8
Calculus of Several Variables
 Functions of Several Variables
 Partial Derivatives
 Maxima and Minima of Functions of
Several Variables
 The Method of Least Squares
 Constrained Maxima and Minima and
the Method of Lagrange Multipliers
 Double Integrals
8.1
Functions of Several Variables
z
f(x, y) = x2 + y2
x2 + y2 = 16
x2 + y2 = 9
x2 + y2 = 4
x2 + y2 = 1
x2 + y2 = 0
x
y
Functions of Two Variables
A real-valued function of two variables f, consists of
1. A set A of ordered pairs of real numbers (x, y)
called the domain of the function.
2. A rule that associates with each ordered pair in
the domain of f one and only one real number,
denoted by z = f(x, y).
Examples
 Let f be the function defined by
f ( x, y )  x  xy  y 2  2
 Compute f(0, 0), f(1, 2), and f(2, 1).
Solution
f (0,0)  0  (0)(0)  02  2  2
f (1,2)  1  (1)(2)  22  2  9
f (2,1)  2  (2)(1)  12  2  7
 The domain of a function of two variables f(x, y), is a set of
ordered pairs of real numbers and may therefore be
viewed as a subset of the xy-plane.
Example 1, page 536
Examples
 Find the domain of the function
f ( x, y )  x 2  y 2
Solution
 f(x, y) is defined for all real values of x and y, so the domain
of the function f is the set of all points (x, y) in the xy-plane.
Example 2, page 536
Examples
 Find the domain of the function
2
g ( x, y ) 
x y
Solution
 g(x, y) is defined for all x ≠ y, so the domain of the function
g is the set of all points (x, y) in the xy-plane except those
lying on the y = x line.
y
y=x
x
Example 2, page 536
Examples
 Find the domain of the function
h ( x, y )  1  x 2  y 2
Solution
 We require that 1 – x2 – y2  0 or x2 + y2  1 which is the set
of all points (x, y) lying on and inside the circle of radius 1
with center at the origin:
y
1
–1
1
–1
Example 2, page 536
x2 + y2 = 1
x
Applied Example: Revenue Functions
 Acrosonic manufactures a bookshelf loudspeaker system
that may be bought fully assembled or in a kit.
 The demand equations that relate the unit price, p and q,
to the quantities demanded weekly, x and y, of the
assembled and kit versions of the loudspeaker systems are
given by
1
1
1
3
p  300  x  y and q  240  x  y
4
8
8
8
a. What is the weekly total revenue function R(x, y)?
b. What is the domain of the function R?
Applied Example 3, page 537
Applied Example: Revenue Functions
Solution
a. The weekly revenue from selling x units assembled speaker
systems at p dollars per unit is given by xp dollars.
Similarly, the weekly revenue from selling y speaker kits at
q dollars per unit is given by yq dollars.
Therefore, the weekly total revenue function R is given by
R( x, y )  xp  yq
1
1 

 x  300  x  y  
4
8 

1
3 

y  240  x  y 
8
8 

1 2 3 2 1
  x  y  xy  300 x  240 y
4
8
4
Applied Example 3, page 537
Applied Example: Revenue Functions
Solution
b. To find the domain of the function R, note that the
quantities x, y, p, and q must be nonnegative, which leads
to the following system of linear inequalities:
300 
1
1
x y0
4
8
1
3
240  x  y  0
8
8
x0
Thus, the graph of the domain is:
y
2000
300 
1
1
x y0
4
8
1000
1
3
240  x  y  0
8
8
y0
D
1000
Applied Example 3, page 537
2000
x
Graphs of Functions of Two Variables
 Consider the task of locating P(1, 2, 3) in 3-space:
 One method to achieve this is to start at the origin and
measure out from there, axis by axis:
z
P(1, 2, 3)
3
y
1
2
x
Graphs of Functions of Two Variables
 Consider the task of locating P(1, 2, 3) in 3-space:
 Another common method is to find the xy coordinate and
from there elevate to the level of the z value:
z
P(1, 2, 3)
3
2
1
x
(1, 2)
y
Graphs of Functions of Two Variables
 Locate the following points in 3-space:
Q(–1, 2, 3), R(1, 2, –2), and S(1, –1, 0).
Solution
z
Q(–1, 2, 3)
3
y
S(1, –1, 0)
x
–2
R(1, 2, –2)
Graphs of Functions of Two Variables
 The graph of a function in 3-space is a surface.
 For every (x, y) in the domain of f, there is a z value on the
surface.
z
z = f(x, y)
(x, y, z)
y
(x, y)
x
Level Curves
 The graph of a function of two variables is often difficult
to sketch.
 It can therefore be useful to apply the method used to
construct topographic maps.
 This method is relatively easy to apply and conveys
sufficient information to enable one to obtain a feel for the
graph of the function.
Level Curves
 In the 3-space graph we just saw, we can delineate the
contour of the graph as it is cut by a z = c plane:
z
z = f(x, y)
z=c
y
f(x, y) = c
x
Examples
 Sketch a contour map of the function f(x, y) = x2 + y2.
Solution
 The function f(x, y) = x2 + y2 is a revolving parabola called
a paraboloid.
z
f(x, y) = x2 + y2
y
x
Example 5, page 540
Examples
 Sketch a contour map of the function f(x, y) = x2 + y2.
Solution
 A level curve is the graph of the equation x2 + y2 = c, which
describes a circle with radius c .
 Taking different
values of c we obtain:
z
f(x, y) = x2 + y2
y
4
2
x2 + y2 = 16
–4
x2 + y2 = 9
x2
x2
+ =4
+ y2 = 1
x2 + y2 = 0
x
y2
y
–2
x2 + y2 = 16
x2 + y2 = 9
x2 + y2 = 4
x2 + 2y2 = 1
2
x +y =0
2
4
–2
–4
Example 5, page 540
x
Examples
 Sketch level curves of the function f(x, y) = 2x2 – y
corresponding to z = –2, –1, 0, 1, and 2.
Solution
 The level curves are the graphs of the equation 2x2 – y = k
or for k = –2, –1, 0, 1, and 2:
y
4
2x2 – y = – 2
2x2 – y = – 1
3
2x2 – y = 0
2
2x2 – y = 1
2x2 – y = 2
1
–2
–1
0
1
–1
–2
Example 6, page 540
2
x
8.2
Partial Derivatives

