Chapter 13
Multiple Integrals
by Zhian Liang
13.1 Double integrals over
rectangles
Recall the definition of definite integrals of
functions of a single variable
Suppose f(x) is defined on a interval [a,b].
Taking a partition P of [a, b] into subintervals:
a  x  x  x  x  b
0
n 1
1
n
Choose the points in [ x , x ] and let xi  xi  xi 1
i 1
i
Using the areas of the small rectangles to approximate the
areas of the curve sided echelons
and summing them, we have
n
(1)
 f ( x )x
i 1
*
i
i
P  max{x }
i
b
(2)
n
 f ( x)dx  lim  f ( x )x
a
P 0 i 1
*
i
i
Double integral of a function of two variables defined
on a closed rectangle like the following
R  [a, b]  [c, d ]  {( x, y )  R | a  x  b, c  y  d }
2
Taking a partition of the rectangle
a  x  x  x  x  b
0
1
m 1
m
c  y  y  y  y  d
0
1
n 1
n
*
*
ij
ij
( x , y ) in Rij and form the
Choosing a point
double Riemann sum
  f ( x , y )A
m
(3)
n
i 1 j 1
*
*
ij
ij
ij
(4) DEFINITION The double integral of f over the
rectangle R is defined as
 f ( x, y)dA  lim  f ( x , y )A
m
n
P 0 i 1 j 1
if this limit exists
*
*
ij
ij
ij
Using Riemann sum can be approximately evaluate a double
integral as in the following example.
EXAMPLE 1 Find an approximate value for the integral
 ( x  3 y )dA, where R  {( x, y) | 0  x  2,0  y  2}, by computing
2
R
the double Riemann sum with partition pines x=1 and x=3/2
*
*
ij
ij
and taking ( x , y ) to be the center of each rectangle.
Solution The partition is shown as above Figure. The area of
each subrectangle is A  12 , ( x , y ) is the center R ,
ij
*
*
ij
ij
ij
and f(x,y)=x-3y2. So the corresponding Riemann sum is
2
2
  f ( x , y )A
i 1 j 1
*
*
ij
ij
ij
 f ( x , y )A  f ( x , y )A  f ( x , y )A  f ( x , y )A
*
*
11
11
11
*
*
12
12
12
*
*
21
21
21
 f ( 12 , 54 )A  f ( 12 , 74 )A  f ( 32 , 54 )A  f ( 32 , 74 )A
11
12
67 1
51 1
123 1
1
 2   139


