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12.7 - 1
Triple Integrals
12.7 - 2
z-Simple, y-simple, z-simpleApproach
12.7 - 3
z-Simple solids (Type 1)
Definition:nA solid region
E is said to bez-Simple if itis bounded bytwo surfacesz=z1(x,y) andz=z2(x,y)(z1 ≤ z2)
x
y
z
12.7 - 4Iterated Triple Integralsover z-Simple solid E
When you project a z-Simple solid E ontothe xy-plane you obtain a planar region D.n 1st you integrate wrt z (the simple variable)
from z=z1(x,y) to z=z2(x,y).n You obtain some function of x and y to
integrate over the region D in the xy-plane.l If D is Type I you have y=y1(x) to y=y2(x) and you
integrate over y.l Finally you integrate over the constant limits from x=x1
to x=x2 and this integration is wrt x.l If D is Type II you have x=x1(y) to x=x2(y).
l Finally you integrate over the constant limits from y=y1to y=y2 and this integration is wrt y.
∫∫∫E
dV(x,y,z)ρ
12.7 - 5
∫∫
∫∫ ∫
∫∫∫
=
=
=
=
xy
xy
Dxy
xyD
yxzz
yxzz
dAf(x,y)
dAdz(x,y,z)
dV(x,y,z)
),(
),(
E
1
1
ρ
ρ
Triple to Double Integralfor z-Simple solid
12.7 - 6
∫∫
∫∫ ∫
∫∫∫
=
=
=
=
xy
xy
Dxy
xyD
yxzz
z
dA(x,y)z
dAdz
dV
2
),(
0
E
2
z-Simple solid: special caseρ=1, z1=0, z2>0.
This gives thevolume V overthe region Dxyin the xy-plane of thesurfacez=z2(x,y)
12.7 - 7
y-Simple solids (Type 2)
Definition:nA solid region
E is said to bey-Simple if itis bounded bytwo surfacesy=y1(x,z) andy=y2(x,y)(y1 ≤ y2)
xy
z
12.7 - 8
Example: Paraboloids
n Find the volume ofthe solid enclosedby the twosurfacesy= 0.5(x2+z2) andy=16-x2-z2.
nWe need to definea region Dxz in thexz-plane.
xy
z
12.7 - 9
Example
n Eliminate y by equating0.5(x2+z2) =16-x2-z2.
n This gives x2+z2 =32/3
yzD
12.7 - 10n It may be helpful to
recall the singleintegral calculusmethod for findingarea between twocurves y=y1(x) andy=y2(x) -- just think ofz as constant, e.g., on ahorizontal trace (sayz=0)
n Next we evaluatethe integral -- dy goesfirst since it is y-simple.
332
12.7 - 11
( )
1.2683
256
)5.116(2
5.15.116
332
0
2
22
16
)(5.0E
22
22
==
−=
−−=
=
∫
∫∫
∫∫ ∫∫∫∫
=
=
−−=
+=
π
πr
r
Dxz
yzD
zxy
zxy
drrr
dAzx
dAdydV
yz
yz
Polarcoordinatesx=r cos θ,z=r sin θ, wereused so thatdAxz can bewritten asr dr dθinstead ofdx dz.
Compare to a cylinder of radius and height 16 which hasdouble this volume (anyone know why?) and contains our solid E inside it.
332
12.7 - 12
xy-Simple, yz-simple, xz-simple Approach
Often a solid is simple in more than one variable.An alternate approach is to look for the one variable that it is not simple in, and make that the outer limitof integration. The inner limit is then a double integral.
12.7 - 13
If not z-simple, try:
dz(x,y,z)dA
dV(x,y,z)
zz
zz zDxy
xy
∫ ∫∫
∫∫∫=
=
=
2
1 )(
E
ρ
ρ
where Dxy(z) is the trace of thesolid in the plane z=constant.
12.7 - 14
Example (Text, page 892#7)
n Sketch the domain of integrationof the triple integral
where
Then evaluate the integral.
∫∫∫E
dV yz
{ }20,20,10|),,( +≤≤≤≤≤≤= zxzyzzyxE
12.7 - 15
Solution:n Perhaps the easiest way to see the solid E
which is our domain of integration is to firstconsider z fixed.
n Then x varies from 0 to (z + 2) and y varies from 0 to 2z.
n This defines a rectangle in the plane z unitsabove the xy-plane.
n Question: What are its vertices?n Answer:
(0,0,z), (z + 2, 0, z), (0, 2z, z) and (z+2, 2z,z).
