In Case 1 where D   x, y  | a  x  b, g1  x   y  g 2  x  the integral is defined to
be,
b
g2  x 
a
g1 x
 f  x, y  dA      f  x, y  dydx
D
In Case 2 where D   x, y  | h1  y   x  h2  y  , c  y  d  the integral is defined to
be,
d
h2  y 
c
h1  y 
 f  x, y  dA   
D
f  x, y  dxdy
The following are some properties of the double integral:
Properties:
1.
  f  x, y   g  x, y   dA   f  x, y  dA   g  x, y  dA
D
D
D
99
 cf  x, y  dA  c  f  x, y  dA where c is any constant.
2.
D
3.
D
If the region D can be split into two separate regions D1 and D2 then the integral can be written as
 f  x, y  dA   f  x, y  dA   f  x, y  dA
D
D1
D2
Example4.5 Evaluate each of the following integrals over the given region D.

y
 e dA , D   x, y  |1  y  2, y  x  y 3
x
(a)

D
3
4
xy

y
dA , D is the region bounded by y  x and y  x3 .

(b)
D
 6 x  40 y dA , D is the triangle with vertices  0,3 , 1,1 , and 5,3 .
(c)
D
Solution:
(a) Applying the formula above
2
y3
1
y
 e dA    e dxdy   ye
x
y
x
y
2
x
y
1
y3
y
D
dy  
2
1
 ye  ye  dy   e  y e   e  2e
y2
1
1
2
y2
1
2
2 1
2
1
1
2
4
1
(b) In this case we need to determine the two inequalities for x and y. The best way
to do this is the graph the two curves.
0  x 1
x3  y  x
Therefore,
3
3
2
4
 4 xy  y dA    3 4 xy  y dydx    2 xy  14 y  3 dx
2
1
x
x
2
x
1
D
x
55
   74 x 2  2 x 7  14 x12  dx   127 x3  14 x8  521 x13   156
1
2
0
1
(c) In this case the region would be given by D  D1  D2 where,
100

D   x, y  1  x  5,

D1   x, y  0  x  1,  2 x  3  y  3
1
2
2

x  12  y  3
or


D   x, y   12 y  32  x  2 y  1, 1  y  3
Solution 1
 6 x  40 ydA   6 x  40 ydA   6 x  40 ydA
D
D1
D2
 6 x  40 y  dydx  1  x  6 x  40 y  dydx
0 2 x  3
 
1
3
5
3
1
2
   6 xy  20 y 2 
1
3
0
2 x  3

1
2
dx    6 xy  20 y 2  1
3
5
1

2
x  12
dx


  12 x3  80 x 2  240 x dx   3 x3  20 x 2  10 x  175 dx
1
0

5
1
    x  x  5 x 175 x   
1
 3 x 4  803 x3  120 x 2
4
3
4
0
20
3
3
5
2
1
1799
6
Solution 2
3
2 y 1
1
 2 y  32
2
 6 x  40 ydA    1
D
 6 x  40 y  dxdy    2 x  40 xy 
3
2
3
1
2 y 1
 12 y  32
dy
2
325
35
65 4
295 3
325 2
35
   654 y 3  295
4 y  4 y  4  dy   16 y  12 y  8 y  4 y 
3
1

6580
16

29526
12

3258
8

35 2
4
 325 
29513
6
 325 
35
2

3
1
2729
3
Note: The last part of the previous example has shown us we can integrate these
integrals in either order (i.e. x followed by y or y followed by x ), although
often one order will be easier than the other. There will be times when it will
not even be possible to do the integral in one order while it will be possible to
do the integral in the other order.
Example4.6
Evaluate the following integrals by first reversing the order of integration.
(a)

3
9
0
2
x
3
x3e y dydx
101
(b)

8
2
0
3 y
x 4  1 dxdy
Solution:
(a) The region is defined as
D  ( x, y ) 0  x  3, x 2  y  9
or


D  ( x, y ) 0  x  y , 0  y  9
Hence
  x e dydx   
3
9
0
x2
9
3 y3
0
y
0
y
9
91
3
3
1
1 3
x e dxdy   x 4e y
dy   y 2e y dy  e y
0 4
0 4
12
0
0
9
3 y3
 121  e729  1
(b) The region is defined as


D  ( x, y ) 3 y  x  2, 0  y  8
D  ( x, y ) 0  x  2, 0  y  x3 
Hence

8
2
0
3 y
2
x3
0
0
x 4  1 dxdy   
2
x3
0
0
x 4  1 dydx   y x 4  1 dx


  x3 x 4  1 dx  16 17  1
2
0
3
2
The first geometrical interpretation of the double integral is that it is the
volume of the solid that lies below the surface given by z  f  x, y  and
102
above the region D in the xy  plane and given by
V   f  x, y  dA
D
Example 4.7
1.
Find the volume of the solid that lies below the surface given by
z  16xy  200 and lies above the region in the xy  plane bounded by
y  x2 and y  8  x2 .
Solution:
The inequalities that define the region D in the xy-plane is given by
2  x  2,
x2  y  8  x2
The volume is then given by,
V   16 xy  200  dA  
2

8 x2
2 x 2
D

2
2
2.
16 xy  200  dydx  2 8 xy 2  200 y  x
2
 128x  400x  512x  1600 dx   32x 
3
2
4
400
3
8 x 2
2
x3  256 x 2  1600 x
dx
 
2
2
Find the volume of the solid enclosed by the planes z  4 x  2 y  10,
y  3x, x  0, and z  0 .
Solution:
The region D will be the region in the xy  plane (i.e. z  0 ) that is
bounded by y  3x , x  0 , and the line where z  4 x  2 y  10 intersects
the xy  plane by substituting z  0 into it, i.e.,
0  4 x  2 y  10  2 x  y  5  y  2x  9
103
12800
3
The region D is bounded on the xy-plane by the inequalities
0  x  1,
3x  y  2 x  5
Hence the volume of this solid is
V   10  4 x  2 y  dA
D
 
1
2 x 5
0 3x
10  4 x  2 y  dydx  0 10 y  4 xy  2 y 2  3x
1
2 x 5
dx
   25 x 2  50 x  25 dx   253 x3  25 x 2  25 x   253
1
1
0
0
The second geometric interpretation of a double integral is the following.
Area of D   dA
D
Note that the area D bounded by the curves y  g1  x  and y  g 2  x  ,
between a  x  b is given by
Area of D    g 2  x   g1  x   dx
b
a
In terms of a double integral we have,
b
g2  x 
a
g1  x 
Area of D   dA   
D
4.4
dydx   y g2 x  dx    g2  x   g1  x   dx
b
g  x
b
a
1
a
Triple Integrals
We use triple integrals to find the volumes of three – dimensional shapes, the
masses and moments of solids, and the average values of functions of three variables. Triple integrals have the same algebraic properties as double integral and
single integral.
Triple integrals for a function f of three variables x, y, and z can also be
defined in a manner to that of a function in two variables.
The notation for the general triple integrals is,
 f  x, y, z dV   f  x, y, z dxdydz
G
G
104
where G is a solid region.
4.4.1 Triple Integrals Over A Rectangular Coordinate(Box)
Suppose the integration is over the box,
G   a, b  c, d    r , s 
Note that when using this notation we list the x ' s first, the y ' ’s second
and the z ’s third.
The triple integral in this case is,
 f  x, y, z dV     f  x, y, z dxdydz
G
s
d
b
r
c
a
Note that we integrated with respect to x first, then y , and finally z here, but
in fact there is no reason to the integrals in this order. There are 6 different possible orders to do the integral in and which order you do the integral in will depend upon the function and the order that you feel will be the easiest.
Example Evaluate the following integral.
 8xyzdV , B  2,3  1, 2  0,1
B
Solution
Just to make the point that order doesn’t matter let’s use a different order from
that listed above. We’ll do the integral in the following order.
2
 8xyzdV     8xyzdzdxdy    4 xyz dxdy    4 xydxdy
2
3
1
2
3
1
2
3
1
2
0
1
2
0
1
2
B
  2 x 2 y dy   10 ydy  5 y 2  15
2
3
2
2
1
2
1
1
Alternatively,
105
 8xyzdV  8 xdx ydy  zdz  x
3
2
1
2 3
2
1
0
2
2
1
1
0
y 2 z 2  5  3 1  15
B
i.e., Using the fact that
 f ( x, y, z)dV   h ( x)h ( y)h ( z)dV   h ( x)dx h ( y)dy  h ( z)dz
b
1
B
2
3
a
d
1
c
s
2
r
3
B
A geometric interpretation of the triple integral is that the volume of the threedimensional region E is given by the integral,
Volume of E   dV
E
4.4.2 Triple Integrals Over General Regions
Consider the triple integral  f ( x, y, z )dV where E is a general threeE
dimensional region. In this case we have three different possibilities
In the first case the region E can be defined as follows,
E   x, y, z  |  x, y   D, u1  x, y   z  u2  x, y 
where  x, y   D . In this case the triple integral is evaluated as follows,
 u2 ( x , y ) f  x, y, z  dz  dA
f
x
,
y
,
z
dV