  16
    
16 2
16 2
16 2
  95
 11.875
8
Thus we have
 ( x  3 y )dA  11.875
2
R
21
22
*
*
22
22
22
Interpretation of double integrals as volumes
  v    f ( x , y )A
m
n
i 1 j 1
m
ij
n
i 1 j 1
*
ij
ij
ij
V    f ( x , y )A
i 1 j 1
(5)
*
m
n
ij
ij
*
*
ij
(6) THEOREM If f ( x, y)  0 and f is continuous on
the rectangle R, then the volume of the solid that
z  f ( x, y)
lies above R and under the surface
is
V   f ( x, y )dA
R
EXAMPLE 2
Estimate the volume of the solid that lies above the square
R  [0,2]  [0,2] and below the elliptic paraboloid z  16  x  2 y .
2
Use the partition of R into four squares and choose
( x , y ) to be the upper right corner of R.
Sketch the solid and the approximating rectangle boxes.
*
*
ij
ij
ij
2
Solution The partition and the graph of the function are
as the above. The area of each square is 1.Approximating
the volume by the Riemann sum, we have
V  f (1,1)A  f (1,2)A  f (2,1)A  f (2,2)A
11
12
21
22
 13(1)  7(1)  10(1)  4(1)  34
This is the volume of the approximating rectangular
boxes shown as above.
The properties of the double integrals
（7） [ f ( x, y )  g ( x, y )]dA   f ( x, y )dA   g ( x, y )dA
R
R
（8） cf ( x, y )dA c  f ( x, y )dA
R
R
（9） If f ( x, y)  g ( x, y) for all x, y  R,
then
 f ( x, y ) dA   g ( x, y ) dA
R
R
R
EXERCISES 13.1
Page 837
1. 3. 15.16
13.2 Iterated Integrals
The double integral can be obtained by evaluating two
single integrals.
The steps to calculate  f ( x, y )dA , where R  [a, b]  [c, d ] :
R
Fix x to calculate A( x)   f ( x, y )dy with respect to y.
d
c
Then calculate
 A( x )dx
b
a
(1)  A( x)dx   [  f ( x, y )dy ]dx (called iterated integral)
b
b
d
a
a
c
(2)   f ( x, y )dydx   [  f ( x, y )dy ]dx
b d
b
a c
a
d
c
(3) Similarly
  f ( x, y )dxdy   [  f ( x, y )dx ]dy
d
c
b
a
d
c
b
a
EXAMPLE 1 Evaluate the iterated integrals
(a)   x ydydx
3 2
2
0 1
(See the blackboard)
(b)   x ydxdy
2
1
3
0
2
(4) Fubini’s Theorem If f is continuous on the
rectangle R  {( x, y) | a  x  b, c  y  d}, then
 f ( x, y )dA    f ( x, y )dydx    f ( x, y )dxdy
b d
d
b
a c
c
a
R
More generally, this is true if we assume that f
is bounded on R , f is discontinuous only on
a finite number of smooth curves, and the iterated
integrals exist.
Interpret the double integral  f ( x, y )dA as the
volume V of the solid
R
where
V   A( x)dx
b
a
A( x)   f ( x, y )dy
d
c
A(x) is the area of a cross-section of S in the
plane through x perpendicular to the x-axis.
Similarly
EXAMPLE 2
Evaluate the double integral
where
 ( x  3 y )dA
2
R
R  {( x, y) | 0  x  2, 1  y  2}.
(See the blackboard)
EXAMPLE 3
Evaluate  y sin( xy)dA ,where R  [1,2]  [0,  ]
R
Solution 1 If we first integrate with respect to x,
we get
 y sin( xy)dA    y sin( xy)dxdy

2
0
R
1
  [ cos( xy)] dy

x2
0
x 1
  ( cos 2 y  cos y )dy

0
  sin 2 y  sin y 0  0
1
2

Solution 2 If we first integrate with respect to y, then
 y sin( xy)dA    y sin( xy)dydx
2
R
1

0
    1x yd (cos( xy))dx

1
1 
  [ x y cos( xy) |  x  cos( xy)dy ]dx

2
1
0
2
1
0
0

1
1
[


cos(

x
)

1 x
x2

sin( x )
1 [ x 2
2
2

 cos(x )
x

sin( xy) | ]dx
0
]dx  1
2 d sin( x )
sin( x )
 1 x 2 dx  1 x
2
2 sin( x )
sin( x ) 
 1 x 2 dx  x  
1
2
2  cos(x )
sin( x )
dx

dx
2

1
x
x
2

2
1
sin( x )
dx
2
x

2
sin( x ) 
 x 
1
0
EXAMPLE 4 Find the volume of the solid S that is
bounded by the elliptic paraboloid x  2 y  z  16 , the
plane y  2 and x  2 , and three coordinate planes.
2
2
Solution We first observe that S is the solid that lies
under the surface z  16  x  2 y and the above the
2
2
Square R  [0,2]  [0,2]. Therefore,
V   (16  x  2 y )dA
2
2
R
   (16  x  2 y )dxdy
2
2
0
0
2

2
  16 x  x  2 y x
2
0

1
3
2
0



88
3
88
y
3
3

y
2

 4 y dy
2

4
3
3
2
0
 48
x2
x 0
dy
If f ( x, y)  g ( x)h( y) on R  [a, b]  [c, d ] , then
 g ( x)h( y )dA   g ( x)dx  h( y )dy
R
b
d
a
c
EXAMPLE 5 If R  [0,  2]  [0,  2] , then
 sin x cos ydA   sin xdx cos ydy


