Multiple Integrals Chapter 13 by Zhian Liang
Dec 27, 2015
13.1 Double integrals over rectangles
Suppose f(x) is defined on a interval [a,b].
Recall the definition of definite integrals offunctions of a single variable
Taking a partition P of [a, b] into subintervals:
bxxxxann
110
let and ],[in points theChoose1 ii
xx 1
iii
xxx
Using the areas of the small rectangles to approximate the areas of the curve sided echelons
Double integral of a function of two variables defined on a closed rectangle like the following
},|),{(],[],[ 2 dycbxaRyxdcbaR
Taking a partition of the rectangle
dyyyyc
bxxxxa
nn
mm
110
110
m
i
n
jijijij
Ayxf1 1
** ),(
),( **
ijijyxChoosing a point in Rij and form the
double Riemann sum
(3)
(4) DEFINITION The double integral of f over the rectangle R is defined as
m
i
n
jijijijP
AyxfdAyxf1 1
**
0),(lim),(
if this limit exists
Using Riemann sum can be approximately evaluate a double integral as in the following example.
EXAMPLE 1 Find an approximate value for the integral},20,20|),{( where,)3( 2 yxyxRdAyx
R by computing
the double Riemann sum with partition pines x=1 and x=3/2
and taking ),( **
ijijyx to be the center of each rectangle.
Solution The partition is shown as above Figure. The area of each subrectangle is ,2
1ij
A ),( **
ijijyx is the center Rij,
and f(x,y)=x-3y2. So the corresponding Riemann sum is
875.11
),(),(),(),(
),(),(),(),(
),(
895
21
16123
21
1651
21
16139
21
1667
47
23
45
23
47
21
45
21
22211211
22
*
22
*
2221
*
21
*
2112
*
12
*
1211
*
11
*
11
2
1
2
1
**
AfAfAfAf
AyxfAyxfAyxfAyxf
Ayxfi j
ijijij
R
dAyx 875.11)3(
have weThus2
m
i
n
j
m
i
n
jijijijij
Ayxfv1 1 1 1
** ),(
Interpretation of double integrals as volumes
m
i
n
jij ij ijA y x f V11
* *) , (
(5)
(6) THEOREM If and f is continuous on the rectangle R, then the volume of the solid that lies above R and under the surface is
0),( yxf
) , (y x f z
R
dAyxfV ),(
EXAMPLE 2
Estimate the volume of the solid that lies above the square
and below the elliptic paraboloid .
Use the partition of R into four squares and choose
to be the upper right corner of .
Sketch the solid and the approximating rectangle boxes.
]2,0[]2,0[ R 22 216 yxz
),( **
ijijyx
ijR
the volume by the Riemann sum, we have
34)1(4)1(10)1(7)1(13
)2,2()1,2()2,1()1,1(22211211
AfAfAfAfV
This is the volume of the approximating rectangular boxes shown as above.
Solution The partition and the graph of the function are
as the above. The area of each square is 1.Approximating
RRR
dAyxgdAyxfdAyxgyxf ),(),()],(),([
RR
dAyxfcdAyxcf ),(),(
),(),( yxgyxf ,, Ryx
RR
dAyxgdAyxf ),(),(
( 7)
( 8)
( 9)If for all
The properties of the double integrals
then
13.2 Iterated Integrals
to calculate
R
dAyxf ),( :],[],[ dcbaR
x dyyxfxA d
c ),()( .y
dxxAb
a )(
The double integral can be obtained by evaluating two single integrals.
The steps to calculate , where
Then calculate
with respect toFix
dxdyyxfdxxA b
a
d
c
b
a ]),([)(
dxdyyxfdxdyyxf b
a
d
c
b
a
d
c ]),([),(
dydxyxfdydxyxf d
c
b
a
d
c
b
a ]),([),(
(1) (called iterated integral)
(2)
(3) Similarly
EXAMPLE 1 Evaluate the iterated integrals
(See the blackboard)
dydxyxdxdyyxa 2
1
3
0
23
0
2
1
2 (b) )(
(4) Fubini’s Theorem If is continuous on the rectangle then
More generally, this is true if we assume that
is bounded on , is discontinuous only on
a finite number of smooth curves, and the iterated
integrals exist.
f},,|),{( dycbxayxR
dydxyxfdxdyyxfdAyxf d
c
b
a
b
a
d
cR
),(),(),(
f
R f
where
is the area of a cross-section of S in the plane through x perpendicular to the x-axis.
