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UNIT 3. MULTIPLE INTEGRALS

Objectives: Upon the completion of the course,

the student must be able to evaluatemultiple integrals using rectangular, polar,cylindrical and spherical coordinates

Review !olar "oordinates

Outline:#ouble $ntegrals %in Rectangular"oordinates&

#ouble $ntegrals %in !olar "oordinates&

Triple $ntegrals %in Rectangular"oordinates&

Triple $ntegrals %in 'ther "oordinate(ystems&

Area and )olume in Rectangular,"ylindrical and (pherical "oordinates

Reference: *3+* to *3+, T"-

$n MATH 3, single integrals were used todetermine area of planar regions+ $n MATH 3-, itwas e.tended to determining lengths of planarcurves, volumes of solids of revolution and center

of mass of planar regions+ $n MATH 38, the theory

*

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and applications of single %Riemann& integrals willbe generali/ed to multiple integrals. $n its basicform, multiple integral is integration in the n

dimensional number space done by successionofsingle integrals+

Looin! b"c #$or%in! t&e sin!le inte!r"l'

"onsider a single0variable functionf

continuous over some closed interval[ ]b,a + 1orm a partition of the intervalinto nsub0intervals of length xi ,

...,,,i 32*= Then, pic a sample point foreach of the sub0intervals+ "hoose ix forthe i th sub0interval+ The 4area4 ( ) xxf ii

is then formed for each of the sub0interval+

( ) ( )=

+=

n

i

iin

b

a

xxflimdxxf

*

This construction can also be done forfunctions of two variables which is defined oversome region on the xy plane+ The region willdivided into sub0regions and 4volumes4 will beconsidered for each of theses sub0regions+

2

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3.( )ouble Inte!r"ls #in Rect"n!ul"r*oor+in"tes'

"onsider a function fof two variables xand y

which is continuous over a rectangular region( ){ }dxc;bxay,xR = +

1orm a partition of Rby means of hori/ontal orvertical lines+ Hence, the region is subdivided intosay nsub0rectangles each denoted by iR ,

n,.,.,,,i 32*= +

5et xi and yi be the lengths of the sides of iR

+ $f Ai is the area of the i th sub0rectangle, thenyxA iii = +

'n each sub0rectangle, choose a sample pointsay ( )ii y,x on iR + "onsider the 4volume4 formed by

( ) Ay,xf iii for each of the sub0rectangle and form theRiemann sum

( )= n

i

iii Ay,xf* +

$f ( ) 6y,xf over the region, ( )=

n

i

iii Ay,xf

*

appro.imates the area of the region+

(Illustrations will be provided in the class.)

3

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)efinition #of +ouble inte!r"l'.

5et fbe a function of two variables xand ydefined over a closed rectangular

region R+ $f ( )=

+

n

i

iiin

Ay,xflim

*e.ists, then

fis said to be inte!r"bleon R +Moreover, the +ouble inte!r"l of foverR , denoted by ( )

R

dAy,xf , is given by( )

R

dAy,xf - ( )=

+

n

i

iiin

Ay,xflim

*

+

The rectangular region R can bee.tended to any closed and bounded regionon the xy plane+ 5ater on, the regions thatwill be considered will be bounded by somecontinuous curves over the plane+

Re%"rs: *+ $f ( ) 6y,xf over the region R ,( )

R

dAy,xf is the volume of the solid under

the surface ( )y,xfz = and above the region R

+2+

R

dAis the area of region R+

T&eore% #on inte!r"bilit'.

$f fis bounded on the closed and

bounded region R and if it is continuous

7

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there e.cept on a finite number ofsmooth curves, then fis integrable onR+ $n particular, if fis continuous on all

of R, then fis integrable there+

Pro/erties of t&e )ouble Inte!r"l

i+ ( ) ( ) =RR

dAy,xfkdAy,xfk

ii+ ( ) ( )[ ] ( ) ( )

+=+

RRR

dAy,xgdAy,xfdAy,xgy,xf

iii+ $f RRR = 2* such that 2* RR is only acurve,

( )[ ] ( ) ( ) +=+2* RRR

dAy,xfdAy,xfdAy,xg

iv+ $f ( ) ( )y,xgy,xf over R, then( ) ( )

RR

dAy,xgdAy,xf +

Ev"lu"tin! )ouble Inte!r"ls #overrect"n!ul"r re!ion'

