Bob Brown, CCBC Dundalk
Math 253 Calculus 3, Chapter 14 Sections 1 and 2
1
Exercise 1: The government has set aside a penal colony in the North Atlantic for
Calculus III students who are repeatedly late to class. A drawing of this island, with the
x-axis and y-axis measured in kilometers, is given below. The contours of a population
density function D = f(x , y) are also given.
Note 1: D is measured in
Note 2: D is a
function.
Make a lower and upper estimate for the population of Calculus III students on this
island. Note that the area of each rectangle is
Lower—find the least value
Upper—find the greatest
of D on each rectangle
value of D on each rectangle
The lower estimate for the population is
The upper estimate for the population is
0  40  0  40  0  40
 0  40  2  40  0  40
 0  40  1  40  0  40
 0  40  0  40  0  40
= (0+0+0+0+2+0+0+1+0+0+0+0)
 40
=
The average of the lower estimate and the upper estimate is
Thus, we might estimate that there are
Calculus III students on this island.
Bob Brown, CCBC Dundalk
Math 253 Calculus 3, Chapter 14 Sections 1 and 2
2
The Definite Integral on a Rectangle
Let f(x , y) be an integrable function defined on the rectangle [a , b] x [c , d].
What is [a , b] x [c , d] in relation to f?
We construct a Riemann sum by subdividing the rectangle into small subrectangles. We
do this by subdividing the closed interval a  x  b into n subintervals and the closed
interval c  y  d into m subintervals.
The length of each subinterval along the x-axis is
=
The length of each subinterval along the y-axis is
=
Thus, the area of each of the nm subrectangles is
=
=
To compute a Riemann sum, we 1) multiply the value of the function at any one point in
each subrectangle by the area of the subrectangle, and 2) sum up those products.
Choosing the point in each subrectangle that gives the maximum value, M ij , of f
in that subrectangle, we get the upper sum
Choosing the point in each subrectangle that gives the minimum value, m ij , of f
in that subrectangle, we get the lower sum
Any other Riemann sum, calculated by determining the function value at any
point xi , y j  in the i, j  th subrectangle, satisfies
Def.: We define the definite integral of the integrable function f over the rectangle
[a , b] x [c , d] by
Note: For a general Riemann sum, the subrectangles do not all have to have the same area.
Bob Brown, CCBC Dundalk
Math 253 Calculus 3, Chapter 14 Sections 1 and 2
The Definite Integral on a Region
Let R be a closed, bounded region in the domain of the function f(x , y). (Thus, R is a
subset of the domain of f.) In much the same way as in the case of a rectangular region,
we can partition R into small subrectangles and compute a Riemann sum.
Def.: We define the definite integral of the integrable function f over the region R by
 f ( x, y) dA =
R
Geometric Interpretation
If f is integrable over a plane region R and f ( x, y )  0 for all (x , y) in R, then
 f ( x, y) dA is the
R
Properties of Double Integrals
Theorem:
1.
 cf ( x, y) dA =
2.
R
3.
  f ( x, y)  g ( x, y)dA =
R
 f ( x, y) dA
 0 if
R
4.
 f ( x, y) dA
R

 g ( x, y) dA if
R
Def.: Two regions, R1 and R2 , are said to be non-overlapping if their intersection
is a set that
5. Let R = R1  R2 be the union of two non-overlapping regions, R1 and R2 .
Then,
 f ( x, y) dA =
R
3
Bob Brown, CCBC Dundalk
Math 253 Calculus 3, Chapter 14 Sections 1 and 2
How Do You Actually Integrate?
 3x

x
Exercise 2: Evaluate
2
y 2  2 x dy .
2
4 y 1
Exercise 3a: Evaluate
 3x
y
2
2

y 2  2 x dx .
4
Bob Brown, CCBC Dundalk
Math 253 Calculus 3, Chapter 14 Sections 1 and 2
5
4 y 1 2 2

Exercise 3b: Evaluate    3x y  2 x dx dy .