Similarly
dyyxfxA d
c ),()(
)(xA
dxxAV b
a )(
Solution 1 If we first integrate with respect to x,we get
R
dAxyy )sin( dxdyxyy
0
2
1)sin(
0
2
1)]cos([ dyxy x
x
0)cos2cos( dyyy
0sin2sin 021 yy
Solution 2 If we first integrate with respect to y, then
R
dAxyy )sin( dydxxyy2
1 0)sin(
dxxyydx 2
1 0))(cos(1
dxdyxyxyy xx 2
1 00])cos(|)cos([ 11
dxxyxxx 2
1 02 ]|)sin()cos([ 11
dxxx
x
x 2
1 2 ][ )cos()sin( 2
1
2
1 2
)cos()sin( dxdx xx
x
x
2
1
2
1 2
)sin()sin(x
xd
x
x dx
2
1 2
2
1 2
)sin(2
1
)sin()sin( dxdxx
x
xx
x
x 02
1
)sin(
xx
EXAMPLE 4 Find the volume of the solid S that is bounded by the elliptic paraboloid , the plane and , and three coordinate planes.2x2y
162 22 zyx
We first observe that S is the solid that lies under the surface 22 216 yxz and the above the
Square ].2,0[]2,0[ R Therefore,
R
dAyxV )216( 22
2
0
2
0
22 )216( dxdyyx
2
0
232
03
1 216 dyxyxxx
x
2
0
243
88 dyy
482
0 3
4
3
88 3 yy
Solution
13.3 Double integrals over general regions
0
),(),(
yxfyxF
Dyx ),(
DRyx in not but in is ),(
R
To integrate over general regions like
which is bounded, being enclosed in a rectangular region R .Then we define a new function F with domain R by
(1)if
if
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
where is given by Equation 1.
dAyxFdAyxfRD ),(),(
F
(2)
Geometric interpretation 0),( yxfWhen
The volume under f and above D equals to that under F and above R.
R
Type I regions )}()(,|),{(
21xgyxgbxayxD
(3) If f is continuous on a type I region D such that
)}()(,|),{(
21xgyxgbxayxD
dydxyxfdAyxf b
a
xg
xgD
)(
)(
2
1),(),(
then
)}()(,|),{(21
yhxyhdycyxD
(5) dxdyyxfdAyxf
d
c
yh
yhD
)(
)(
2
1
),(),(
Type II regions
(4)
where D is a type II region given by Equation 4
dxdyyxx
x
1
1
1
2
2
2 )2(
dxyxyxy
xy
2
2
1
2
1
1
2
dxxxxxx
1
1
43222 42)1()1(
dxxxxx
1
1
234 123
1
1232
453
345
x
xxxx
15
32
it is Type I region!
22 yxz xy 2 .2xy
xyExample 2 Find the volume of the solid that lies under the
paraboloid and above the region D in the
-plane bounded by the line and the parabola
}2,20|),{(
I Type 2 xyxxyxD }
2,40|),{(
II Type
yxy
yyxD
Solution 1
D
dAyx )( 22 2
0
2 222 ))((x
xdxdyyx
2
0
232
23
1dxyyx
xy
xy 2
0
6433 )2(3
1
3
8 dxxxxx
2
0
346 )(3
14
3
1 dxxxx 206
7
5
1
21
1 457 xxx 35
216
Solution 2 D
dAyx )( 22 4
0 2/
22 )(y
ydxdyyx
4
0 2/
23
3
1 dyxyxyx
yx
4
0
332/52/3
2
1
24
1
3
1 dyyyyy
4096
13
7
2
15
2 42/72/5 yyy 35
216
Type I
Type II
Example 3 Evaluate , where D is the region bounded by the line and the parabola
D xydA1xy .622 xy
D as a type I D as a type II
13 ,42|),(2
2
yxyyxD y
Solution We prefer to express D as a type II
D xydA
4
2
1
32
2
y
yxydxdy
4
2
1
32
2
2
2dyy
xyx
yx
4
2
22 )3()1(2
1
2
1 dyyyy
4
2
245 8244
5
2
1 dyyyyy
4
23
2
24
12
1 2346 4
yyyy
36
.0 and ,0 ,2 ,22 zxyxzyx
21
2,10|),(
xy
xxyxD
Example 4 Find the volume of the tetrahedron bounded by the planes
D
D
Solution
dAyxVD )22( 1
02
1
2
)22(x
x dydxyx
1
0
21
2
22 dxyxyyx
y
xy
1
0
222
4222 112 dxxxx xxxx
1
0
2 12 dxxx
1
0
23
3
xxx
3
1
Here is wrong in the book!
dydxyx1
0
1 2 )sin(
1,10|),( yxxyxD yxyyx 0,10|),(
D as a type I D as a type II
Example 5 Evaluate the iterated integral
Solution If we try to evaluate the integral as it stands, we
.)sin( 2 dyyimpossible to do so in finite terms since dyy )sin( 2
elementary function.(See the end of Section 7.6.) So we must
are faced with the task of first evaluating But it is
is not an
change the order of integration. This is accomplished by first
expressing the given iterated integral as a double integral.