x
f
x

x
  f   2 f
 
x  x  x 2

y
  f   2 f
 
y  x  yx

x
  f   2 f



x  y  xy
f

y
f
y

y
  f   2 f

y  y  y 2
2 f
2 f

yx xy
When both are
continuous
First Partial Derivatives
First Partial Derivatives of f(x, y)
 Suppose f(x, y) is a function of two variables x and y.
 Then, the first partial derivative of f with respect to x
at the point (x, y) is
f
f ( x  h, y )  f ( x , y )
 lim
x h0
h
provided the limit exists.
 The first partial derivative of f with respect to y at the
point (x, y) is
f
f ( x, y  k )  f ( x, y )
 lim
y k 0
k
provided the limit exists.
Geometric Interpretation of the Partial Derivative
z
f
What does
mean?
x
f(x, y)
y
x
Geometric Interpretation of the Partial Derivative
z
f
What does
mean?
x
f(x, y)
f
 slope of f ( x, b)
x
f(x, b)
y = b plane
b
a
(a, b)
x
y
Geometric Interpretation of the Partial Derivative
z
f
What does
mean?
y
f(x, y)
y
x
Geometric Interpretation of the Partial Derivative
z
f
What does
mean?
y
f(x, y)
f
 slope of f (c, y )
y
f(c, y)
x = c plane
d
c
(c, d)
x
y
Examples
 Find the partial derivatives ∂f/∂x and ∂f/∂y of the function
f ( x, y )  x 2  xy 2  y 3
 Use the partials to determine the rate of change of f in the
x-direction and in the y-direction at the point (1, 2) .
Solution
 To compute ∂f/∂x, think of the variable y as a constant and
differentiate the resulting function of x with respect to x:
f ( x, y )  x 2  y 2 x  y 3
f
 2x  y2
x
Example 1, page 546
Examples
 Find the partial derivatives ∂f/∂x and ∂f/∂y of the function
f ( x, y )  x 2  xy 2  y 3
 Use the partials to determine the rate of change of f in the
x-direction and in the y-direction at the point (1, 2).
Solution
 To compute ∂f/∂y, think of the variable x as a constant and
differentiate the resulting function of y with respect to y:
f ( x, y )  x 2  xy 2  y 3
f
 2 xy  3 y 2
y
Example 1, page 546
Examples
 Find the partial derivatives ∂f/∂x and ∂f/∂y of the function
f ( x, y )  x 2  xy 2  y 3
 Use the partials to determine the rate of change of f in the
x-direction and in the y-direction at the point (1, 2).
Solution
 The rate of change of f in the x-direction at the point (1, 2)
is given by
f
 2(1)  22  2
x (1,2)
 The rate of change of f in the y-direction at the point (1, 2)
is given by
Example 1, page 546
f
 2(1)(2)  3(2)2  8
y (1,2)
Examples
 Find the first partial derivatives of the function
w( x, y ) 
xy
x2  y2
Solution
 To compute ∂w/∂x, think of the variable y as a constant and
differentiate the resulting function of x with respect to x:
xy
w( x, y )  2
x  y2
w ( x 2  y 2 ) y  xy (2 x )

x
( x 2  y 2 )2
y( y 2  x 2 )
 2
( x  y 2 )2
Example 2, page 547
Examples
 Find the first partial derivatives of the function
w( x, y ) 
xy
x2  y2
Solution
 To compute ∂w/∂y, think of the variable x as a constant and
differentiate the resulting function of y with respect to y:
xy
w( x, y )  2
x  y2
w ( x 2  y 2 ) x  xy (2 y )

y
( x 2  y 2 )2
x( x 2  y 2 )
 2
( x  y 2 )2
Example 2, page 547
Examples
 Find the first partial derivatives of the function
g ( s, t )  ( s 2  st  t 2 )5
Solution
 To compute ∂g/∂s, think of the variable t as a constant and
differentiate the resulting function of s with respect to s:
g ( s, t )  ( s 2  st  t 2 )5
g
 5( s 2  st  t 2 ) 4  (2 s  t )
s
 5(2s  t )( s 2  st  t 2 )4
Example 2, page 547
Examples
 Find the first partial derivatives of the function
g ( s, t )  ( s 2  st  t 2 )5
Solution
 To compute ∂g/∂t, think of the variable s as a constant and
differentiate the resulting function of t with respect to t:
g ( s, t )  ( s 2  st  t 2 )5
g
 5( s 2  st  t 2 ) 4  (  s  2t )
t
 5(2t  s)( s 2  st  t 2 )4
Example 2, page 547
Examples
 Find the first partial derivatives of the function
h(u, v)  e
u 2 v 2
Solution
 To compute ∂h/∂u, think of the variable v as a constant and
differentiate the resulting function of u with respect to u:
h(u, v)  e
u 2 v 2
h
u2 v2
e
 2u
u
 2ue
Example 2, page 547
u2 v 2
Examples
 Find the first partial derivatives of the function
h(u, v)  e
u 2 v 2
Solution
 To compute ∂h/∂v, think of the variable u as a constant and
differentiate the resulting function of v with respect to v:
h(u, v)  e
u 2 v 2
h
u2 v2
e
 ( 2v )
u
 2ve
Example 2, page 547
u 2 v 2
Examples
 Find the first partial derivatives of the function
w  f ( x, y, z )  xyz  xe yz  x ln y
Solution
 Here we have a function of three variables, x, y, and z, and
we are required to compute
f f f
,
,
x y z
 For short, we can label these first partial derivatives
respectively fx, fy, and fz.
Example 3, page 549
Examples
 Find the first partial derivatives of the function
w  f ( x, y, z )  xyz  xe yz  x ln y
Solution
 To find fx, think of the variables y and z as a constant and
differentiate the resulting function of x with respect to x:
w  f ( x, y, z )  xyz  xe yz  x ln y
f x  yz  e yz  ln y
Example 3, page 549
Examples
 Find the first partial derivatives of the function
w  f ( x, y, z )  xyz  xe yz  x ln y
Solution
 To find fy, think of the variables x and z as a constant and
differentiate the resulting function of y with respect to y:
w  f ( x, y, z )  xyz  xe yz  x ln y
x
f y  xz  xze 
y
yz
Example 3, page 549
Examples
 Find the first partial derivatives of the function
w  f ( x, y, z )  xyz  xe yz  x ln y
Solution
 To find fz, think of the variables x and y as a constant and
differentiate the resulting function of z with respect to z:
w  f ( x, y, z )  xyz  xe yz  x ln y
f z  xy  xye yz
Example 3, page 549
The Cobb-Douglas Production Function
 The Cobb-Douglass Production Function is of the form
f(x, y) = axby1– b
(0 < b < 1)
where
a and b are positive constants,
x stands for the cost of labor,
y stands for the cost of capital equipment, and
f measures the output of the finished product.
The Cobb-Douglas Production Function
 The Cobb-Douglass Production Function is of the form
f(x, y) = axby1– b
(0 < b < 1)
 The first partial derivative fx is called the marginal
productivity of labor.
✦ It measures the rate of change of production with respect
to the amount of money spent on labor, with the level of
capital kept constant.
 The first partial derivative fy is called the marginal
productivity of capital.
✦ It measures the rate of change of production with respect
to the amount of money spent on capital, with the level of
labor kept constant.
Applied Example: Marginal Productivity
 A certain country’s production in the early years following
World War II is described by the function
f(x, y) = 30x2/3y1/3
when x units of labor and y units of capital were used.
 Compute fx and fy.
 Find the marginal productivity of labor and the marginal
productivity of capital when the amount expended on
labor and capital was 125 units and 27 units, respectively.
 Should the government have encouraged capital
investment rather than increase expenditure on labor to
increase the country’s productivity?
Applied Example 4, page 550
Applied Example: Marginal Productivity
f(x, y) = 30x2/3y1/3
Solution
 The first partial derivatives are
1/3
2 1/3 1/3
 y
f x  30  x y  20  
3
x
x
1
f y  30 x 2/3  y 2/3  10  
3
 y
Applied Example 4, page 550
2/3
Applied Example: Marginal Productivity
f(x, y) = 30x2/3y1/3
Solution
 The required marginal productivity of labor is given by
1/3
 27 
f x (125, 27)  20 