"onsider ( ) 6y,xf over a rectangular region( ){ }dxc;bxay,xR = + ( )

R

dAy,xf maybe

interpreted as the volume of the solid under thesurface ( )y,xfz = and above the region R+ Hence,

( )=R

dAy,xfV

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"onsider the given solid+ Alternatively, thevolume of the solid can be calculated by 4slicing4 itinto %vertical& slabs parallel to the xz plane+ The

area of the face of each of these slabs depends onhow far it is from the xz plane %or on y&+ Hence,the area of each slab can be denoted by ( )yA +

The volume of each slab % V & can beappro.imated by ( )dyyAV where dyis the

thicness of the slab+ Hence,( )

=

d

c

dyyAV

+

'n the other hand, for particular values of y,( )yA can be evaluated by using single integrals+ $n

fact, ( ) ( )=b

a

dxy,xfyA where ( )y,xfz = and dxare theheight and width of each, respectively+

(Illustrations will be provided in the class.)

Thus, ( ) ( )

==

d

c

b

aR

dydxy,xfVdAy,xf +

Also, if slabs parallel to the yz plane will be

used,( ) ( )

==

b

a

d

cR

dxdyy,xfVdAy,xf +

The e.pressions ( )

d

c

b

a

dydxy,xf and( )

b

a

d

c

dxdyy,xf are called as iter"te+ inte!r"ls.

These are used to evaluate double integrals+ 9ote

that the bounds of each of the integral depend on

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R

1

MATH 38

the bounds of the region and variable ofintegration+

*AUTION0

The bracets within the iterated integrals can

be omitted allowing dxdy or dydx to specify theorder of integration to be done+

:hen evaluating single integral of the form( ) dxy,xf , treat yas a constant while for ( ) dyy,xf ,

treat xas a constant+

Ev"lu"tin! )ouble Inte!r"ls #over

nonrect"n!ul"r re!ion'

"onsider a function( )y,xfz = defined over a

region Rboundedabove by ( )xgy= , belowby ( )xhy= , on the left by

-

E1"%/le.

;valuate( ) +

R

dAyx 32where R is the region given by

( ){ }362* x;xy,x +

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R2

MATH 38

ax= and on the right bx= +Then,

( ) ( )( )

( )

=b

a

xg

xhR

dxdyy,xfdAy,xf

"onsider a function( )y,xfz = defined over a

region R bounded onthe right by ( )ygx= , onthe left by ( )yhx= , aboveby dy= and below by

cy= + Then,

( ) ( )

( )

( )

=

d

c

yg

yhR

dydxy,xfdAy,xf

8

E1"%/les.

*+ ;valuate( ) +

R

dAyx7if R is the region bounded by 2xy= ,

xy = , and the lines *=x and 7=x +

Try both dxdy and dydx as the order of integration+

2+ ;valuate( )

+

R

dAyx7

if R is the region bounded by3

xy= and

xy= +

3+ (et0up the iterated integral that will solve for the volume of the

tetrahedron bounded by the plane 6,23 =++ zyx and the

coordinate planes+7+ (et0up the iterated integral that will solve for the volume of the solid in

the first octantbounded by the paraboloid 22 yxz += and the

cylinder 722 =+ yx +

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There are iterated integrals that can beefficiently evaluated by changing the order of

integration+

*AUTION0 The process of changing the order ofintegration in NOTa simple process ofinterchanging variables+ $t always goes bacto the region and setting0up a new iteratedintegral+

uations in Cartesian

coordinates+ ?ut, ( )R

dAy,xf can also be defined

over regions bounded by curves inpolarcoordinates.

Revie4:

$fP

is a point with "artesian coordinates( )y,x

and polar coordinates ( ),r , then

222 yxr += = cosrx

x

ytan = = sinry

$f a region Risbounded by the curve

( )= fr , , thenthe area of the regionis given by

( )[ ]

= dfA 2

2

*

+

**

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( )( )[ ]

= dfddrr

f2

6 2

*

5et ( )y,xfz = be continuous and nonnegativeover a region R+ Then, the volume of the solidunder the surface ( )y,xfz = and above Ris given by

( )=R

dAy,xfV +

9ow, let R apolar rectangle given by R=( ) ;bra,r , where 6a and 2 + Also,

the e>uation of the surface can be written as

( ) ( ) ( )=== ,rgsinr,cosrfy,xfz

$f R is partitioned into nsmaller polar

rectangles, then the area of thei

th polarrectangle is given by iii rr where ir is theaverage radius of the polar rectangle and ri and

i are its dimensions+

(Illustrations will be provided in the class.)