{ }20,20,10|),,( +≤≤≤≤≤≤= zxzyzzyxE
12.7 - 16Horizontal trace of domain E (z=constant)Rectangle (0,0,z), (z + 2, 0, z), (0, 2z, z), (z+2, 2z,z).
n This rectangle lieson a plane which islocated z unitsabove the xy-plane.
n Let’s graph thefamily ofhorizontal traces,for several valuesof z between 0and 1.
z
xy
2+= zx
0=x0=y
zy 2=
12.7 - 17
Domain E is not z-simple
n Next, as z increases, therectangles become larger(and higher)
n If we stack them, oneabove the other, we getthe solid domain ofintegration
n Let’s assume ρ(x,y,z)=1, sothat the triple integralwill give us a volumeinstead of a mass.
z
xy
zy 2=
0=y 0=x2+= zx
12.7 - 18Cross-section Rxy(keeping z constant)
n We integrate ρ(x,y,z)=1 over Rxytreating z as constant to get the areaof Rxy
l Its horizontal traces defineregions R=Rxy(z) above the xy-plane. Each of these planar regionsshould be Type I (or Type II) sothat the areas of the cross-sections can be evaluated as doubleintegrals
n Between z and z+dz the total volumeis dV=Axydz. Sum from z=0 to 1 toget the total volume: that sumconverges to the integral over z.
z
xy
12.7 - 19
Evaluation of the triple integral
31
0
1
0
2
0
1
0
2
0
2
0
1
0E
38
)2(2
)2(
mdzzz
dzdyz
dzdydx
dzdAdV
z
z
z
z
z
z
z
zz
z
z Dxy
xy
=+=
+=
=
=
∫
∫ ∫
∫ ∫ ∫
∫ ∫∫∫∫∫
=
=
=
=
=
=
+
=
=
kgdzzz
dzdyzyz
dzdydxyz
dzdAyzdVyz
z
z
z
z
z
z
z
zz
z
z Dxy
xy
57
)2(2
)2(
1
0
3
1
0
2
0
1
0
2
0
2
0
1
0E
=+=
+=
=
=
∫
∫ ∫
∫ ∫ ∫
∫ ∫∫∫∫∫
=
=
=
=
=
=
+
=
=
12.7 - 20
ChallengeRewrite
as
or
using the fact that E is y-simple (i.e., dy on the inside)
∫∫∫+2
0
2
0
1
0
zz
dzdydxyz
∫∫∫?
?
?
?
?
?
dzdxdyyz
∫∫∫?
?
?
?
?
?
dxdzdyyz
12.7 - 21
Answers: dy dx dz is easy
but dy dz dx has to be split into 2.
∫ ∫∫∫∫ ==++ z zz
dzdxzdzdxdyyz2
0
2
0
31
0
2
0
1
0 57
2
52
1221
2
33
2
1
0
32
0
+=+ ∫∫∫∫−x
dxdzzdxdzz
12.7 - 22Example: Tetrahedron
(x,y, and z-simple :-)n Evaluate the triple integral
n where E is the solid tetrahedronwith vertices (0,0,0), (1,0,0),(0,2,0), and (0,0,3).
∫∫∫E
dVxy
12.7 - 23
Example: Step 1
n Visualize the solid. You need to getequations of the 4 planar sides of thetetrahedron.
n Consider P(0,0,0), Q(1,0,0), R(0,2,0)Convince yourself -- or show using
that the equation of the plane throughthese 3 points is z=0.
0,,, =−−−⋅×= zyx PzPyPxnPRPQn
12.7 - 24
n Consider P(0,0,0), Q(1,0,0), S(0,0,3)lSimilarly the equation of the plane through
these 3 points is y=0.n Consider P(0,0,0), R(0,2,0), S(0,0,3)lThis corresponds the plan x=0.
nNow consider Q(1,0,0), R(0,2,0), S(0,0,3)
06236,,2,3,6
2,3,63,0,10,2,1
=−++=−−−⋅
=−×−=×=
zyxQzQyQx
QSQRn
zyx
12.7 - 25
Example: Step 2
n The region is described by0 ≤ x, 0 ≤ y, 0 ≤ z, and 6x+3y+2z ≤ 6.
n This solid is x-simple, y-simple and z-simple.
n To describe it as z-simple, we let z1=0 andz2=3-3x-1.5y.
n Equating z=z1=z2 we obtain the region Dxydescribed by 3x+1.5y=3 in the xy-plane.
n Since 0 ≤ y ≤ 2-2x for 0 ≤ x ≤ 1 ...
12.7 - 26
Example: Step 3
nWe now are in the position to set up thetriple integral with limits.
( ) dxdyxyz
dxdydzxy
dVxyI
yxzz
x
yxx
E
5.1330
22
0
1
0
5.133
0
22
0
1
0
−−==
−
−−−
∫∫∫∫∫
∫∫∫
=
=
=
12.7 - 27
…continued...
∫
∫∫∫∫
−=
=
−
−
−−=
−−=
−−=
1
0
22
0
3222
22
0
221
0
22
0
1
0
21
23
23
)5.133(
)5.133(
dxyxyxxy
dxdyxyyxxy
dxdyyxxyI
xy
y
x
x
12.7 - 28
…and finis.
( )
101
52
233
26621
0
5432
1
0
432
=
−+−=
−+−==
=
∫x
xxxxx
dxxxxxI