Multiple Integrals

Chapter 13

by Zhian Liang

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

i

n

ii

xxf 1

* )(

}max{i

xP

i

n

ii

b

aP

xxfdxxf 1

*

0)(lim)(

and summing them, we have

(1)

(2)

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

EXERCISES 13.1

Page 837

1. 3. 15.16

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

Interpret the double integral as the volume V of the solid

R

dAyxf ),(

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 )(

EXAMPLE 2 Evaluate the double integral dAyxR )3( 2

where }.21 ,20|),{( yxyxR

(See the blackboard)

EXAMPLE 3

R dAxyy )sin( ],0[]2,1[ REvaluate ,where

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

If on , then )()(),( yhxgyxf ],[],[ dcbaR

dyyhdxxgdAyhxg d

c

b

aR

)()()()(

]2,0[]2,0[ R

111

][sin]cos[

cossincossin

22

2 2

00

0 0

yx

ydyxdxydAxR

EXAMPLE 5 , thenIf

EXERCISES 13.2

Page 842 1(2), 6, 10, 16, 17,

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:

(10)

(11) D DAdA )(1

then, ),( allfor ),( DyxMyxfm

)(),()( DMAdAyxfDmAD

, the area of region D.

If

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

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

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(),(

Caution: Do not forget the factor r in (2)!

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

Exercises 13.4

Page 856: 1, 4, 6, 7, 22