2
0
2
0
R
 [  cos x] [sin y ]

0
 1 1  1
2

0
2
EXERCISES 13.2
Page 842
1(2), 6, 10, 16, 17,
13.3 Double integrals over general regions
To integrate over general regions like
R
which is bounded, being enclosed in a rectangular region R .
Then we define a new function F with domain R by
(1)
 f ( x, y ) if ( x, y)  D
F ( x, y )  
0
if ( x, y) is in R but not in D
If F is integrable over R , then we say f is integrable
over D and we define the double integral of f over D by
(2)
where F
 f ( x, y )dA   F ( x, y )dA
D
R
is given by Equation 1.
Geometric interpretation
When
f ( x, y)  0
R
The volume under f and above D equals to that under
F and above R.
Type I regions
D  {( x, y ) | a  x  b, g ( x)  y  g ( x)}
1
2
(3) If f is continuous on a type I region D such that
D  {( x, y ) | a  x  b, g ( x)  y  g ( x)}
1
then
 f ( x, y )dA   
b g2 ( x )
D
a g1 ( x )
2
f ( x, y )dydx
Type II regions
(4)
(5)
D  {( x, y ) | c  y  d , h ( y )  x  h ( y )}
1
d
h2 ( y )
c
h1 ( y )
 f ( x, y)dA   
D
2
f ( x, y)dxdy
where D is a type II region given by Equation 4


   ( x  2 y)dy dx
it is Type I region!
1
1
1 x 2
2 x2
   xy  y
1
1


2

y 1 x 2
y 2 x 2
dx
   x(1  x )  (1  x )  2 x  4 x dx
1
2
2
2
3
4
1
   3x  x  2 x  x  1dx
1
4
3
2
1
1
x x
x x


  3   2   x
5 4
3 2

1
5
32

15
4
3
Example 2 Find the volume of the solid that lies under the
paraboloid z  x  y and above the region D in the xy
2
y

x
.
y

2
x
-plane bounded by the line
and the parabola
2
2




Type II
Type I
D  {( x, y ) | 0  x  2, x  y  2 x}
2
y
D  {( x, y ) | 0  y  4,  x 
2
y}
Type I
Solution 1
 ( x  y )dA   (  ( x  y )dy )dx
8
1
dx
  (2 x  3 x  x  3 x )dx
2
y 2 x
D
2
 2 1 3
  x y  y 
0
0
3  y x2

2
1 6
14 3
4
 0 ( 3 x  x  3 x )dx
2
Solution 2
2
2x
0
x2
2
3
3

1 7
x
21
Type II4
y
0
y/2
0
1 3
x
3
2
15
5/ 2
2
4

1 5
x
5
6
 x
7
6
4

2
216

35
0
 ( x  y )dA    ( x  y )dxdy
2
2
D
    y x dy    y  y
216
 y  y  y   35
4
2
x y
4
2
0
x y / 2
2
7
7/2
13
96
4
4
0
1
3
3/ 2
2
5/ 2
2

1
24

y  y dy
3
1
2
3
Example 3 Evaluate D xydA , where D is the region bounded
2
y
by the line y  x  1 and the parabola  2 x  6.



D as a type I
D as a type II
Solution We prefer to express D as a type II

y2
2
D  ( x, y ) | 2  y  4,
 xydA   
D
4
y 1
2
y2
2
xydxdy    x y 
 2 
2
x  y 1
4
3
dy
2


 3  x  y 1
x
y2
3
2


1
2
 y ( y  1)  ( 2 y  3) dy

1
2


4
2
1
2
4
2


5
4
 1


 24
 36
1
2
2

y  4 y  2 y  8 y dy
5
y6  y4  2
3
4
2
4

y 3 4 y 2 
 2
Example 4 Find the volume of the tetrahedron bounded by
the planes x  2 y  z  2, x  2 y, x  0, and z  0.
x
x