D  u1 ( x, y )

E
Example 4.8
1. Evaluate  2 xdV , where E is the region under the plane 2x  3 y  z  6 that
E
lies in the first octant.
Solution:
In this case D will be the triangle with vertices at  0, 0  ,  3, 0  , and  0, 2  and z
has the following limits
0  z  6  2x  3 y
and so
106
 2 xdV   
6  2 x 3 y
0
E
D
2 xdz  dA

and the double integral over D can be evaluated using either of the following two
sets of inequalities.
0 x3
0  x   32 y  3
0  y   23 x  2
0y2
 2 x2
6  2 x 3 y
3  3 x2
3
2 3
2 xdV    
2 xdz  dA   
2 x  6  2 x  y  dydx    x  6  2 x  y 

 0

0 0
0
0
E
D
2
2
2
  x  6  2 x    4  43 x   dx   95 x 4  403 x3  10 x   285
0


0
3
3
In the second case the region E may be defined as follows
E   x, y, z  |  y, z   D, u1  y , z   x  u2  y , y 
So, the region D will be a region in the yz-plane.
 f  x, y, z  dV    
E
D
u2 ( y , z )
u1 ( y , z )
f  x, y, z  dx  dA

2. Determine the volume of the region that lies behind the plane x  y  z  8
and in front of the region in the yz  plane that is bounded by
z  32
y and z  34 y .
Solution:
Here the limits for the variables are
0  y  4,
3
4
y  z  32 y ,
The volume is then,
107
0  x 8 y  z
dx
 2 xdV    
8 y  z
0
E
D

dx  dA   
8  y  z  dzdy  0 8z  yz  12 z 2  3 y 4 dy

0 3y 4
4
3 y 2
3 y 2
4


33 2
57 2
11 3
  12 y 2  578 y  23 y 2  32
y dx  8 y 2  16
y  53 y 2  32
y
4
0
1
1
3
5
  495
4
0
In this final case E is defined as,
E   x, y, z  |  x, z   D, u1  x, z   y  u2  x, z 
and here the region D will be a region in the xz-plane.
 u2  x , z  f  x, y, z  dy  dA
f
x
,
y
,
z
dV




D  u1 x, z 

E
3. Evaluate  3x 2  3z 2 dV where E is the solid bounded by y  2x2  2z 2 and
E
the plane y  8 .
Solution:
In the xz  plane we use the following transformations
x  r cos , z  r sin  , x2  z 2  r 2
and get the following limits of the variables for this integral.
2x2  2z 2  y  8,
0    2
0  r  2,
The integral is then,

E

8
3x 2  3z 2 dV     2 2 3x 2  3z 2 dy  dA   y 3x 2  3z 2
 2 x  2 z

D
D


y 8
dA
2 x2  2 z 2

  3 x 2  3 z 2 8   2 x 2  2 z 2  dA
D
The integral is then,

E
3x 2  3z 2 dV   3 8r  2r 3  dA  3 
2
0
D
 3
2
0
 8r  2r  rdrd
2
 r  r  d  3 
8
3
3
2
5
108
5
3
0
2
2
0
0
128
15
d  25615 3
4.5
Applications of Double and Triple Integrals
In this section some applications of double and triple integrals are considered.
4.5.1 Calculating Areas and Volumes using Double and Triple Integrals
The area A of the region R is given by the double integral
A   dA
R
The volume V of the solid between the surface f ( x, y) and the region R is given by the double integral:
V   f ( x, y )dA
R
The volume V of the solid E is given by the triple integral
V   dV
E
4.5.2 The Moment/Center of Mass of a Lamina
The mass M of the lamina R of area density  ( x, y) is given by
M    ( x, y )dA
R
The moment of masses M x and M y with respect to the x  axis and y  axis of
the lamina R with area density  ( x, y) are given by
109
M x   y  ( x, y )dA
and
M y   x  ( x, y )dA
R
R
respectively.
The centre of mass of the lamina R is denoted by ( x , y ) where
x
4.6
My
M
, and y 
Mx
,
M
Substitutions in Multiple Integrals
This section shows how to evaluate multiple integrals by substitution. As in single
integration, the goal of substitution is to replace complicated integrals by ones
that are easier to evaluate. Substitutions accomplish this by simplifying the integrand, the limits of integration, or both.
4.6.1 Transformations
Let the equations x  g (u, v) and y  h(u, v) define a transformation of points from
xy  coordinate into the uv  coordinate system. R  is said to be the image of
R under the given transformation.
The Jacobian of the transformation x  g  u , v  , y  h  u , v  is defined as:
x
  x, y  u

  u, v  y
u
x
v x y x y


y u v v u
v
Example 4.9
Determine the new region that we get by applying the given transformation to
the region R .
(a) R is the ellipse x 2 
y2
u
 1 and the transformation is x  , y  3v .
2
36
110
(b) R is the region bounded by y   x  4, y  x  1 and y 
formation is x 
1
u  v  ,
2
y
x 4
 and the trans3 3
1
u  v  .
2
Solution
(a) Substituting for x and y we have
 u   3v 
1
  
36
2
2
2

u 2 9v 2

1
4 36

u 2  v2  4
which is circle of radius 2.
(b) Substituting for x and y in each of the three equations we have
Using y   x  4 gives,
1
2
 u  v    12  u  v   4
 u  v  u  v  8
 u4
 u v  u v2
 v  1
and y  x  1 gives
1
2
Finally, y 
 u  v   12  u  v   1
x 4
 gives
3 3
1
1 1
4
 u  v     u  v   
2
3 2
 3

3u  3v  u  v  8  4v  2u  8 
v
u
4
2
4.6.2 Change of Variables for a Double Integral
Suppose we are required to evaluate  f ( x, y )dA . Under the transformation
D
x  g  u , v  , y  h  u , v  the given integral can be transformed as follows:
 f ( x, y)dA   f ( x, y)dxdy   f ( g (u, v), h(u, v))
D
D
S
111
 ( x, y )
 ( u ,v )
dudv
where S is the image of D under the transformation.
Example 4.10
Show that when changing to polar coordinates we have dA  rdrd
Solution:
The transformation here is the standard conversion formulas,
x  r cos
y  sin 
The Jacobian for this transformation is,
x
  x, y  r

  r ,  y
r
x
  cos 
y
sin 

r sin 
 r cos 2   r sin 2   r
r cos 
and so
dA 
  x, y 
drd  r drd  rdrd
  r , 
Example 4.11
Evaluate   x  y  dA where R is the trapezoidal region with vertices given by
R
 0, 0  ,  5, 0  ,  52 , 52  and  52 ,  52  using the transformation x  2u  3v and
y  2u  3v .
Solution:
The region R is bounded by the following equations
y  x, y   x, y   x  5 and y  x  5
Using the transformation with y  x, y   x, y   x  5 and y  x  5 we have
2u  3v  2u  3v  6v  0  v  0
2u  3v  2u  3v 
112
4u  0

u0
2u  3v    2u  3v   5

4u  5
2u  3v  2u  3v  5

6v  5

u

v
5
4
5
6
respectively.
The region S is then a rectangle whose sides are given by
u  0, v  0, u 
5
5
and v  and so the ranges of u and v are,
4
6
0u
5
and
4
0v
5
6
The Jacobian for the transformation is
  x, y  2 3

 6  6  12
  u, v  2 3
and the integral is then,
2
  x  y  dA  0 0  2u  3v    2u  3v   12 dudv  0 0 48ududv  0 24u 0 dv
5
6
5
4
5
6
5
4
5
6
5
4
R
5
75
75 6 125

dv  v 
0 2
2 0
4
5
6
Example 4.12 Evaluate   x 2  xy  y 2  dA where R is the ellipse given by
R
x2  xy  y 2  2 using the transformation x  u 
2
3
v, y  u 
2
3
v.
Solution:
Substituting for x and y we have:
2  x 2  xy  y 2 