125


 3
 20    12
 5
or 12 units of output per unit increase in labor expenditure
(keeping capital constant).
 The required marginal productivity of capital is given by
 125 
f y (125, 27)  10 

27


2/3
 25 
 10    27 79
 9 
or 27 7/9 units of output per unit increase in capital
expenditure (keeping labor constant).
Applied Example 4, page 550
Applied Example: Marginal Productivity
f(x, y) = 30x2/3y1/3
Solution
 The government should definitely have encouraged capital
investment.
 A unit increase in capital expenditure resulted in a much
faster increase in productivity than a unit increase in labor:
27 7/9 versus 12 per unit of investment, respectively.
Applied Example 4, page 550
Second Order Partial Derivatives
 The first partial derivatives fx(x, y) and fy(x, y) of a function
f(x, y) of two variables x and y are also functions of x and y.
 As such, we may differentiate each of the functions fx and
fy to obtain the second-order partial derivatives of f.
Second Order Partial Derivatives
 Differentiating the function fx with respect to x leads to the
second partial derivative
2 f

f xx  2  ( f x )
x
x
 But the function fx can also be differentiated with respect
to y leading to a different second partial derivative
2 f

f xy 
 ( fx )
yx y
Second Order Partial Derivatives
 Similarly, differentiating the function fy with respect to y
leads to the second partial derivative
2 f

f yy  2  ( f y )
y
y
 Finally, the function fy can also be differentiated with
respect to x leading to the second partial derivative
2 f

f yx 
 ( fy)
xy x
Second Order Partial Derivatives
 Thus, four second-order partial derivatives can be
obtained of a function of two variables:

x
f
x

x
  f   2 f
  2
x  x  x

y
  f   2 f
 
y  x  yx

x
  f   2 f



x  y  xy
f

y
f
y

y
  f   2 f
 2


y  y  y
2 f
2 f

yx xy
When both are
continuous
Examples
 Find the second-order partial derivatives of the function
f ( x, y )  x 3  3x 2 y  3xy 2  y 2
Solution
 First, calculate fx and use it to find fxx and fxy:
fx 
 3
( x  3x 2 y  3xy 2  y 2 )
x
 3x 2  6 xy  3 y 2

f xx  (3x 2  6 xy  3 y 2 )
x

f xy  (3x 2  6 xy  3 y 2 )
y
 6x  6 y
 6 x  6 y
 6( x  y )
 6( y  x )
Example 6, page 552
Examples
 Find the second-order partial derivatives of the function
f ( x, y )  x 3  3x 2 y  3xy 2  y 2
Solution
 Then, calculate fy and use it to find fyx and fyy:
 3
f y  ( x  3x 2 y  3xy 2  y 2 )
y
 3x 2  6 xy  2 y

f yx  ( 3x 2  6 xy  2 y )
x

f yy  ( 3x 2  6 xy  2 y )
y
 6 x  6 y
 6x  2
 6( y  x )
 2(3 x  1)
Example 6, page 552
Examples
 Find the second-order partial derivatives of the function
f ( x, y )  e
xy 2
Solution
 First, calculate fx and use it to find fxx and fxy:
 xy 2
f x  (e )
x
ye
2 xy 2
 2 xy 2
f xx  ( y e )
x
ye
4 xy 2
 2 xy 2
f xy  ( y e )
y
 2 ye
xy 2
 2 xy e
3 xy 2
 2 ye (1  xy 2 )
xy 2
Example 7, page 553
Examples
 Find the second-order partial derivatives of the function
f ( x, y )  e
xy 2
Solution
 Then, calculate fy and use it to find fyx and fyy:
 xy 2
f y  (e )
y
 2 xye

