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