2u 
2
3
 
2
v 
2u 
 2u 2  2v 2
and so R is transformed into the unit circle
113
2
3
v

2u 
2
3
 
v 
2u 
2
3
v

2
u 2  v2  1
The Jacobian for the transformation is
  x, y 

  u, v 
2 
2
3
2
2
3

2
2
4


3
3
3
and the integral is then,
  x  xy  y  dA   2 u  v 
2
2
2
R
2
4
3
dudv
S
Converting the above into polar coordinates we have
  x  xy  y  dA   2 u  v 
2
2
2
R
2
4
3
dudv 
S
 83 
2
0
1
4
1
2
0
0
8 2 1 2
 r  rdrd
3 0 0
r 4 d  23  d  43
4.6.3 Double Integrals in Polar Coordinates
Consider the double integral  f  x, y  dA . In some cases the region D cannot be
D
easily described in terms of simple functions in Cartesian coordinates where D a
disk, ring, or a portion of a disk is or ring.
The following transformations are used to convert the double integral in Cartesian coordinates to polar coordinates.
x  r cos ,
y  r sin  ,
r 2  x2  y 2
Suppose the inequalities that describe D is given by:
     , h1    r  h2   or r1  r  r2 ,
g1  r     g 2  r 
Then

h2  
r2
g2  r 
h1
r1
g1 r
 f  x, y  dA      f  r cos , r sin   rdrd      f  r cos , r sin   rddr
D
114
It is important to not forget the added r and don’t forget to convert the Cartesian
coordinates in the function over to polar coordinates.
Example 4.13
1. Evaluate the following integrals by converting them into polar coordinates.
(a)
 2 xydA , D is the portion of the region between the circles of radius 2 and raD
dius 5 centered at the origin that lies in the first quadrant
(b)
 e
x2  y 2
dA , D is the unit circle centered at the origin.
D
Solution:
(a) The circle of radius 2 is given by r  2 and the circle of radius 5 is given by
r  5 . The inequalities that determine D is given by
2  r  5, 0    2
Hence



3
4
 2 xydA    2  r cos   r sin   rdrd    r sin  2  drd   14 r sin  2  d
2
5
0
2
2
5
0
2
2
5
0
2
D


2
609
609
  609
4 sin  2  d   8 cos  2  0  4
2
0
(b) The inequalities that determine D is given by
0    2 ,
0  r 1
In terms of polar coordinates the integral is then,
 e
D
 e
D
x2  y 2
dA  
2
0
x2  y 2
dA  
0
 re drd  
1
0
r2
2
2
0
1
2
er
 re drd
1
r2
0
2
115
1
0
d  
2
0
1
2
 e  1 d
   e  1
2. Determine the area of the region that lies inside r  3  2sin  and outside
r  2.
.
Solution:
To determine the area D we solve r  3  2sin  and r  2 to find the value
of  for which the two curves intersect. i.e.,
3  2sin   2
sin   21

  76 ,  6

The ranges that define the region D are:
 6    76 ,
2  r  3  2sin 
The area of the region D is then:
7
6
3 2sin 
6
2
A   dA    
D
7
6
3 2sin 
6
2
rdrd    12 r 2
d     52  6sin   2sin 2   d
7
6
6
7
6
7
6
6
6
    72  6sin   cos  2   d   72   6 cos   sin  2    
14
3
3. Determine the volume of the region that lies under the sphere x2  y 2  z 2  9
, above the plane z  0 and inside the cylinder x2  y 2  5 .
Solution:
The volume of a region is given by
V   f  x, y  dA
D
In this case
z  f  x, y   9  x 2  y 2
116
We only want the portion of the sphere that actually lies inside the cylinder given
by x2  y 2  5 this is also the region D. The region D is the disk x2  y 2  5 in the
xy  plane.
In terms of polar coordinates the limits for the region are
0    2 ,
0r 5
and converting the function to polar coordinates we have
z  9  x2  y 2  9  r 2
The volume is then,
V   9  x 2  y 2 dA  
2
0

5
0
D
r 9  r 2 drd    13  9  r 2 
2
5
32
0
0
d  
2
0
19
3
d  383
4. Find the volume of the region that lies inside z  x2  y 2 and below the plane
z  16 .
Solution:
The volume is given by


V   16dA    x 2  y 2  dA  V   16   x 2  y 2  dA
D
D
D
The inequalities for the region and the integrand in terms of polar coordinates
are
0    2 ,
0  r  4,
z  16  r 2
The volume is then,


V   16   x2  y 2  dA  
D
2
0
 r 16  r  drd   8r  r  d   64d  128
4
2
2
0
0
117
2
1
4
4
4
2
0
0
5. Evaluate the following integral by first converting to polar coordinates.

1
1 y 2
0 0
cos  x 2  y 2  dxdy
Solution:
The inequalities that define the region in terms of Cartesian coordinates are
0  y  1,
0  x  1 y2
Converting into polar coordinates we have
0    2 ,
0  r 1
and so the integral becomes,

1
1 y 2
0 0
cos  x 2  y 2  dxdy   12 sin  r 2  d   12 sin 1 d  4 sin 1

2
0
1

0
0
2
4.6.4 Change of Variables for a Triple Integral
The triple integral  f  x, y, z  dV under the transformation x  g  u , v, w 
R
y  h  u , v, w  and z  k  u , v, w  becomes:
 f  x, y, z  dV   f  g  u, v, w ,h u, v, w ,k u, v, w  
 x , y , z 
 u ,v , w
R
dudvdw
S
x
  x, y, z  uy

where
  u, v, w  uz
u
x
v
y
v
z
v
x
w
y
w
z
w
.
Example 4.14
Verify that dV   2 sin  d  d d when using spherical coordinates.
Solution
The transformation for the spherical coordinates is
118
x   sin  cos
y   sin  sin 
z   cos 
The Jacobian is,
  x, y, z  sin  cos    sin  sin   cos  cos 
 sin  sin   sin  cos   cos  sin 
   , ,  
cos 
0
  sin 
   2 sin 3  cos 2    2 sin  cos 2  sin 2   0
  2 sin 3  sin 2   0   2 sin  cos 2  cos 2 
   2 sin 3   cos 2   sin 2     2 sin  cos 2   sin 2   cos 2  
   2 sin   cos 2   sin 2      2 sin 
Finally, dV becomes,
dV    2 sin  d  d d   2 sin  d  d d
Exercise Verify that dV  rdzdrd when using cylindrical coordinates.
4.6.4 Triple Integrals in Cylindrical Coordinates
The triple integral  f  x, y, z  dV can be converted into cylindrical coordinates
E
using the following conversion formulas:
x  r cos
y  sin 
zz
dV  rdzdrd
The region, E , over which we are integrating becomes,
E   x, y, z  |  x, y   D, u1  x, y   z  u2  x, y 
  r , , z  |      , h1    r  h2   , u1  r cos  , r sin    z  u2  r cos  , r sin  
and we have:

h2  
u2  r cos , r sin 
h1
u1 r cos , r sin 
 f  x, y, z  dV       
E
Example 4.15
119
r f  r cos  , r sin  , z  dzdrd
1. Evaluate  ydV where E is the region that lies below the plane z  x  2
E
above the xy  plane and between the cylinders x2  y 2  1 and x2  y 2  4 .
Solution:
The range for z in cylindrical coordinates is
0 z  x2
0  z  r cos   2

Next, the region D is the region between the two circles x2  y 2  1 and
x2  y 2  4 in the xy  plane and so the ranges for it are,
0    2 and 1  r  2
Hence
2
2
r cos  2
0
1
0
2
2  1
2
1 3


2
r
sin
2


2
r
sin

drd


r 4 sin  2   r 3 sin   d



1  2

0 8
3


1
 ydV    
E

0
2
 r sin   rdzdrd  0 1 r 2 sin   r cos   2  drd
2
2
2
2
2
14
14
 15

 15

   sin  2   sin   d    cos  2   cos    0
0
3
3
8
1
 16
0
2
2. Convert  
1
1 0
1 y 2

x2  y 2
x2  y 2
xyzdzdxdy into an integral in cylindrical coordinates.
Solution:
The ranges of the variables from the iterated integral are
1  y  1,
0  x  1  y2 ,
x2  y 2  z  x2  y 2
And so, the ranges for D in cylindrical coordinates are,
 2    2 , 0  r  1, r 2  z  r
120
Hence
 
1
1 0
1 y 2

x2  y 2
x y
2
2
xyzdzdxdy  
 2
  r  r cos  r sin   zdzdrd
1
r
 2 0 r 2

 2
  zr cos sin  dzdrd
1
r
3
 2 0 r 2
4.6.5 Triple Integrals in Spherical Coordinates
The triple integral  f  x, y, z  dV can be converted into spherical coordinates
E
using the following conversion formulas:
x   sin  cos  , y   sin  sin  , z   cos  ,
x2  y 2  z 2   2
dV   2 sin  d  d d
We also have the following restrictions on the coordinates.
  0 and 0    
For our integrals we are going to restrict E down to a spherical wedge. This will
mean that we are going to take ranges for the variables as follows,
a    b,
   ,
  