xy 2
f yy  (2 xye )
y

xy 2
f yx  (2 xye )
x
 2 ye
xy 2
 2 xy e
3 xy 2
 2 ye (1  xy )
xy 2
Example 7, page 553
xy 2
2
 2 xe
xy 2
 (2 xy)(2 xy)e
 2 xe (1  2 xy 2 )
xy 2
xy 2
8.3
Maxima and Minima
of Functions of Several Variables
z
x
(g, h)
(a, b)
(c, d)
(e, f )
y
Relative Extrema of a Function of Two Variables
 Let f be a function defined on a region R
containing the point (a, b).
 Then, f has a relative maximum at (a, b)
if f(x, y)  f(a, b) for all points (x, y) that are
sufficiently close to (a, b).
✦ The number f(a, b) is called a relative
maximum value.
 Similarly, f has a relative minimum at (a, b)
if f(x, y)  f(a, b) for all points (x, y) that are
sufficiently close to (a, b).
✦ The number f(a, b) is called a relative
minimum value.
Graphic Example
 There is a relative maximum at (a, b).
z
x
(a, b)
y
Graphic Example
 There is an absolute maximum at (c, d).
(It is also a relative maximum)
z
x
(c, d)
y
Graphic Example
 There is a relative minimum at (e, f ).
z
x
y
(e, f )
Graphic Example
 There is an absolute minimum at (g, h).
(It is also a relative minimum)
z
x
(g, h)
y
Relative Minima
 At a minimum point of the graph of a function of two
variables, such as point (a, b) below, the plane tangent to the
graph of the function is horizontal (assuming the surface of
the graph is smooth):
z
y
x
(a, b)
Relative Minima
 Thus, at a minimum point, the graph of the function has a
slope of zero along a direction parallel to the x-axis:
z
f
( a , b)  0
x
y
x
(a, b)
Relative Minima
 Similarly, at a minimum point, the graph of the function
has a slope of zero along a direction parallel to the y-axis:
z
f
( a , b)  0
y
y
x
(a, b)
Relative Maxima
 At a maximum point of the graph of a function of two
variables, such as point (a, b) below, the plane tangent to the
graph of the function is horizontal (assuming the surface of
the graph is smooth):
z
y
(a, b)
x
Relative Maxima
 Thus, at a maximum point, the graph of the function has a
slope of zero along a direction parallel to the x-axis:
z
f
( a , b)  0
x
y
(a, b)
x
Relative Maxima
 Similarly, at a maximum point, the graph of the function
has a slope of zero along a direction parallel to the y-axis:
z
f
( a , b)  0
y
y
(a, b)
x
Saddle Point
 In the case of a saddle point, both partials are equal to zero,
but the point is neither a maximum nor a minimum.
z
y
x
Saddle Point
In the case of a saddle point, the function is at a minimum
along one vertical plane…
z
f
( a , b)  0
x
y
(a, b)
x
Saddle Point
… but at a maximum along the perpendicular vertical plane.
z
f
( a , b)  0
y
y
(a, b)
x
Extrema When Partial Derivatives are Not Defined
 A maximum (or minimum) may also occur when both partial
derivatives are not defined, such as point (a, b) in the graph
below:
z
(a, b, f(a, b))
y
x
(a, b)
Critical Point of a Function
 A critical point of f is a point (a, b) in the domain of f
such that both
f
( a , b)  0
x
and
f
(a, b)  0
y
or at least one of the partial derivatives does not exist.
Determining Relative Extrema
1. Find the critical points of f(x, y) by solving the system of
simultaneous equations
fx = 0
fy = 0
2. The second derivative test: Let
D(x, y) = fxx fyy – f 2xy
3. Then,
a. D(a, b) > 0 and fxx(a, b) < 0 implies that f(x, y) has a
relative maximum at the point (a, b).
b. D(a, b) > 0 and fxx(a, b) > 0 implies that f(x, y) has a
relative minimum at the point (a, b).
c. D(a, b) < 0 implies that f(x, y) has neither a relative
maximum nor a relative minimum at the point (a, b), it
has instead a saddle point.
d. D(a, b) = 0 implies that the test is inconclusive, so some
other technique must be used to solve the problem.
Examples
 Find the relative extrema of the function
f ( x, y )  x 2  y 2
Solution
 We have
fx = 2x
and
fy = 2y.
 To find the critical points, we set fx = 0 and fy = 0 and solve
the resulting system of simultaneous equations
2x = 0
and
2y = 0
obtaining x = 0, y = 0, or (0, 0), as the sole critical point.
 Next, apply the second derivative test to determine the
nature of the critical point (0, 0).
 We compute
 Thus,
fxx = 2,
fyy = 2,
and
fxy = 0,
D(x, y) = fxx fyy – f 2xy = (2)(2) – (0)2 = 4.
Example 1, page 561
Examples
 Find the relative extrema of the function
f ( x, y )  x 2  y 2
Solution
 We have D(x, y) = 4, and in particular, D(0, 0) = 4.
 Since D(0, 0) > 0 and fxx = 2 > 0, we conclude that f has a
relative minimum at the point (0, 0).
 The relative minimum value, f (0, 0) = 0, also happens to
be the absolute minimum of f.
Example 1, page 561
Examples
 Find the relative extrema of the function
f ( x, y )  x 2  y 2
Solution
 The relative minimum
value, f(0, 0) = 0, also
happens to be the
absolute minimum
of f:
Example 1, page 561
z
x
f(x, y) = x2 + y2
Absolute
minimum
at (0, 0, 0).
y
Examples
 Find the relative extrema of the function
f ( x, y )  3x 2  4 xy  4 y 2  4 x  8 y  4
Solution
 We have
f x  6x  4 y  4
and
f y  4 x  8 y  8
 To find the critical points, we set fx = 0 and fy = 0 and solve
the resulting system of simultaneous equations
6x – 4y – 4 = 0
and
– 4x + 8y + 8 = 0
obtaining x = 0, y = –1, or (0, –1), as the sole critical point.
 Next, apply the second derivative test to determine the
nature of the critical point (0, –1).
 We compute
 Thus,
fxx = 6,
fyy = 8,
and
fxy = – 4,
D(x, y) = fxx · fyy – f 2xy = (6)(8) – (– 4)2 = 32.
Example 2, page 562
Examples
 Find the relative extrema of the function
f ( x, y )  3x 2  4 xy  4 y 2  4 x  8 y  4
Solution
 We have D(x, y) = 32, and in particular, D(0, –1) = 32.
 Since D(0, –1) > 0 and fxx = 6 > 0, we conclude that f has a
relative minimum at the point (0, –1).
 The relative minimum value, f (0, –1) = 0, also happens to
be the absolute minimum of f.
Example 2, page 562
Examples
 Find the relative extrema of the function
f ( x, y )  4 y 3  x 2  12 y 2  36 y  2
Solution
 We have
f x  2x
and f y  12 y 2  24 y  36
 To find the critical points, we set fx = 0 and fy = 0 and solve
the resulting system of simultaneous equations
2 x  0 and 12 y 2  24 y  36  0
 The first equation implies that x = 0, while the second
equation implies that y = –1 or y = 3.
 Thus, there are two critical points of f : (0, –1) and (0, 3).
 To apply the second derivative test, we calculate
fxx = 2
fyy = 24(y – 1)
fxy = 0
D(x, y) = fxx · fyy – f 2xy = (2)· 24(y – 1) – (0)2 = 48(y – 1)
Example 3, page 562
Examples
 Find the relative extrema of the function
f ( x, y )  4 y 3  x 2  12 y 2  36 y  2
Solution
 Apply the second derivative test to the critical point (0, –1):
 We have D(x, y) = 48(y – 1).
 In particular, D(0, –1) = 48[(–1) – 1] = – 96.
 Since D(0, –1) = – 96 < 0 we conclude that f has a saddle
point at (0, –1).
 The saddle point value is f (0, –1) = 22, so there is a saddle
point at (0, –1, 22).
Example 3, page 562
Examples
 Find the relative extrema of the function
f ( x, y )  4 y 3  x 2  12 y 2  36 y  2
Solution
 Apply the second derivative test to the critical point (0, 3):
 We have D(x, y) = 48(y – 1).
 In particular, D(0, 3) = 48[(3) – 1] = 96.
 Since D(0, –1) = 96 > 0 and fxx (0, 3) = 2 > 0, we conclude that
f has a relative minimum at the point (0, 3).
 The relative minimum value, f (0, 3) = –106, so there is a
relative minimum at (0, 3, –106).
Example 3, page 562
Applied Example: Maximizing Profit
 The total weekly revenue that Acrosonic realizes in
producing and selling its loudspeaker system is given by
1 2 3 2 1
R( x, y )   x  y  xy  300 x  240 y
4
8
4
where x denotes the number of fully assembled units and y
denotes the number of kits produced and sold each week.
 The total weekly cost attributable to the production of
these loudspeakers is
C ( x, y )  180 x  140 y  5000
 Determine how many assembled units and how many kits
should be produced per week to maximize profits.
Applied Example 3, page 563
Applied Example: Maximizing Profit
Solution
 The contribution to Acrosonic’s weekly profit stemming
from the production and sale of the bookshelf loudspeaker
system is given by
P ( x , y )  R ( x, y )  C ( x, y )
 1 2 3 2 1