D  ( x, y ) | 0  x  1,  y  1  
2
2

D
D
Solution
V   (2  x  2 y )dA 
D
  2 y  xy  y
1
2
0

  2  x  x1 
1
0

y 1
y
x
2
2
0

1

x3  x2  x
 0
3
0 x
2
Here is wrong in the book!
dx
  1    x 
   x  2 x  1dx
1
x
2
x
2
x
2
  (2  x  2 y )dydx
1 1
1

3
x
2
2
x2
x2

2
4
dx
Example 5 Evaluate the iterated integral
  sin( y )dydx
1 1
2
0 x
D  ( x, y ) | 0  x  1, x  y  1  ( x, y ) | 0  y  1,0  x  y
D as a type I
D as a type II
Solution If we try to evaluate the integral as it stands, we
are faced with the task of first evaluating  sin( y )dy. But it is
impossible to do so in finite terms since  sin( y )dy is not an
2
2
elementary function.(See the end of Section 7.6.) So we must
change the order of integration. This is accomplished by first
expressing the given iterated integral as a double integral.
  sin( y )dydx   sin( y )dA
1 1
2
2
0 x
D
Where D  ( x, y ) | 0  x  1, x  y  1
Using the alternative description of D,
we have D  ( x, y ) | 0  y  1,0  x  y
This enables us to evaluate the integral in the reverse
order:
  sin( y )dydx   sin( y )dA
1 1
2
2
0 x
D
   sin( y )dxdy
1 y
2
0 0
  x sin( y ) dy
1
2
x y
x 0
0
  y sin( y )dy
1
0


2

1
1
2
cos(
y
)
2
0
1
(1  cos1)
2
Properties of double integrals
(6)
(7)
(8)
(9)
 [ f ( x, y )  g ( x, y )]dA   f ( x, y )dA   g ( x, y )dA
D
D
D
 cf ( x, y )dA c  f ( x, y )dA
D
D
 f ( x, y )dA   g ( x, y )dA
D
D
if f ( x, y)  g ( x, y) for
all ( x, y) in D.
 [ f ( x, y )dA   f ( x, y )dA   f ( x, y )dA
D
D1
if D  D
1
D2
 D where D and D do not overlap except
2
1
2
perhaps on their boundaries like the following:
(10)  1dA  A( D) , the area of region D.
(11) If m  f ( x, y)  M for all ( x, y)  D, then
D
mA( D)   f ( x, y )dA MA( D)
D
Example 6
Use Property 11 to estimate the integral  e
sin x cos y
D
dA,
where D is the disk with center the region and radius 2.
Solution Since  1  sin x  1, and  1  cos x  1, we have
 1  sin x cos x  1,
and therefore
e  esin x cos x  e  e
1
1
thus, using m=e-1=1/e, M=e, and A(D)=(2)2 in Property 11
we obtain
4
sin x cos y
  e
dA  4πe
e
D
Exercises 13.3
Page 850: 7, 9, 11, 33, 35
13.4 DBOUBLE INTEGRALS IN POLAR
COORDINATE
Suppose we want to evaluate a double integral f ( x, y ) dA,
where R is one of regions shown in the following.
R
(a)
R  {(r , ) | 0  r  1,0    2 }
(b)R  {(r, ) | 1  r  2,0     }
Recall from Section 9.4 that the polar coordinates
of a point related to the rectangular coordinates (r , )
by the equations: r 2  x 2  y 2
x  r cos
y  r sin 
The regions in
the above
figure are
special cases
of a polar rectangle
R  {(r , ) | a  r  b,     }
Do the following partition (called polar partition)
The center of this subrectangle
R  {(r , ) | r  r  r ,     }
i 1
ij
is
1
r  (ri 1  ri )
2
A 
ij