*2

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Hence, the volume Vcan be appro.imated by( )

=

n

i

iiiii rr,rgV

*+ Taing the limit of this

summation as the number sub0polar rectanglesincreases without bound,

( ) ( ) ==RR

ddrr,rgdAy,xfV +

*3

E1"%/les.

Use polar coordinates to evaluate the following+

*+ +R

yx dAe22

where R is the region bounded by *22 =+ yx in

the first >uadrant+

2+ +R

dAyx 22where R is the region bounded by *22 =+ yx ,

and the lines xy= and 6=y +

(;T0U! the iterated integrals in polar coordinates that will solve for thevolume of the following solids+

3+ under the plane yz = and above the region in the first >uadrant

that is outside the circle2

=r and inside the cardioid+= cosr 22

7+ under the surface 22 yxz += above the xy plane and inside thecylinder yyx 222 =+ +

E1ercises.

;valuate the following using polar coordinate+

*+ ( )

+*

6

*

6

22

2y

dxdyxsin

2+ ++R

dAyx 227

*

where R is the region outside the circle

*22 =+ yx and inside the 722 =+ yx

(;T0U! the iterated integrals that will solve for the area of the followingregion+

3+ inside the lemniscate = 2

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222222222222222222222222

3.3 Tri/le Inte!r"ls #in Rect"n!ul"r*oor+in"tes'

*7

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"onsider a function fof three variables

defined over some rectangular parallelepiped %or abo.& with faces parallel to the coordinate planes+!artition the bo. into nsub0bo.es each withdimensions y,x ii and zi + $f Vi is the volume ofthe i th sub0bo., then zyxV iiii = + Then, choosea sample point ( )iii z,y,x from the i th sub0bo.+

(Illustrations will be provided in the class.)

1orm the Riemann sum ( )=

n

i

iiii Vz,y,xf

*+

Taing the limit of this sum as the number ofsub0bo.,

( ) ( )=

+=

n

i

iiiin

S

Vz,y,xflimdVz,y,xf

*

where Sis the given rectangularparallelepiped+

*

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5et ( )z,y,xfw = be defined over some rectangular

parallelepiped Sgiven by( ) fze;dyc;bxaz,y,x + A triple integral can be

evaluated by e.pressing it as an iterated integral( ) ( ) =

b

a

d

c

f

eS

dxdydzz,y,xfdVz,y,xf

9ote that the orders of integration can vary+

1or dx, dyand dz, there are si. possible orders ofintegration+ The bounds of the integrals aredependent on the order of integration+

The illustration above can also be interpretedas

( ) ( ) =f

e

b

a

d

cS

ddxdyz,y,xfdVz,y,xf +

5et fbe defined on some solid S+

(Illustration will be provided in the class.)

*

E1"%/le.

;valuate S

dVzxy 32,if Sgiven is the rectangular parallelepiped

given by ( ) 86722* z;y;xz,y,x

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To evaluate ( )S

dVz,y,xf , it is e.pressed in

terms of an iterated integral each with its own

respective bounds( ) ( )

( )

( )

( )

( )

=

2

*

2

*

2

*

a

a

xh

xh

y,xg

y,xgS

dxdydzz,y,xfdVz,y,xf

+

9ote that the bounds of the integrals aredependent on the order of integration+ $f dzdydx ,then the bounds of the innermost integral shouldbe functions of yand z , the bounds of the middle

integral should be functions ofz

and the outermostintegral should be bounded by constants+

REMAR5: S

dVgives the volume of the solid S+

*-

E1"%/le.

;valuate the following+

*+ 2

6 * 6

2z z

x

dzdxdyxyz

2+ S

dVxyz2where Sis the solid bounded by the parabolic

cylinder2

2

*2 xz = and the planes 6=z , xy= and 6=y

Use dxdydz as the order of integration+

:hat if dzdxdy will be used as the order of integration+

@@@@@@@@@@@@@@

3+ :rite ( ) 2

6