   x  y  xy  300 x  240 y   (180 x  140 y  5000)
8
4
 4

1
3
1
  x 2  y 2  xy  120 x  100 y  5000
4
8
4
Applied Example 3, page 563
Applied Example: Maximizing Profit
Solution
1 2 3 2 1
 We have P( x, y )   x  y  xy  120 x  100 y  5000
4
8
4
 To find the relative maximum of the profit function P, we
first locate the critical points of P.
 Setting Px and Py equal to zero, we obtain
1
1
3
1
Px   x  y  120  0 and Py   y  x  100  0
2
4
4
4
 Solving the system of equations we get x = 208 and y = 64.
 Therefore, P has only one critical point at (208, 64).
Applied Example 3, page 563
Applied Example: Maximizing Profit
Solution
 To test if the point (208, 64) is a solution to the problem, we
use the second derivative test.
 We compute
1
3
1
Pxx  
Pyy  
Pxy  
2
4
4
 So,
2
 1  3   1  3 1 5
D( x, y )             
 2  4   4  8 16 16
 In particular, D(208, 64) = 5/16 > 0.
 Since D(208, 64) > 0 and Pxx(208, 64) < 0, the point (208, 64)
yields a relative maximum of P.
Applied Example 3, page 563
Applied Example: Maximizing Profit
Solution
 The relative maximum at (208, 64) is also the absolute
maximum of P.
 We conclude that Acrosonic can maximize its weekly profit
by manufacturing 208 assembled units and 64 kits.
 The maximum weekly profit realizable with this output is
1 2 3 2 1
P( x, y )   x  y  xy  120 x  100 y  5000
4
8
4
1
3
1
P(208,64)   (208) 2  (64) 2  (208)(64)
4
8
4
 120(208)  100(64)  5000
 $10,680
Applied Example 3, page 563
8.4
The Method of Least Squares
y
d5
L
10
d3
5
d4
d2
d1
5
10
x
The Method of Least Squares
 Suppose we are given the data points
P1(x1, y1), P2(x2, y2), P3(x3, y3), P4(x4, y4), and P5(x5, y5)
that describe the relationship between two variables x
and y.
 By plotting these data points, we obtain a scatter diagram:
y
P5
10
P3
P2
P4
5
P1
5
10
x
The Method of Least Squares
 Suppose we try to fit a straight line L to the data points
P1, P2, P3, P4, and P5.
 The line will miss these points by the amounts
d1, d2, d3, d4, and d5 respectively.
y
d5
L
10
d3
5
d4
d2
d1
5
10
x
The Method of Least Squares
 The principle of least squares states that the straight line L
that fits the data points best is the one chosen by requiring
that the sum of the squares of d1, d2, d3, d4, and d5, that is
d12  d 22  d 32  d 42  d 52
be made as small as possible.
y
d5
L
10
d3
5
d4
d2
d1
5
10
x
The Method of Least Squares
 Suppose the regression line L is y = f(x) = mx + b, where m
and b are to be determined.
 The distances d1, d2, d3, d4, and d5, represent the errors the
line L is making in estimating these points, so that
d1  f ( x1 )  y1 , d 2  f ( x2 )  y2 , d 3  f ( x3 )  y3 , and so on.
y
d5
L
10
d3
5
d4
d2
d1
5
10
x
The Method of Least Squares
 Observe that
d12  d 22  d 32  d 42  d 52
 [ f ( x1 )  y1 ]2  [ f ( x2 )  y2 ]2  [ f ( x3 )  y3 ]2
 [ f ( x4 )  y4 ]2  [ f ( x5 )  y5 ]2
 [mx1  b  y1 ]2  [mx2  b  y2 ]2  [mx3  b  y3 ]2
 [mx4  b  y4 ]2  [mx5  b  y5 ]2
 This may be viewed as a function of two variables m and b.
 Thus, the least-squares criterion is equivalent to minimizing
the function
f (m, b)  (mx1  b  y1 )2  (mx2  b  y2 )2  (mx3  b  y3 )2
 (mx4  b  y4 )2  (mx5  b  y5 )2
The Method of Least Squares
 We want to minimize
f (m, b)  (mx1  b  y1 )2  (mx2  b  y2 )2  (mx3  b  y3 )2
 (mx4  b  y4 )2  (mx5  b  y5 )2
 We first find the partial derivative with respect to m:
f
 2(mx1  b  y1 ) x1  2(mx2  b  y2 ) x2  2(mx3  b  y3 ) x3
m
 2(mx4  b  y4 ) x4  2(mx5  b  y5 ) x5
 2[mx12  bx1  y1 x1  mx22  bx2  y2 x2  mx32  bx3  y3 x3
 mx42  bx4  y4 x4  mx52  bx5  y5 x5 ]
 2[( x12  x22  x32  x42  x52 )m  ( x1  x2  x3  x4  x5 )b
 ( y1 x1  y2 x2  y3 x3  y4 x4  y5 x5 )]
The Method of Least Squares
 We want to minimize
f (m, b)  (mx1  b  y1 )2  (mx2  b  y2 )2  (mx3  b  y3 )2
 (mx4  b  y4 )2  (mx5  b  y5 )2
 We now find the partial derivative with respect to b:
f
 2(mx1  b  y1 )  2(mx2  b  y2 )  2(mx3  b  y3 )
b
 2(mx4  b  y4 )  2(mx5  b  y5 )
 2[( x1  x2  x3  x4  x5 )m  5b  ( y1  y2  y3  y4  y5 )]
The Method of Least Squares
 Setting
f
0
m
and
f
0
b
gives
( x12  x22  x32  x42  x52 )m  ( x1  x2  x3  x4  x5 )b
 y1 x1  y2 x2  y3 x3  y4 x4  y5 x5
and
( x1  x2  x3  x4  x5 )m  5b  y1  y2  y3  y4  y5
 Solving the two simultaneous equations for m and b then
leads to an equation y = mx + b.
 This equation will be the ‘best fit’ line, or regression line for
the given data points.
The Method of Least Squares
 Suppose we are given n data points:
P1(x1, y1), P2(x2, y2), P3(x3, y3), … , Pn(xn, yn)
 Then, the least-squares (regression) line for the data
is given by the linear equation
y = f(x) = mx + b
where the constants m and b satisfy the equations
( x12  x22  x32  ...  xn2 )m  ( x1  x2  x3  ...  xn )b
 y1 x1  y2 x2  y3 x3  ...  yn xn
and
( x1  x2  x3  ...  xn )m  nb  y1  y2  y3  ...  yn
simultaneously.
 These last two equations are called normal equations.
Example
 Find the equation of the least-squares line for the data
P1(1, 1), P2(2, 3), P3(3, 4), P4(4, 3), and P5(5, 6)
y
6
5
4
3
2
1
1
Example 1, page 570
2
3
4
5
x
Example
 Find the equation of the least-squares line for the data
P1(1, 1), P2(2, 3), P3(3, 4), P4(4, 3), and P5(5, 6)
Solution
 Here, we have n = 5 and
x1 = 1
x2 = 2
x3 = 3
x4 = 4
x5 = 5
y1 = 1
y2 = 3