E is really nothing more than the intersection of a sphere and a cone.
Therefore, the integral will become,
121


 f  x, y, z  dV      sin  f   sin  cos ,  sin  sin  ,  cos   d d d
b
2
a
E
Example 4.16
1. Evaluate  16 zdV where E is the upper half of the sphere x2  y 2  z 2  1 .
E
Solution:
Since we are taking the upper half of the sphere the limits for the variables are,
0    1,
0    2 ,
0 

2
The integral is then,
16 zdV  
 2
0
0
E

 2
0
2. Convert  
3
0

2
1
 2
0
0
2
  sin  16 cos   d d d  
2
 2
0
0
 2sin  2  d d  
9 y 2
0

18 x 2  y 2
x2  y 2
2
1
0
0
  8 sin  2  d d d
3
 2
4 sin  2  d  2 cos  2  0  4
 x  y  z  dzdxdy into spherical coordinates.
2
2
2
Solution:
The limits for the variables are
0  y  3,
0  x  9  y2 ,
x 2  y 2  z  18  x 2  y 2
Converting to spherical coordinates we have
0    2 ,
0    18  3 2,
0    4
The integral is then,

3
0
0
9 y 2

18 x 2  y 2
x y
2
2
 x  y  z  dzdxdy    
2
2
2
 4
 2
3 2
0
0
0
122
 4 sin  d d d
4.7
Surface Area
Consider a body whose surface S : z  f  x, y  forms the projection R in the
xy  plane and that f  x, y  has continuous partial derivative on R .
S : z  f  x, y 
1
2
2
2
Then by definition, the area A    f x  f y  1 dA where dA  dxdy or
R
dydx as applicable. Similarly, if the surface S has suitable projection on the yz
or xz  plane, one can find the surface area of such bodies thus if S is the
123
graph of an equation y  g  x, z  whose projection on the xz  plane is R ,
then A 
1
2
  g  g  1 dA , dA  dxdz or dzdx
2
x
R
2
z
Example 4.16
1. Find the surface area of the part of the plane 3x  2 y  z  6 that lies in the
first octant.
Solution:
Solving 3x  2 y  z  6 for z and taking the partial derivatives gives,
z  6  3 x  2 y,
f x  3,
f y  2
The limits defining D are,
0  x  2,
0  y  32 x  3
The surface area is then,
2
 32 x 3
0
0
S    32   22  1 dA   
D
14dydx  14    32 x  3 dx
2
0
2
3
 14   x 2  3x   3 14
 4
0
2. R be the triangular region in the xy  plane with vertices  0,0,1 ,  0,1,0  ,
and 1,1,0  . Find the surface area of that part of graph of z  3x  y that
2
lies over R .
Solution:
1
2
A   9  4 y  1 dA    10  4 y 
2
R
1
y
2 1/2
0 0
143/2  103/2

12
3. Determine the surface area of the part of z  xy that lies in the cylinder given by x2  y 2  1 .
124
Solution:
The surface is z  f  x, y   xy and taking the partial derivatives gives,
f x  y, f y  x
The integral for the surface area is,
S   x 2  y 2  1 dA
D
The region D is the disk x2  y 2  1 and so the surface area is
S   x 2  y 2  1 dA  
2
0
D

2
0
2
12
 r 1  r drd   2  3  1  r  d
1
2
0
2
3
2
0
1 3
2
 2 2  1 d   2 32  1
3
3
Exercise
1. Sketch the region bounded by the graph of the given equation and find its
area by means of double integrals.
a) y 
x , y  x , x  1 , x  4
b) x  y , y  x  2, y  2, y  3
2
c) x  y  1  0,7 x  y  17  0,2 x  y  2  0
d) y  ln x , y  0, y  1
e) y  x , y 
2
f)
1
1  x2
y  cosh x, y  sinh x, x  1, x  1
2. Sketch the solid in the first octant that is bounded by the graph of the given
equations and find its volume.
a) z  4  x , x  y  2, x  0, y  0, z  0
2
b) y  z, y  x, x  4, z  0
2
125
c) z  y , y  x , x  0, z  0, y  1
2
2
d) x  y  16, x  7, y  0, z  0
2
2
3. Find the volume of the solid that lies under the graph of z  x  4 y and
2
2
over the triangular region R in the xy  plane having vertices  0,0,0  ,
1,0,0 and 1,2,0 .
4. Use polar coordinates to evaluate the integrals in the following:
a)
 x  x  y dA where R is the region bounded by the graph of
2
2
2
R
y  1  x 2 and y  0 .
b)
  x  y dA where R is the region bounded by the graph of
R
x2  y 2  2 y  0 . and z  5
5. Evaluate the following integrals by changing to polar coordinates
 x  y dydx
a)

a2  x2
b)

1 x 2
c)
2

2 y y2
0
 2 y y2
d)
 
2 ax  x 2
a
0
0
1
0 0
2a
0
2
2
e x  y dydx
0
2
2
xdxdy
 x  y  dydx
2
2
6. Find the volume of the solid bounded by the graphs of z  a  x  y and
2
z  5.


7. If Q   x, y, z  :1  x  2, 1  y  0,0  z  3 . Evaluate
  x  2 y  4 z dV in six different ways.
Q
8. Evaluate the following iterated integrals:
a)
2
x2
x z
0
0
x z
 
zdydxdz
126
2
b)
3
3y
y2
2
0
1
    2 x  y  z  dxdydz
9. Using cylindrical coordinates
a) Find the volume and the centroid of the solid bounded by the graphs of
x2  y 2  z 2  0 and x 2  y 2  4
b) Let a homogeneous solid be bounded by the graphs of z 
x 2  y 2 and
z  x2  y 2 . Find
(i)
The centre of mass
(ii)
The moment of inertia w.r.t. z  axis
10. Find the area of the region on the plane z  y  1 that lies inside the cylinder
x2  y 2  1 .
11. Set up an integral for finding the surface area of that part of the sphere
x 2  y 2  z 2  4 that lies over the square region in the xy  plane have vertices 1,1,0  , 1, 1,0  ,  1,1,0  and  1, 1,0  .
12. Find the surface area of that part of the cylinder y  z  a that lies inside
2
the cylinder x  y  a
2
2
2
2
2
13. Let R be the triangular region in the xy  plane with vertices
 0,0,0 ,  0,2,0  , and  2,2,0 . Find the surfaces area of that part of the
graph of z  y that lies over R .
2
14. Find the surface area of that part of the hyperbolic paraboloid z  x  y
2
that lies in the first octant and is inside the cylinder x  y  1 .
2
127
2
2
UNIT FIVE
LINE INTEGRAL
5.0
INTRODUCTION
When we integrate a function f  x  of one variable from, say x  a to x  b
we can interpret the integration as taking place over the path on the x  axis
from a to b and the value of the integral as giving the area bounded by the
curve y  f  x  over this path,  a, b  :
y
height  y  f  x 
area  
 a ,b


a
b
fdx
x
But we can also integrate functions over other paths, not just straight – line
paths along the x  axis. The result is called a curve, contour, or path inte128
gral, or, most commonly, a line integral (even though the path does not need
to be a straight line).
CounterClockwise
x2 y 2