2 
 y2
1


Note 1: The integral in Exercise 3b is an iterated integral. In fact, the square brackets
are usually not written. Instead, iterated integrals are usually written simply as a double
integral, like
Very Important Note 2: The inside limits of integration can be variable with respect to
the outer variable of integration. However, the outside limits of integration must be
constant with respect to both variables of integration.
Note 3: After performing the inside integration, you obtain a “standard” definite integral,
like the ones seen in Calculus 1. Then, that second integration produces a real number.
Bob Brown, CCBC Dundalk
Math 253 Calculus 3, Chapter 14 Sections 1 and 2
Exercise 4a: Let R = [-1 , 1] x [0 , 3]. Determine
 4  x
2

 y 2 dA .
R
Exercise 4b: Redo Exercise 4a by interchanging the order of integration.
Important—no, CRUCIAL—Note: Only because the region of integration was a
rectangle could we switch the order of integration (from with respect to y then with
respect to x to from with respect to x then with respect to y) without affecting the
limits of integration. For a general region of integration, switching the order of
integration requires substantial changes to the limits of integration.
6
Bob Brown, CCBC Dundalk
Math 253 Calculus 3, Chapter 14 Sections 1 and 2
7
Exercise 5a: Write an iterated integral for f(x , y) over the shaded region.
2
Integrating over vertical
strips, y goes from
2
Inside integral:
4
4
2
 There is a vertical strip for
each x from
4
Thus, the iterated integral:
Exercise 5b: Write an iterated integral for f(x , y) over the shaded region, reversing the
order of integration from Exercise 5a.
2
Integrating over horizontal
strips, x goes from
2
Inside integral:
4
4
 There is a horizontal strip
for each y from
2
4
The iterated integral:
6
Exercise 6: Sketch the region of integration for the iterated integral
2
 x
0 x
3
y 3  1 dydx .
Bob Brown, CCBC Dundalk
Math 253 Calculus 3, Chapter 14 Sections 1 and 2
8
Area of a Plane Region
Theorem: If R is defined by a  x  b and g1 ( x)  y  g 2 ( x) , where g 1 and g 2 are
continuous on a  x  b , then the area of R is given by…
If R is defined by c  y  d and h1 ( y)  x  h2 ( y) , where h1 and h2 are continuous on
c  y  d , then the area of R is given by…
Exercise 7: Determine the area of the region bounded by the graphs of y  x 2  3 ,
y = -2x, x = 0, and x = 1.
Connection Between Area Integral and a Volume Integral
b g2 ( x )
The formulas at the top of this page, like
 
1 dydx , give the area of a region R.
a g1 ( x )
However, by it we are also integrating the function z = f(x , y) = 1 above the region R,
which gives the volume of the solid region below the plane z = 1 and above R. (See the
Geometric Interpretation on page 3.) Although the units for area and the units for volume
are different, the number-answers are the same.
1
2
3
What is the area of this rectangle?
3
2
What is the volume of this prism?
Bob Brown, CCBC Dundalk
Math 253 Calculus 3, Chapter 14 Sections 1 and 2
If a single definite integral from Calculus 1 or 2 is difficult to evaluate, it is generally because of the
function f and not because of the interval [a , b]. On the other hand, what can make double integrals
challenging to evaluate is the region R.
Exercise 8: Determine the volume of the solid region bounded by the paraboloid
z  4  x 2  y 2 and the xy-plane.
f:=(x,y)->4-x^2-y^2;
g:=(x,y)->0;
plot3d({f(x,y),g(x,y)},x=-2..2,y=-2..2);
By letting z = 0, we see that the base of the region in
the xy-plane is
Integrating over vertical strips, y goes from
Thus, the inside integral is
There is a vertical strip for each x from
Therefore, Volume =
9
Bob Brown, CCBC Dundalk
Math 253 Calculus 3, Chapter 14 Sections 1 and 2
4 x 2
  4  x
2
From the last page, the Volume =
2

2

 y 2 dy dx .
4 x 2
y  4 x 2  M

y3 
2
Inside Integral =  4 y  x y  
3  y 

4 x 2   M
= 4( M   M )  x 2 ( M   M ) 
= 8M  2 x 2 M 
2 3
M
3
= 8M  2 x 2 M 
2
4  x2 M
3


8 2 

= M 8  2x 2   x 2 
3 3 

=
4  x2 
Therefore, Volume =
Trigonometric Substitution:
Thus, Volume =

4
4  x2
3


1
M 3  (  M )3
3

since
=
=
 16 4 
4  x2   x2 
 3 3 

4
4  x2
3

3
2
10