y3 = 4
y4 = 3
y5 = 6
 Substituting in the first equation we get
( x12  x22  x32  ...  xn2 )m  ( x1  x2  x3  ...  xn )b
 y1 x1  y2 x2  y3 x3  ...  yn xn
(12  22  32  42  52 )m  (1  2  3  4  5)b
 (1)(1)  (3)(2)  (4)(3)  (3)(4)  (6)(5)
55m  15b  61
Example 1, page 570
Example
 Find the equation of the least-squares line for the data
P1(1, 1), P2(2, 3), P3(3, 4), P4(4, 3), and P5(5, 6)
Solution
 Here, we have n = 5 and
x1 = 1
x2 = 2
x3 = 3
x4 = 4
x5 = 5
y1 = 1
y2 = 3
y3 = 4
y4 = 3
y5 = 6
 Substituting in the second equation we get
( x1  x2  x3  ...  xn )m  5b  y1  y2  y3  ...  yn
(1  2  3  4  5)m  5b  1  3  4  3  6
15m  5b  17
Example 1, page 570
Example
 Find the equation of the least-squares line for the data
P1(1, 1), P2(2, 3), P3(3, 4), P4(4, 3), and P5(5, 6)
Solution
 Solving the simultaneous equations
55m  15b  61
15m  5b  17
gives m = 1 and b = 0.4.
 Therefore, the required least-squares line is
y = x + 0.4
Example
 Find the equation of the least-squares line for the data
P1(1, 1), P2(2, 3), P3(3, 4), P4(4, 3), and P5(5, 6)
Solution
 Below is the graph of the required least-squares line
y = x + 0.4
y
L
6
5
4
3
2
1
1
Example 1, page 570
2
3
4
5
x
Applied Example: Maximizing Profit
 A market research study provided the following data
based on the projected monthly sales x (in thousands) of
an adventure movie DVD.
p
38
36
34.5
30
28.5
x
2.2
5.4
7.0
11.5
14.6
 Find the demand equation if the demand curve is the least-
squares line for these data.
 The total monthly cost function associated with producing
and distributing the DVD is given by
C(x) = 4x + 25
where x denotes the number of discs (in thousands)
produced and sold, and C(x) is in thousands of dollars.
 Determine the unit wholesale price that will maximize
monthly profits.
Applied Example 3, page 572
Applied Example: Maximizing Profit
Solution
 The calculations required for obtaining the normal
equations may be summarized as follows:
x2
xp
38.0
4.84
83.6
5.4
36.0
29.16
194.4
7.0
34.5
49.00
241.5
11.5
30.0
132.25
345.0
14.6
28.5
213.16
416.1
40.7
167.0
428.41
1280.6
x
p
2.2
 Thus, the nominal equations are
5b  40.7m  167
Applied Example 3, page 572
and
40.7b  428.41m  1280.6
Applied Example: Maximizing Profit
Solution
 Solving the system of linear equations simultaneously, we
find that
m  0.81 and b  39.99
 Therefore, the required demand equation is given by
p  f ( x )  0.81x  39.99
Applied Example 3, page 572
(0  x  49.37)
Applied Example: Maximizing Profit
Solution
 The total revenue function in this case is given by
R( x )  xp
 x ( 0.81x  39.99)
 0.81x 2  39.99 x
 Since the total cost function is
C(x) = 4x + 25
we see that the profit function is
P( x )  R  C
 0.81x 2  39.99 x  (4 x  25)
 0.81x 2  35.99 x  25
Applied Example 3, page 572
Applied Example: Maximizing Profit
Solution
 To find the absolute maximum of P(x) over the closed
interval [0, 49.37], we compute
P( x )  1.62 x  35.99
 Since P(x) = 0 , we find that x ≈ 22.22 as the only critical
point of P.
 Finally, from the table
x
0
22.22
49.37
P(x)
–25
374.78
– 222.47
we see that the optimal wholesale price is
p  0.81(22.22)  39.99  21.99
or $21.99 per disc.
Applied Example 3, page 572
8.5
Constrained Maxima and Minima and
the Method of Lagrange Multipliers
z
f(x, y) = 2x2 + y2
h(x) = 3x2 – 2x + 1
(a, b, f(a, b))
(a, b)
g(x, y) = 0
x
y
Constrained Maxima and Minima
 In many practical optimization problems, we must
maximize or minimize a function in which the independent
variables are subjected to certain further constraints.
 We shall discuss a powerful method for determining
relative extrema of a function f(x, y) whose independent
variables x and y are required to satisfy one or more
constraints of the form g(x, y) = 0.
Example
 Find the relative minimum of f(x, y) = 2x2 +y2 subject to
the constraint g(x, y) = x + y – 1 = 0.
Solution
 Solving the constraint equation for y explicitly in terms
of x, we obtain
y=–x+1
 Substituting this value of y into f(x, y) results in a
function of x,
h( x )  2 x 2  (  x  1)2  3x 2  2 x  1
Example 1, page 580
Example
 Find the relative minimum of f(x, y) = 2x2 +y2 subject to
the constraint g(x, y) = x + y – 1 = 0.
Solution
 We have
h(x) = 3x2 – 2x + 1.
 The function h describes
the curve lying on the
graph of f on which the
constrained relative
minimum point (a, b)
of f occurs:
z
h(x) = 3x2 – 2x + 1
f(x, y) = 2x2 + y2
(a, b, f(a, b))
(a, b)
g(x, y) = 0
Example 1, page 580
x
y
Example
 Find the relative minimum of f(x, y) = 2x2 +y2 subject to
the constraint g(x, y) = x + y – 1 = 0.
Solution
 To find this point (a, b), we determine the relative extrema
of a function of one variable:
h( x )  6 x  2  2(3x  1)
1
 Setting h′ = 0 gives x = 3
as the sole critical point of h.
1
 Next, we find h″(x) = 6 and, in particular, h″( ) = 6 > 0.
3
 Therefore, by the second derivative test, the point gives
rise to a relative minimum of h.
1
 Substitute x = into the constraint equation x +y – 1 = 0 to
3
1
get y = .
3
Example 1, page 580
Example
 Find the relative minimum of f(x, y) = 2x2 +y2 subject to
the constraint g(x, y) = x + y – 1 = 0.
Solution
1 2
 Thus, the point ( 3 , 3 ) gives rise to the required constrained
relative minimum of f.
2
2
2
1 2
1  2
f
,