1
a 2 b2
 Ellipse 
x2  y 2  r 2
 circle 
x  a cos t
y  b sin t
0  t  2
x  r cos t
y  r sin t
0  t  2
Clockwise
x  a cos t
y  b sin t
0  t  2
x  r cos t
y   r sin t
0  t  2
H
e
r
e
a
r
e some of the basic curves that we will need to know how to do as well as limits
on the parameter if they are required.
129
y  f  x
x t
y  f t 
x  g  y
x  g t 
y t
r  t   1  t  x0 , y0 , z0  t x1, y1, z1 , 0  t  1
Line
Segment
from
 x0 , y0 , z0  to  x1, y1, z1 
or
x  1  t  x0  tx1
y  1  t  y0  ty1
, 0  t 1
z  1  t  z0  tz1
5.1
Line Integrals With Respect to Arc Length
To see how these integrals can be defined, let us begin by considering a function
f  x, y  and a smooth curve, C , in the xy  plane. [A curve given parametrical130


ly by the vector equation r  r  t   x  t  , y  t  is said to be smooth if the derivative r  t  is continuous and nonzero. To say that C is piecewise smooth
means that it’s composed of a finite number of smooth curves joined at consecutive corners]. Imagine breaking C into n tiny segments, of arc length si
y
si
Pi
curve C
x
On each of these tiny pieces, choose any point Pi   xi , yi  , then multiply
f  xi , yi   the value of the function at Pi  by the length si and form the following sum:
n
 f  x , y s
i 1
i
i
i
If the value of this sum approaches a finite, limiting value as n   , then the
result is called the line integral of f along C with respect to arc length,
and is written as:
 fds
C
What does the value of the integral mean geometrically? It is the area of the region whose base is the curve C and whose height above each point  x, y  is
given by the value of the function, f  x, y  . That is, it is the area of the vertical
131
curtain or fence (portion of the vertical cylinder) whose base is C and whose
height at each point  x, y  is f  x, y  :
z
height of curtain  f  x, y 
area of curtain =  fds
C
y
curve C
x
To actually evaluate this line integral, first parametrize the curve C :
 x  x t 
t
C:
for a  t  b or a 
b
y

y
t





We consider the curve C to be directed; that is, we think of the curve as being
traced in a definite direction, which we call the positive direction. By writing
t
a 
b in the parametrization, we are saying that the parameter t runs from
a to b , so A   x  a  , y  a   is the initial point and B  x  b  , y  b   is the final
point
of
the
curve,
which
gives
C
a
positive.
Now,
since
 ds 2   dx 2   dy 2 , we can write
2
2
2
2
ds
 dx   dy 
          x  t    y  t 
dt
 dt   dt 
where we use the  sign if the parameter t increases in the positive direction
on C and the  sign if t decreases in the positive direction on C . We then
have:
132
 fds   f  x  t  , y  t  
b
C
a
ds
dt
dt
Example 5.1
Determine the value of the line integral of the function f  x, y   x  y
2
over
the quarter circle x  y  4 in the first quadrant, from  2,0  to  0,2  .
2
2
Solution:
First we parametrize the curve. Since C is directed counterclockwise, we can
write
 x  2cos t
t
C :
for 0 
 12 
 y  2sin t
arrow specifies
positive direction
t  / 2
y
 0,2 
C
 2cos t,2sin t 
t0
 2,0 
x
Next, we use the parametrization to find ds / dt :
2
2
ds
   x  t     y  t  
dt

 2sin t 2   2cos t 2  2
where we choose the  sign, since in our parametrization, t
 from 0 to  / 2 in the positive direction on C . This gives us:
133
increases
 fds   f  x  t  , y  t  
ds
dt
dt


b
C
a

 /2
0
2cos t  4sin 2 t  2dt  4
 /2
0
 cos t  2sin t dt
2
 4sin t  t  12 sin 2t 0  2  2   
 /2
Let us see what the value of this line integral would have been if we had traversed the curve C in the opposite direction, that is, if we had chosen the positive direction for C to be clockwise, from  0,2  to  2,0  . In this case, we could
parametrize the curve as follows:
 x  2cos t
t
C :
for 12  
0
 y  2sin t
y
 0,2 
t  / 2
 2cos t,2sin t 
t0
 2,0 
x
Notice that the parameter t decreases  from  / 2 to 0  in the positive direction
on C , so we would choose the minus sign when finding ds / dt . So we would
have had
2
2
ds
   x  t     y  t    2
dt
and our integral would have been:
 fds   f  x  t  , y  t  
b
C
a
ds
dt
dt
134

0
 /2
0
0
2
 /2
 4
 2cos t  4sin t    2 dt    2cos t  4sin t    2  dt
2
 /2
 cos t  2sin t dt  4sin t  t  sin 2t 
2
1
2
 /2
0
 2 2   
just as we found before. This illustrates an important property of line integrals
with respect to arc length: The value of
 fds does not depend on the orientaC
tion of C .
Line integrals with respect to arc length can also be easily defined over curves in
3  space. If f  x, y, z  is a function defined on a smooth curve
 x  x t 

t
C :  y  y  t  for a 
b
z  z t


then
 fds   f  x  t  , y t  , z t  
b
C
a
ds
dt
dt
where
2
2
2
ds
   x  t     y  t     z  t  
dt
In this case, one interpretation of the value of the line integral involves thinking
of C as a curved wire and f  x, y, z  as its linear density (mass per unit
length);
5.2
 fds then gives the total mass of the wire.
C
The Line Integral of a Vector Field
135
Now we’ll look at another kind of line integral; the line integral of a vector field.
We’ll define this line integral in two dimensions; the generalization to three dimensions follows as easily as it did in the case of a line integral with respect to
arc length.
A vector field is a vector-valued function. Let D be a region of the xy  plane
on which a pair of continuous functions, M  x, y  and N  x, y  , are both defined. Then the function F that assigns to each point  x, y  in D the vector
F  x, y   M  x, y  i  N  x, y  j   M  x, y   N  x, y  
is a continuous vector field on D . Now, if C is an oriented, piecewise smooth
curve in D , we want to define the line integral of F along C . Let
t
r  t   x  t  i  y  t  j   x  t  , y t   for a 
b
be a parametrization of the curve C . Then the line integral of the vector field F
along C is defined as:
 F dr   F  r  t   r  t  dt
b
C
a
What is the interpretation of the value of this integral? Consider F a force field,
and imagine a particle moving along the curve C . Along each infinitesimal piece
– specified by the vector dr - of the curve C , the field F is approximately constant, so the work, dW , done by F as the particle undergoes the displacement
dr is equal to the dot product F dr .
y
136
F
B
r
C
A
x
Adding up (that is, integrating) all these contributions gives the total work done
by F as the particle moves along the entire curve:
W   dW   F dr
C
C
To actually evaluate this integral, we proceed as before. First we parametrize the
curve,
t
r  t   x  t  ia  y  t  j   x  t  , y  t   , for a 
b
then we substitute, to write everything in terms of t :
 F  dr   F  r  t   r  t  dt
b
C
a
  F  x  t  , y  t    x  t  , y  t   dt
b
a
   M  x  t  , y  t   , N  x  t  , y  t     x  t  , y  t   dt
b
a
   M  x  t  , y  t   x  t   N  x  t  , y  t   y  t  dt
b
a
Since dx  xdt and dy  ydt , the last equation is also commonly written in the
abbreviated form:
 F  dr   Mdx  Ndy
C
C
Example 5.2
137
Evaluate the line integral
  7 y  x dx   x  2 y  1 dy
3
2
C
a) If C is the portion of the parabola y  x from  0,0  to 1,1 ;
2
b) If C is the straight – line path from  0,0  to 1,1 .
Solution:
a) We can parametrize the curve as follows:
 xt
t
C :
for 0 
1
2
y  t
With this parameterization, we have dx  dt and dy  2tdt , so:
  7 y  x dx   x  2 y  1 dy    7 t   t  dt  t  2t  1  2tdt
3
1
2
C
2 3
2
2
0
1


  7t 6  6t 3  3t dt  t 7  32 t 4  32 t 2   1
0
1
0
b) We can parametrize the curve as follows:
x  t
t
C :
for 0 
1
y  t
With this parameterization, we have dx  dt and dy  dt , so:
  7 y  x dx   x  2 y  1 dy    7t  t  dt  t  2t  1 dt
3
1
2
C
3
2
2
0
1


  7t 3  t 2  t  1 dt
0
1
 74 t 4  13 t 3  12 t 2  t   19
12
0
This example shows that, for a given vector field F , the value of
 F dr
C
depends, in general, not just on the endpoints of C , but also on the choice
of the path.
138
Example 5.3
Evaluate the line integral
 y dx   2 xy  1 dy
2
C
along each of the paths from A   2,0  to B   0,2  shown in this figure:

C1
C3
C2


C3
Solution:
a) We can parametrize the circular arc C1 as we did earlier:
 x  2cos t
t
C1 : 
for 0 
 12 
y

2sin
t

Since dx   2sin t  dt and dy   2cos t  dt , the line integral along C1
becomes:
 y dx   2 xy  1 dy    4sin t   2sin tdt   8cos t sin t  1 2cos t dt
2
C
 /2
2
0

 /2
0


 8 1  cos 2 t sin t  16cos 2 t sin t  2cos t dt


 8cos t  8cos 2 t  2sin t 
 /2
0
 2
b) We can parametrize the straight line C2 as follows:
x  2  t
t
C2 : 
for 0 
2
 y t
139
Because we now have dx  dt and dy  dt , the line integral along C2
is:
 
