MATH 38 UNIT 3
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Transcript of MATH 38 UNIT 3
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MATH 38
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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MATH 38
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+
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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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MATH 38
)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
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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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MATH 38
"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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MATH 38
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