2


 Since 

   
3
3
3


3  3
2
3
, the required constrained
1 2
 , .
3 3
relative minimum value of f is at the point
2
 It may be shown that is in fact a constrained absolute
3
minimum value of f.
Example 1, page 580
The Method of Lagrange Multipliers
To find the relative extrema of the function f(x, y)
subject to the constraint g(x, y) = 0 (assuming that
these extreme values exist),
1. Form an auxiliary function
F ( x, y ,  )  f ( x, y )   g ( x, y )
called the Lagrangian function (the variable λ is
called the Lagrange multiplier).
2. Solve the system that consists of the equations
Fx = 0
Fy = 0
Fλ = 0
for all values of x, y, and λ.
3. The solutions found in step 2 are candidates for
the extrema of f.
Example
 Using the method of Lagrange multipliers, find the
relative minimum of the function f(x, y) = 2x2 +y2 subject
to the constraint x + y = 1.
Solution
 Write the constraint equation
x+y=1
in the form
g(x, y) = x + y – 1 = 0
 Then, form the Lagrangian function
F ( x, y ,  )  f ( x, y )   g ( x, y )
 (2 x 2  y 2 )   ( x  y  1)
Example 2, page 582
Example
 Using the method of Lagrange multipliers, find the
relative minimum of the function f(x, y) = 2x2 +y2 subject
to the constraint x + y = 1.
Solution
 We have
F  2 x 2  y 2   ( x  y  1)
 To find the critical point(s) of the function F, solve the
system composed of the equations
Fx  4 x    0
Fy  2 y    0
F  x  y  1  0
 Solving the first and second equations for x and y in terms
of λ, we obtain
1
x 
4
and
1
y 
2
 Which, upon substitution into the third equation yields
Example 2, page 582
1
1
    1  0
4
2
or
4

3
Example
 Using the method of Lagrange multipliers, find the
relative minimum of the function f(x, y) = 2x2 +y2 subject
to the constraint x + y = 1.
Solution
4
1
1
 Substituting   
into x    and y   
3
4
2
yields
1 4
1
x   ( ) 
4 3
3
 Therefore, x  1 and y 
2
3
and
1 2
1 4
2
y   ( ) 
2 3
3
, and  3 , 3  results in a
3
constrained minimum of the function f.
Example 2, page 582
Applied Example: Designing a Cruise-Ship Pool
 The operators of the Viking Princess, a luxury cruise liner,
are contemplating the addition of another swimming pool
to the ship.
 The chief engineer has suggested that an area of the form
of an ellipse located in the rear of the promenade deck
would be suitable for this purpose.
 It has been determined that the shape of the ellipse may be
described by the equation
x2 + 4y2 = 3600
where x and y are measured in feet.
 Viking’s operators would like to know the dimensions of
the rectangular pool with the largest possible area that
would meet these requirements.
Applied Example 5, page 582
Applied Example: Designing a Cruise-Ship Pool
Solution
 We want to maximize the area of the rectangle that will fit the
ellipse:
y
(x, y)
x
x2 + 4y2 = 3600
 Letting the sides of the rectangle be 2x and 2y feet, we see that
the area of the rectangle is A = 4xy.
 Furthermore, the point (x, y) must be constrained to lie on the
ellipse so that it satisfies the equation x2 + 4y2 = 3600.
Applied Example 5, page 582
Applied Example: Designing a Cruise-Ship Pool
Solution
 Thus, the problem is equivalent to the problem of maximizing
the function
f(x, y) = 4xy
subject to the constraint
g(x, y) = x2 + 4y2 – 3600 = 0
 The Lagrangian function is
F ( x, y,  )  f ( x, y )   g ( x, y )  4 xy   ( x 2  4 y 2  3600)
 To find the critical points of F, we solve the system of
equations
Fx  4 y  2 x  0
Fy  4 x  8 y  0
F  x 2  4 y 2  3600  0
Applied Example 5, page 582
Applied Example: Designing a Cruise-Ship Pool
Solution
 We have
Fx  4 y  2 x  0
Fy  4 x  8 y  0
F  x 2  4 y 2  3600  0
2y
x
 Which, substituting into the second equation, yields
 Solving the first equation for λ, we obtain
 2y 
4x  8 
y0
 x 
 Solving for x yields x = ± 2y.
Applied Example 5, page 582
or