2
2
 y dx   2 xy  1 dy   t  dt   2  2  t  t  1 dt
2
C
0
2


  3t 2  4t  1 dt  t 3  2t 2  t   2
0
2
0
c) The path C3 is a piecewise smooth curve, composed of two smooth
curves that meet at a corner. The first, which we will call C31 , is the
straight – line path along the x  axis from A to the origin; the second, C32 , is the straight – line path along the y  axis from the origin
to B . The line integral along C3 is the sum of the line integrals along
these two paths, which we can parametrize as follows:
x  t
x  0
t
t
C31 : 
for 2 
for 0 
 0 C32 : 
2
y  t
y t
Along C31 , we have dx  dt and dy  0 ; along C3 2 , we have
dx  0 and dy  dt , so:
 y dx   2xy  1 dy  
2
C3
C31
y 2dx   2 xy  1 dy  
C3 2
y 2dx   2 xy  1 dy
 0    1dt  t 0  2
2
2
0
Notice that although we had a different path in each part, all three of
these line integrals had the same value. We might guess that – unlike the
line integral in Example 5.2 – the line integral from A to B in this example does not depend on the choice of path. This guess would turn out to
be correct, and in the next section, we will learn why.
140
5.3
The Fundamental Theorem of Calculus for Line Integrals
Simply put, the reason that the line integral in Example 5.3 did not depend on
the path was because the vector field we were integrating,
F  x, y   Mdx  Ndy  y 2 dx   2 xy  1 dy
is a gradient field. That is, F is equal to the gradient of a scalar field (that is,
of a real – valued function). In this case, it is easy to see that F is equal to f
where
f  x, y   xy 2  y
(we could also add any constant to this function and not change the fact that
F  f ). A function f , such that F  f , is called a potential for F . The
fundamental theorem of calculus for line integrals says that the line integral of a gradient field depends only on the endpoints of the path, not the choice
of the path. More precisely, it says that if C is any piecewise smooth curve oriented from the initial point A to the final point B , and f is a continuously differentiable function defined on C , then:
 f dr  f  B   f  A
C
Notice the similarity between this equation and the corresponding equation for
the fundamental theorem of calculus which states that
dy
 dx dx  f  b   f  a 
b
a
Using the fundamental theorem, we could evaluate the line integral of Example
5.3 very easily. Notice that every path in that example connected the initial
point, A   2,0  , to the final point, B   0,2  ; so, because F  f , where we
can take f  x, y   xy  y , we have:
2
2
 y dx   2 xy  1 dy  
C
B 0,2 

 
A 2,0

 xy 2  y dr   xy 2  y 
  0  2    0  0   2
141
 0,2
 2,0
which is the value we obtained for every path in that example. A vector field F
that has the property that the value of
 F dr depends only on the initial and
C
final points of C (but not on the choice of the path C ), is called conservative.
So, if F is a gradient field, then F is conservative.
But how so we know if F is a gradient field? Remember that, given f  x, y  , its


gradient is f  f x , f y . So, if F   M , N  is a gradient field, then M  f x
and N  f y for some function f . If we were to differentiate M with respect to
y , we would get M y  f xy ; and if we were to differentiate N with respect to
x , we would get N x  f yx . Assuming that these mixed partials are continuous,
we know they are identical; that is, f xy  f yx , so true M y  N x . Therefore, we
can say that if F   M , N  is gradient of a scalar field f  x, y  , then it must be
true that M y  N x . This condition is necessary for F to be a gradient field,
meaning that if M y  N x , then F is definitely not a gradient field. However, the
condition M y  N x is not sufficient to conclude that F is a gradient field; an additional, mild hypothesis is required, one that we will give in the last section.
The path over which a line integral is evaluated can be closed; that is, its final
point can also be its initial point. When a line integral is taken around a closed
curve C , the notation for the integral is changed slightly. To indicate that C is
closed, a small circle is drawn on the integral sign, like this:
 F dr
or
C
 Mdx  Ndy
C
If F is a gradient field, the value of this line integral is easy to figure out. Using
the fundamental theorem, we know that:
 F dr   f dr  f  B   f  A
C
C
142
But since the curve is closed, the final point, B , is the same as the initial point,
A . Because A  B , the right – hand side of the equation above, f  B   f  A
, is equal to zero. Therefore, we can say that if F is a gradient field, then
 F dr equals zero for every closed path C . The converse is also true.
C
Example 5.4
Let C be the boundary of the square whose vertices are  0,0  ,  2,0  ,  2,2  ,
and  0,2  , taken counterclockwise. Evaluate each of the following line integrals
around C .
a)
  x  2 y dx   x  3 y  dy
b)
  x  y dx   x  3 y  dy
2
C
2
C
Solution:
Notice that
My  2
and
N y  1 ; since M y  N x , we know that
 M , N    x  2 y, x  3 y 2  is not a gradient field, so we cannot expect its
integral around the closed path, C , simply to be zero. We have to do the
calculation (since we cannot apply the fundamental theorem), so we begin by
parametrizing the four sides of the square, as shown:
xt
t
C3 : 
for 2 
0
y

2

 0,2 B
 2,2 
C3
y
x  0
t
C4 : 
for 2 
0 C
4
y t
 0,0 
C2
x  2
t
C2 : 
for 0 
2
y t
C1
 2,0 
A
xt
t
C1 : 
for 0 
2
y  0
143
x
Since
       
C
C1
C2
C3
, we have:
C4
  x  2 y  dx   x  3 y dy   t  dt    2  3t  dt
2
C
2
2
0
0
2
over C1
over C2


   t  4  dt   3t 2 dt
0
2
2
2
over C3

over C4



  t  2  3t 2   t  4   3t 2 dt
2
0
   2 dt  4
2
0
It is worth noting that, if the positive direction of this path C instead had
been clockwise, the value of the line integral would have been 4 . For line integrals of vector fields, the orientation of the curve is important. Let C be a
path (closed or not) with a given orientation; if C denotes the same path
with the opposite orientation, then the following equation is always true:
 F dr   F dr
C
C

b) This part is easy.  M , N   x  2 y, x  3 y
2
 is a gradient field (notice that
M y  N x  1 ) because it is equal to the gradient of
f  x, y   12 x 2  xy  y 3
Since C is a closed path, the integral of the gradient field  M , N  around
C is zero. No calculation is needed.
144
5.4
Green’s Theorem
Now that we have looked at line integrals and double integrals, we will introduce
an equation that connects them. The type of line integral that we will be using is
that of a vector field around a simple closed curve – is one that does not cross
itself between its endpoints. A circle, an ellipse, and the boundary of a rectangle
are all examples of simple closed curves; a figure-8 is not (it’s closed but not
simple).
Next, we will always assume that the closed curve is oriented; the positive direction of the closed curve is defined as the direction you would have to walk in order to keep the region enclosed by the curve on your left. Consider a simple
closed curve C enclosing a region R , so that C is the boundary of R . If
M  x, y  and N  x, y  are functions that are defined and have continuous partial derivatives both on C and throughout R , then Green’s Theorem says that:
 N M 
Mdx

Ndy

C
R  x  y dA
Since these two integrals are identical, if you’re asked to evaluate one of them, it
might be easier to figure out the other one instead; the answer will be the same.
145
Here’s an example in which your first thought might be to evaluate a double integral, but it may be easier to work out the corresponding line integral. The double integral over R of the function f  x, y  that is identically equal to 1:
1 dA   dA
R
R
gives the area of the region R . If we take M  x, y    y and N  x, y   x , then
the line integral
1
1
Mdx  Ndy    ydx  xdy

2 C
2 C
will, by Green’s Theorem, be equal to:
1  N M 
1
1

dA

1


1
dA

2dA   dA










2 

x

y
2
2

R 
R
R
R
So if is a region of the xy  plane whose boundary, C , is a simple closed curve,
then the area of R can be determined by integrating the vector field F    y, x 
around C and multiplying by
1
:
2
area of R= 12   ydx  xdy
C
It’s also easy to see that we could compute the area of R by using either of
these two shorter formulas:
area of R= 12   ydx   xdy
C
C
146
Example 5.5
Use Green’s Theorem to find the area enclosed by the ellipse:
x2 y 2

1
a 2 b2
Solution:
we parameterize the ellipse by the equations:
 x  a cos t
t
for 0 
 2

 y  a sin t
Green’s Theorem tells us that the area of R , the region enclosed by this simple
closed curve, is:
area of R= 12  xdy
C