D
xdAydydxy )sin()sin( 21
0
1 2
Where 1,10|),( yxxyxD
Using the alternative description of D,
we have yxyyxD 0,10|),(
This enables us to evaluate the integral in the reverse
order:
D
xdAydydxy )sin()sin( 21
0
1 2
dxdyyy
1
0 0
2 )sin(
1
0 0
2 )sin( dyyxyx
x
1
0
2 )sin( dyyy
1
0 2
1 )cos( 2y
)1cos1(2
1
Properties of double integrals
D DD
dAyxgdAyxfdAyxgyxf ),(),()],(),([
DD
dAyxfcdAyxcf ),(),(
DD
dAyxgdAyxf ),(),( .in ),( allfor ),(),( Dyxyxgyxf
1 2
),(),(),([D DD
dAyxfdAyxfdAyxf
2121 and where DDDDD
(6)
(7)(8) if
(9)
if do not overlap except
perhaps on their boundaries like the following:
Example 6
2. radius andregion center thedisk with theis where
, integral theestimate to11Property se U cossin
D
dAeD
yx
Solution Since ,1cos1 and ,1sin1 xx we have
,1cossin1 xx and therefore
eeee xx 11 cossin
thus, using m=e-1=1/e, M=e, and A(D)=(2)2 in Property 11
we obtain
πedAeD
yx 4e
4 cossin
13.4 DBOUBLE INTEGRALS IN POLAR COORDINATE
Suppose we want to evaluate a double integral
where is one of regions shown in the following.
,),(R dAyxfR
}20,10|),{( rrR(a) }0,21|),{( rrR(b)
Recall from Section 9.4 that the polar coordinates of a point related to the rectangular coordinates ),( r
siny cos 222 rrxyxr
},|),{( brarR
Do the following partition (called polar partition)
The regions in
the above
figure are
special cases
of a polar rectangle
by the equations:
},|),{(11 jjiiij
rrrrR
)(2
1 )(
2
11
*1
*jjjiii rrr
jii
jiiii
jii
jijiij
rr
rrrr
rr
rrA
*
11
2
1
2
2
1
2
))((
)(
21
21
21
21
1
iiirrr
The center of this subrectangle
is
and the area is
where
jii
m
i
n
jjiji
ij
m
i
n
jjiji
rrrrf
Arrf
*
1 1
****
1 1
****
)sin,cos(
)sin,cos(
The typical Riemann sum is
(1)
If we write , then the
above Riemann sum can be written as
which is the Riemann sum of the double integral
Therefore we have
)sin,cos(),( rrrfrg
ji
m
i
n
jji
rrg 1 1
** ),(
ddrrgb
a ),(
rdrdrrf
drdrg
rrg
ArrfdAyxf
b
a
b
a
ji
m
i
n
jjiP
ij
m
i
n
jjijiR P
)sin,cos(
),(
),(lim
)sin,cos( lim),(
1 1
**
0
1 1
****
0
(2) Change to polar coordinates in a double integral If is continuous on a polar rectangle given by
where then
f
R ,,0 bra
,20
rdrdrrfdAyxf b
aR )sin,cos(),(
Example 1 Evaluate , where is the region in the upper half-plane bounded by the circles
dAyxR )43( 2 R
.4 and ,1 2222 yxyx
Solution The region R can be described as
}41 ,0|),{( 22 yxyyxR }0 ,21|),{( rr
dAyxR )43( 2 rdrdrr 0
2
1
22 )sin4cos3(
drdrr 0
2
1
232 )sin4cos3(
drrrr
210
243 sincos d
0
2sin15cos7
d
0)2cos1(
2
15cos7
2
152sin
4
15
2
15sin7
0
Example 2 Find the volume of the solid bounded
by the xy-plane and the paraboloid 221 yxz
}1|),{( 22 yxyxD}20 ,10|),{( rr
dAyxVD )1( 22
rdrdr 2
0
1
0
2 )1(
drrrd 2
0
1
0
3 )(
2422
1
0
42
rr
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)}()(,|),{(
21 hrhrD
rdrdrrf
dAyxf
h
h
D
)(2
)(1)sin,cos(
),(
then
Example 3 Use a double integral to find the area enclosed by one loop of the four-leaved rose 2cosr
}2cos0 ,44
|),{( rrD
4/
4/
2cos
0)(
rdrddADAD
4/
4/
2cos
0
2
2
1
dr
4/
4/
2 2cos2
1
d
4/
4/4cos1
4
1
d
84sin
4
1
4
14/
4/
Example 4 Find the volume of the solid that lies under the paraboloid , above the plane, and inside the cylinder
22 yxz xy.222 xyx
Solution The solid lies above the disk, whose boundary circle
}cos20 ,22
|),{( rrD
D
dAyxV )( 22
2/
2/
cos2
0
2
rdrdr
2/
2/
cos2
0
4
4
dr
2/
2/
4cos4
d
2/
0
4cos8 d
2/
0
2
2
2cos18
d
2/
0]4cos12cos21[2
2
1 d
2
3
22
324sin2sin
2
32
2/
08
1