x2  4 y2  0
Applied Example: Designing a Cruise-Ship Pool
Solution
 We have
Fx  4 y  2 x  0
Fy  4 x  8 y  0
F  x 2  4 y 2  3600  0
 Substituting x = ± 2y into the third equation, we have
4 y 2  4 y 2  3600  0
 Which, upon solving for y yields
y   450  15 2
 The corresponding values of x are
x  2 y  2(15 2)  30 2
 Since both x and y must be nonnegative, we have
x  30 2 and y  15 2
or approximately 42 ☓ 85 feet.
Applied Example 5, page 582
8.6
Double Integrals
z
z = f(x, y) = y
R
x
y
y  1  x2
A Geometric Interpretation of the Double Integral
 You may recall that we can do a Riemann sum to
approximate the area under the graph of a function of one
variable by adding the areas of the rectangles that form
below the graph resulting from small increments of x (x)
within a given interval [a, b]:
y
x
y= f(x)
a
b
x
A Geometric Interpretation of the Double Integral
 Similarly, it is possible to obtain an approximation of the
volume of the solid under the graph of a function of two
variables.
z
y= f(x, y)
d
c
a
R
b
x
y
A Geometric Interpretation of the Double Integral
 To find the volume of the solid under the surface, we can
perform a Riemann sum of the volume Si of parallelepipeds
with base Ri = x ☓ y and height f(xi, yi):
z
z = f(x, y)
d
c
a
y
b
x
x
R
y
A Geometric Interpretation of the Double Integral
 To find the volume of the solid under the surface, we can
perform a Riemann sum of the volume Si of parallelepipeds
with base Ri = x ☓ y and height f(xi, yi):
z
z = f(x, y)
Si
d
c
a
y
b
x
x
R
y
A Geometric Interpretation of the Double Integral
 To find the volume of the solid under the surface, we can
perform a Riemann sum of the volume Si of parallelepipeds
with base Ri = x ☓ y and height f(xi, yi):
z
z = f(x, y)
c
a
b
x
d
y
A Geometric Interpretation of the Double Integral
 The limit of the Riemann sum obtained when the number of
rectangles m along the x-axis, and the number of subdivisions n
along the y-axis tends to infinity is the value of the double
integral of f(x, y) over the region R and is denoted by
z
  f ( x, y)dA
R
z = f(x, y)
y ·n
c
a
x ·m
b
x
y
x
R
d
y
Theorem 1
Evaluating a Double Integral Over a Plane Region
a. Suppose g1(x) and g2(x) are continuous functions on [a, b] and
the region R is defined by R = {(x, y)| g1(x)  y  g2(x); a  x  b}.
Then,
b
g2 ( x )

 dx
f
(
x
,
y
)
dA

f
(
x
,
y
)
dy
R 
a  g1 ( x )

y
y = g2(x)
R
y = g1(x)
x
a
b
Theorem 1
Evaluating a Double Integral Over a Plane Region
b. Suppose h1(y) and h2(y) are continuous functions on [c, d] and
the region R is defined by R = {(x, y)| h1(y)  x  h2(y); c  y  d}.
Then,
d
h2 ( y )

 dy
f
(
x
,
y
)
dA

f
(
x
,
y
)
dx
R 
c  h1 ( y )

y
d
R
x = h1(y)
c
x = h2(y)
x
Examples
 Evaluate ∫R∫f(x, y)dA given that f(x, y) = x2 + y2 and R is the
region bounded by the graphs of g1(x) = x and g2(x) = 2x
for 0  x  2.
Solution
 The region under consideration is:
y
g2(x) = 2x
4
g1(x) = x
3
2
R
1
1
Example 2, page 593
2
3
4
x
Examples
 Evaluate ∫R∫f(x, y)dA given that f(x, y) = x2 + y2 and R is the
region bounded by the graphs of g1(x) = x and g2(x) = 2x
for 0  x  2.
Solution
 Using Theorem 1, we find:

R
2x

2 
2
2x
1 3 
2
2
2


f ( x, y )dA    ( x  y )dy dx    x y  y   dx
0

0 
 x
3  x 


2
0
 3 8 3   3 1 3  
 2 x  3 x    x  3 x   dx
 


2
10 3
5
x dx  x 4  13 13
0 3
6 0

Example 2, page 593
2
Examples
 Evaluate ∫R∫f(x, y)dA, where f(x, y) = xey and R is the plane
region bounded by the graphs of y = x2 and y = x.
Solution
 The region under consideration is:
 The points of intersection of
the two curves are found by
y
2
solving the equation x = x,
1
giving x = 0 and x = 1.
g2(x) = x
R
g1(x) = x2
1
Example 3, page 593
x
Examples
 Evaluate ∫R∫f(x, y)dA, where f(x, y) = xey and R is the plane
region bounded by the graphs of y = x2 and y = x.
Solution
 Using Theorem 1, we find:

R
x
x
1
y
y



f ( x, y )dA    2 xe dy dx    xe  2  dx

0
0
x 
 x


1
1
1
  ( xe  xe )dx  0 xe dx  0 xe dx
1
x
x2
x
x2
0
1
1 

 ( x  1)e x  e x 
2 0

2
Integrating by parts on
the right-hand side
1
1 1

  e   1    (3  e)
2
2 2

Example 3, page 593
The Volume of a Solid Under a Surface
 Let R be a region in the xy-plane and let f be
continuous and nonnegative on R.
 Then, the volume of the solid under a surface
bounded above by z = f(x, y) and below by R is
given by
V 
R
 f ( x, y)dA
Example
 Find the volume of the solid bounded above by the plane
z = f(x, y) = y and below by the plane region R defined by
y  1  x2
(0  x  1)
Solution
 The graph of the region R is:
 Observe that f(x, y) = y > 0 for (x, y) ∈ R.
y
1
y  1  x2
R
1
Example 4, page 594
x
Example
 Find the volume of the solid bounded above by the plane
z = f(x, y) = y and below by the plane region R defined by
y  1  x2
(0  x  1)
Solution
 Therefore, the required volume is given by
 1 x2

V    ydA    
ydy  dx
R
0
 0

1
1 x 2 

1 1
11
2
 dx   (1  x 2 )dx
  y
0 2
0 2


0


1
1 1 3
 x x 
2
3 0
Example 4, page 594
1

3
Example
 Find the volume of the solid bounded above by the plane
z = f(x, y) = y and below by the plane region R defined by
y  1  x2
Solution
 The graph of the solid
in question is:
(0  x  1)
z
z = f(x, y) = y
R
x
Example 4, page 594
y
y  1  x2
End of
Chapter