2
0
 a cos t  b cos tdt 
 ab 
2
 ab 
2
0
0
 cos t  dt
2
1
2
1  cos 2t  dt
 12 ab t  12 sin 2t 0
2
  ab
Notice that if b is equal to a , then the ellipse is actually a circle of radius a , and
the expression for the area of the ellipse becomes  a , which we know is the ar2
ea of the circle.
147
Example 5.6
What is the value of the line integral
  sin  log  x  1   4 y  dx   6 x  y arctan e  dy
5
3
y
C
if C is the rectangle shown below?
y
 2,5
 0,4 
C
 4,1
 2,0 
x
Solution:
This line integral would be a nightmare to try to calculate directly, but we do not
have to do so. We simple notice that


N 

6 x  y 3 arctan e y  6
x x
M 

sin 5  log  x  1   4 y  4
y y


So by Green’s Theorem, the given line integral is equal to
 N
M 
  x  y dA    6  4dA  2 dA  2  area of R 
R
R
R
Where R is the rectangular region enclosed by C . The area of the rectangle is
equal to
 5  2 5   10 , so the value of the double integral, and thus the given
line integral, is 2 10   20 .
148
5.5
Path Independence and Gradient Fields
The statement of Green’s theorem allows us to complete some unfinished business from the last section. We know that the condition M y  N x is necessary for
the vector field F   M , N  to be a gradient field, that is, for the line integral
 F dr
C
to be path independent. The question we left unresolved was, is this condition
also sufficient to establish that F is a gradient field? By itself, the answer is
“no”, as the following important (and classic) example illustrates. Consider the
vector field:
 y
x 
F  x, y    2
, 2
2
2 
x y x y 
This vector field passes the test M y  N x , since
2
2
   y   x  y   1    y  2 y 
y 2  x2
My   2


2
2
y  x  y 2 
 x2  y 2 
 x2  y 2 
and
  x   x  y  1   x  2 x 
y 2  x2
Nx   2


2
2
2 2
x  x  y 2 
x

y


 x2  y 2 
2
2
But is the line integral of F path independent? If it is, then the integral around
any closed path would have to be zero. Let us integrate this field F around the
unit circle. We parameterize the circle by the equations x  cos t and y  sin t ,
with the parameter t increasing from 0 to 2 . Then:
 F dr    Mdx  Ndy 
C
C
 y

x
  2
dx  2
dy 
2
2
C x  y
x y


149
2
    sin t   sin tdt    cos t  cos tdt  
0
 2
Since this is not zero, we must conclude that F is not conservative.
But if M y  N x , is not the integrand of the double integral in Green’s Theorem
equal to zero, which means that the value of the line integral around any closed
curve (such as the unit circle) must also be zero? What went wrong in this example? Well, notice that the hypothesis of Green’s Theorem says that if M  x, y 
and N  x, y  are functions that are defined and have continuous partial derivatives both on C and throughout R (the region enclosed by C ), then equality of
the line integral and double integral is guaranteed. But for the vector field F defined above, the functions M and N are not defined and continuously differentiable throughout R ; the unit circle encloses the origin, and neither M nor N is defined at this point. Since this field F does not satisfy the hypothesis of the theorem, it should not be surprising that the conclusion of the theorem does not follow either.
To make sure this kind of behaviour is not seen, we can restrict the regions we
study to be convex, which means that for any pair of points in the region, every
point on the line segment joining the pair also lies in R . If R is the interior of the
unit circle with the origin excluded, then R is not convex, since, for example,
the line segment that joins the points   12 ,0  and  12 ,0  would have to pass
through the origin, but this point is excluded from R . A less restrictive condition
is to require that the domain of F merely be simply connected. This means that
the interior of every simple closed curve in the domain of F  which we are calling R  is contained in R ; intuitively, it means that R does not contain any
holes. Every convex set is simply connected but not every simply connected set
150
is convex, so requiring that the domain of F be simply connected is a weaker
(and more easily satisfied) condition than requiring that it be convex.
simply connected
and convex
simply connected
but not convex

neither simply connected
nor convex

So, if F  x, y   M  x, y  , N  x, y  is a vector field defined and continuously
differentiable throughout a simple connected region R of the plane, then all of
the following statements about F are equivalent:
M y  Nx
the line integral of F around
the line integral of F
 F is a gradient field 
every closed path equals zero
is path independent
Exercise
1. Determine whether or not each vector field is conservative and if it is, find a
scalar potential.
a) y i  2 xyj
2
b) 2 xy i  3 y x j
3
2 2
d)
xe xy sin yi   e xy cos y  y  j
e)
 y  e x sin yi   x  2  e xy cos yj
151
c)
 y  x  i   2x  y  j
2
2
e2 x sin yi  e2 x cos yj
f)
2. Show that a vector field F is conservative, find a scalar potential for F and
 F dr where C is any path connecting  0,0  to
evaluate the line integral
C
1,1 .
a) F  x, y    x  2 y  i   2 x  y  j
b) F  x, y   2 xyi  x j
2


c) F  x, y   2cos x  e  cos xy  y sin xy  i  xe sin xy j
x
3. Evaluate the line integral

x
 F dr where
C
 

a) F  e sin y  y i  e cos y  x  2 j and C is the path given by
x
x
t
r  t    t 3 sin 2t  i  2 cos  2t  2  j for 0 
1 .

3
xy   x 2
y
F

t
sin
y
/
x

i


e
1

y
b)





 j where C is the

2
2
2
2
x

y
x

y

 

curve given by r  t   e
t 1t 
t
 cos ti  t sin tj for 0 
1 .
4. Verify that each of the following vector field F is conservative.
a) yz i  xz j  2 xyzk
2
2
1
1
2
b) yz i  xz j  xyz k
c)
 y sin z  i   x sin z  2 y  j   xy cos z  k
d)
 xy  yz  i   x y  xz  3y z  j   xy  y k
2
2
2
3
152
5. The gravitational force field F between two particles of mass M and m
separated by a distance r is given by:
G  x, y, z   
kmMR
,
r3
where R  xi  yj  zk and k is the gravitational constant.
Show that F is conservative and find a scalar potential for F . Compute the
amount of work done against G by an object from  a1 , b1 , c1  to  a2 , b2 , c2  .
GREEN’S
  2 xy  x dx   x  y  dy where C
2
1. Verify Green’s theorem in plane for
2
C
is the closed curve of the region bounded by y  x and y  x .
2
2
2. Use Green’s theorem to evaluate
  x  xy dx   x  y  dy where C is
2
2
2
C
the square formed by lines y  1 .
Let C be any sample closed curve bounding a region having area A and
prove that if a1 , a2 , a3 , b1 , b2 , b3 are constants.
 a x  a y  a dx  b x  b x  b  dy  b  a 
1
C
2
3
1
2
3
1
2
3. Under what conditions will the line integral around any path be zero? Find the
area bounded by the hypocycloid x
2/3
 y 2/3  a 2/3 .
[Hint: Parametric equations are x  a cos t , y  a sin t ,0  t  2 ]
3
4. Verify Green’s Theorem in the plane for
3
  x  x y dx  xy dy where C is
3
2
2
C
the boundary of the regions enclosed by the curves x  y  4 and
2
2
x2  y 2  16 .
a) Evaluate
  2 xy  y cos x  dx  1  2 y sin x  3x y dy along the pa3
2
2
C
rabola 2x   y from  0,0  to  / 2,1
2
153
.
2
b) Evaluate the integral around a parallelogram with vertices at
 3,0 ,  5,2 ,  2,2 
154
 0,0  ,
UNIT SIX
SURFACE INTEGRL
6.0
INTRODUCTION
It is now time to think about integrating functions over some surface, S , in
three-dimensional space. Let’s start off with a sketch of the surface S since the
notation can get a little confusing once we get into it. Here is a sketch of some
surface S .
The region S will lie above (in this case) some region D that lies in the xy 
plane. We used a rectangle here, but it does not have to be of course. Also note
that we could just as easily looked at a surface S that was in front of some region D in the yz  plane or the xz  plane. Do not get so locked into the xy 
plane that you can’t do problems that have regions in the other two planes.
155
Terminology
A surface is smooth if it has a unit normal vector n  x, y, z  that varies continuously as the point x, y , z ranges over a surface and it is piecewise –smooth if it
consist of a finite number of smooth pieces.
A surface is orientable if it is possible to choose a unit normal n at each point on
the surface S so that n varies continuously over S . A closed surface is one that
bounds a solid region.
Definition 6.1
Let S be a surface defined by z  f  x, y  and D it’s projection on the xy 
plane. If f , f x , f y are continuous in D and g  x, y, z  is continuous on S , then
the surface integral of g over S is:
 g  x, y, z  dS    g  x, y, z  f  f  1dA
2
x
D
2
y
xy
S
dAxy  dxdy or dxdy
Example 6.1
Evaluate the surface integral
 g  x, y, z ds where g  x, y, z   xz  2 x  3xy
2
S
and S is that portion of the plane 2 x  3 y  z  6 that lies over the unit square:
Rxy : 0  x  1,0  y  1.
Solution:
ds 
f x2  f y2  1dAxy
z  6  2 x  3 y  f  x, y  ; f x  2, f y  3
So that
f x2  f y2  1dAxy  4  9  1dAxy  14dAxy
156
  g  x, y, z ds   g  x, y, z  14dAxy
S
Rxy
 