*
j
1 2
r  j
2 i

2
1
r  j
2 i 1
 r )
1 2
(r
2 i
2
i 1
j
 12 (r  r )(r  r )
i 1
i
 r r 
where
*
i
r  r  r
i
i
i 1
j
1
  ( j 1   j )
2
*
i
and the area is
j 1
i
i
j
i
i 1
j
The typical Riemann sum is
(1)
  f ( r cos , r sin  )A
m
n
i 1
j 1
*
*
i
*
j
*
i
j
ij
   f (r cos , r sin  )r r 
m
n
i 1
j 1
*
i
*
j
*
i
*
j
*
i
i
j
If we write g (r , )  rf (r cos , r sin  ) , then the
above Riemann sum can be written as
  g (r , )r 
m
n
*
i
i 1 j 1
*
j
i
j
which is the Riemann sum of the double integral
  g (r , )drd

b
a
Therefore we have
 f ( x, y ) dA  lim   f ( r cos , r sin  )A
R
P 0
m
n
i 1
j 1
*
*
i
j
*
i
 lim   g (r , )r 
P 0
m
n
i 1
j 1
*
i
*
j
*
j
i
j
   g (r , )drd

b

a
   f (r cos , r sin  )rdrd

b

a
ij
(2) Change to polar coordinates in a double
integral If f is continuous on a polar
rectangle R given by 0  a  r  b,     ,
where 0      2 , then
 f ( x, y )dA    f (r cos , r sin  )rdrd
 b
R
a
Caution: Do not forget the factor r in (2)!
Example 1 Evaluate (3 x  4 y )dA , where R is the region
in the upper half-plane bounded by the circles
2
R
x  y  1, and x  y  4.
2
Solution
2
2
2
The region R can be described as
R  {( x, y ) | y  0, 1  x  y  4}  {(r , ) | 1  r  2, 0     }
2
2
2
2
 (3 x  4 y )dA  0 1 (3r cos  4r sin  )rdrd

2
2
R
   (3r cos  4r sin  )drd

0
2
2
3
2
1
  r cos  r sin  
 7 cos  15 sin  d
15
 

  7 cos  (1  cos 2 )d
2



0
3
4
2
r 2

r 1 d
0

15 15
15

 7 sin  
 sin 2  
2
4
2
0

0
2
Example 2 Find the volume of the solid bounded
2
2
by the xy-plane and the paraboloid z  1  x  y
D  {( x, y ) | x  y  1}
2
2
 {(r , ) | 0  r  1, 0    2 }
V   (1  x  y )dA
2
D
2
   (1  r )rdrd

  d  (r  r )dr
2
1
0
0
2
2
1
0
0
3
r r  
 2    
 2 4 0 2
2
4
1
What we have done so far can be extended to the
complicated type of region shown in the following.
(3) If f is continuous on a polar
region of the form
D  {( r , ) |      , h ( )  r  h ( )}
1
2
then
 f ( x, y )dA 
D
 h2 (  )
 
h1 (  )
f (r cos , r sin  )rdrd
Example 3 Use a double integral to find the area enclosed
by one loop of the four-leaved rose r  cos 2
D  {(r , ) | 

4
 

4
, 0  r  cos 2 }
A( D)   dA    
 /4
cos 2
 /4 0
rdrd
D
 
 /4
1 
 2 r 
cos 2 
d
2
 /4
0
1 
   cos 2d
2
1 
   1  cos 4 d
4

1
1


   sin 4  
4
4
8

/4
2
 /4
/4
 /4
/4
 /4
Example 4 Find the volume of the solid that lies under the
paraboloid z  x  y , above the xy plane, and inside
the cylinder x  y  2 x.
2
2
2
2
Solution The solid lies above the disk, whose boundary circle
D  {(r , ) | 

2
 

2
, 0  r  2 cos }
 /2
2 cos 
V   ( x  y )dA   / 2 0
2
2
r rdrd
2
D
2 cos 

r 
     d  4   cos d
4

  1  cos 2 
 8 cos d  8 
 d
2


4
 /2
/2
 /2
 /2
0
/2
0
4
2
4
/2
0
 2  [1  2 cos 2  1  cos 4 ]d
 /2
0
1
2
 /2
3

1
 2   sin 2  8 sin 4 
2
0
 3    3
 2   
 2  2  2
Exercises 13.4
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