 xz  2 x  3xy  14dA
 
x  6  2 x  3 y   2 x  3xy 14dA  14   6 xdA
2
Rxy
xy
1 1
2
Rxy
xy
By type (1)
 g  x, y, z ds  14   6 xdydx  3 14
By type (2)
 g  x, y, z ds  14   6 xdxdy  3 14
0 0
xy
1 1
S
0 0
1 1
S
0 0
If g  x, y, z   9 then the surface integral
 ds  Surface Area of any
S
surface S .
6.1
Surface Integrals of Vector Fields
Application of surface integrals require the integral of the normal component of a
given vector field F that is the integral of the form
 F ndS where n is the unit
S
normal to the surface.
Example 6.2
Compute
 F ndS where F  xi  5 yj  4zk and S is the union of the two
S
squares: S1 : x  0,0  y  1,0  z  1 and S2 : z  1,0  x  1,0  y  1 .
Solution:
S1 : N  i, S2 : N  k
On S1 , F n   x  0  x  0 on S1  ,
 F n dS  0
S1
1
1
On S2 , n2 =k , F n2  4 z  4  z  1 on S2 
 F ndS  F n dS   F n dS  0  4 dS  4
S
S1
1
1
S2
2
2
157
S2
2
For an orientable surface defined S by z  g  x, y  let G  x, y, z   z  g  x, y   0
Then n (unit normal to S ) 
G
.
G
Example 3
Compute
 F ndS where F  xyi  zj   x  y  k and S is the triangular region
S
cut off from the plane x  y  z  1 by the positive coordinates.
Solution:
F   xyi  zj   x  y  k
z  1  x  y, G  z  g  x   x  y  z  1  0
i jk
n
12  1  1

1
i  j  k 
3
1
1
1
 xy  z  x  y    xy  1  x  y  x  y    xy  1
3
3
3
F n
S : z  g  x, y   1  x  y 
Also ds 
1
 xy  1
3
g x2  g y2  1dA where dA  dydx or dxdy
 1  1  1dA  3dA
1
 xy  1 3dA
R
3
  F ndS  
S
1 1 x
   xy  1 dA   
R
0 0
1 1 y
   xy  1 dA   
R

0 0
 xy  1 dydx (Type 1)
 xy  1 dxdy (Type 2)
13
24
158
6.3
Stokes Theorem
In this section we are going to take a look at a theorem that is a higher dimensional version of Green’s Theorem. In Green’s Theorem we related a line integral to a double integral over some region. In this section we are going to relate
a line integral to a surface integral. However, before we give the theorem we
first need to define the curve that we’re going to use in the line integral
Let’s start off with the following surface with the indicated orientation.
Around the edge of this surface we have a curve C . This curve is called the
boundary curve. The orientation of the surface S will induce the positive
orientation of C . To get the positive orientation of C think of yourself as
walking along the curve. While you are walking along the curve if your head is
pointing in the same direction as the unit normal vectors while the surface is on
the left then you are walking in the positive direction on C .
159
Definition 6.2:
The orientation of a closed path C on the surface S is compatible with the orientation on S if the positive direction is counter clockwise in relation to the outward normal of the surface.
Theorem 6.1
Let S be an orientable surface with unit normal n and assume that S is bounded by closed piecewise – smooth curve C whose orientable is compatible with
that of S . If F is a vector field that is continuously differentiable on S , then:
 F dr    F ndS
C
S
Example 6.2
1. Let F  x, y, z    32 y i  2 xyj  yzk whose S is that part of the surface of
2
the plane x  y  z  1 contained within the triangle C with vectors
1,0,0 ,  0,1,0 ,  0,0,1 traversed counter clockwise as viewed above. Verify
Stoke’s Theorem.
Solution:
 0,0,1
 0,1,0
1,0,0
We wish to verify that
 F dr     F ndS
C
C
160
Consider the LHS=
 F dr
C
E1 : x  y  1, z  0 ,
E2 : y  z  1, x  0
In the counter clockwise direction
E3 : x  z  1, y  0
On E1 : let x  1  t , y  t , 0  t  1 ,  dx   dt , dy  dt
3
F   y 2i  2 xyj  yzk dr  idx  jdy  kdz
2
3
 F dr   y 2dx  2 xydy  yzdz
2
1
2
2
 F dr    32 t  2t  2t dt  16
E1
0
On E2 : x  0 , y  1  s , z  s , 0  s  1 dy  ds, dz  ds then,

E2
F dr   1  s ds  16
1
0
On E3 : y  0 , x  r and z  1  r , 0  r  1 dx  dr , dz  dr , then
2
 F dr   0 dr  0
E3
0
1
1
1
 F dr   F dr   F dr   F dr  6  6  0  3
C
E1
Consider the RHS = 
E2
E3
   F ndS
C
i

Clearly,   F 
x
 32 y 2
j

y
2 xy
k

 zi  yk
z
yz
The triangular region on the surface of the plane x  y  z  1 has unit normal:
161
1
i  j  k 
3
n
1
1
 z  y   1  x  since z  1  x  y
3
3
  F n 
Finally, ds 
zx 2  zy 2  1dAxy
    F ndS  
1
R 3
C
1  x 
1 1 x
1  x dydx 
0 0
1
or
3
1 1 y
1
3
 
1  x dxdy 
0 0
 
2. Evaluate
3dAxy
 y dx  zdy  xdz where C is the curve of the intersection of
2
1
C2
the plane x  z  1 and the ellipsoid x  2 y  z  1 .
2
2
2
Solution:
F  12 y 2i  zj  xk where dr  idx  jdy  dzk
Then, 12 y dx  zdy  xdz is the required line integral. We use Stoke’s theo2
rem to evaluate the above.
Then F  12 y i  zj  xk
2
Then  F  i  j  yk
n  12  i  k     F n  12  1  y 
Also, ds 
z x2  z y2  1dAxy 
 12  1dAxy
Finally to describe C , let z  1  x with the ellipsoid to obtain
x 2  y 2  1  x   1   x  12   y 2 
2
2
with centre 1 / 2,0  and radius 1 / 2 .
162
1
on the xy  plane is a circle
4
We finally require R of dAxy to be the circle, but we wish to employ polar
coordinates to resolve the double integral, then we have the following
 /2
cos 
 1  y dA     1  r sin   rdrd   / 4
R
xy
 /2 0
163
6.4
Divergence Theorem
In this section we are going to relate surface integrals to triple integrals. We will
do this with the
Definition 6.3:
The Divergence Theorem says that under suitable conditions, the outward flux of
a vector field across a closed surface (oriented outward) equals the triple integral
of the divergence of the field over the region enclosed by the surface.
Theorem 6.2:
The flux of a vector field F across a closed oriented surface S in the direction of
the surface of the surface’s outward unit normal field n equals the integral of
 F over the region D enclosed by the surface:
 F ndS    FdV
S
Outward
flux
D
Divergence
integral
Example 6.3
Evaluate both sides of the theorem for the field F  xi  yj  zk over the sphere
x2  y 2  z 2  a2 .
Solution:
The outer unit normal to S , calculated from the gradient of
f  x, y, z   x 2  y 2  z 2  a 2 is
n
2  xi  yj  zk 

4 x2  y 2  z 2


xi  yj  zk
a
Hence,
164
x2  y 2  z 2
a2
F ndS 
dS  dS  adS
a
a
Because x  y  z  a on the surface. Therefore,
2
2
2
2
 F ndS   adS  a dS  a  4 a   4 a
2
S
S
3
S
The divergence of F is:
 F
So



 x   y   z   3
x
y
z
  FdV   3dV  3  a   4 a
D
4
3
D
3
3
Exercise 6.1`(Surface)
1. Let S be the hemisphere x  y  z  4 with z  0 . Evaluate the sur2
2
face integral
a)
 zdS b)
S
2. Evaluate
  x  2 y  dS
S
c)
 xydS where
S
a) S : z  2  y 0  x  z,0  y  z
b) S : z  5 , x  y  1
2
3. Evaluate
2
  x  y dS where
2
2
S
a) S : z  4  x  3 y;0  x  4,0  y  2
b)
S : z  4; x 2  y  1
165
  x  y  zdS
2
S
2