12.7 - 1

Triple Integrals

12.7 - 2

z-Simple, y-simple, z-simple Approach

12.7 - 3

z-Simple solids (Type 1)

Definition: A solid region E

is said to be z-Simple if it is bounded by two surfaces z=z1(x,y) and z=z2(x,y)(z1 zz2)

x

y

z

12.7 - 4Iterated Triple Integrals over z-Simple solid E

When you project a z-Simple solid E onto the xy-plane you obtain a planar region D. 1st you integrate wrt z (the simple variable) from

z=z1(x,y) to z=z2(x,y). You obtain some function of x and y to integrate

over the region D in the xy-plane. If D is Type I you have y=y1(x) to y=y2(x) and you

integrate over y. Finally you integrate over the constant limits from x=x1 to

x=x2 and this integration is wrt x.

If D is Type II you have x=x1(y) to x=x2(y). Finally you integrate over the constant limits from y=y1 to

y=y2 and this integration is wrt y.

E

dV(x,y,z)

12.7 - 5

xy

xy

D

xy

xy

D

yxzz

yxzz

dAf(x,y)

dAdz(x,y,z)

dV(x,y,z)

),(

),(

E

2

1

Triple to Double Integral for z-Simple solid

12.7 - 6

xy

xy

D

xy

xy

D

yxzz

z

dA(x,y)z

dAdz

dV

2

),(

0

E

2

z-Simple solid: special case =1, z1=0, z2>0.

This gives the volume V over the region Dxy in the xy-plane of the surface z=z2(x,y)

12.7 - 7

y-Simple solids (Type 2)

Definition: A solid region E

is said to be y-Simple if it is bounded by two surfaces y=y1(x,z) and y=y2(x,y)(y1 y y2)

xy

z

12.7 - 8

Example: Paraboloids

Find the volume of the solid enclosed by the two surfacesy= 0.5(x2+z2) and y=16-x2-z2.

We need to define a region Dxz in the xz-plane.

xy

z

12.7 - 9

Example

Eliminate y by equating0.5(x2+z2) =16-x2-z2.

This gives x2+z2

=32/3yzD

12.7 - 10 It may be helpful to recall

the single integral calculus method for finding area between two curves y=y1(x) and y=y2(x) -- just think of z as constant, e.g., on a horizontal trace (say z=0)

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

r

r

D

xz

yz

D

zxy

zxy

drrr

dAzx

dAdydV

yz

yz

Polar coordinates x=r cos ,z=r sin , wereused so that dAxz can be written asr dr d instead ofdx dz.Compare to a cylinder of radius and height 16 which has

double 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.

This approach is also helpful in sketching the solid of integration, because as we will see the outer limit ofintegration corresponds to constant values on whichcontour regions in the simple plane lie

12.7 - 13

If not z-simple, try:

dz(x,y,z)dA

dV(x,y,z)

zz

zz zD

xy

xy

2

1 )(

E

where Dxy(z) is the trace of the solid (a trace of a solid is a region instead of a curve) in

the plane z=constant.

12.7 - 14

Example (Text, page 892#7)

Sketch the domain of integration of the triple integral

where

Then evaluate the integral.

E

dV yz

20,20,10|),,( zxzyzzyxE

12.7 - 15

Solution: Perhaps the easiest way to see the solid E which is

our domain of integration is to first consider z fixed. Then x varies from 0 to (z + 2)

and y varies from 0 to 2z. This defines a rectangle in the plane z units above the

xy-plane. Question: What are its vertices? 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).

This rectangle lies on a plane which is located z units above the xy-plane.

Let’s graph the family of horizontal traces, for several values of z between 0 and 1.

z

xy

2zx

0x0y

zy 2

12.7 - 17

Domain E is not z-simple

Next, as z increases, the rectangles become larger (and higher)

If we stack them, one above the other, we get the solid domain of integration

Let’s assume (x,y,z)=1, so that the triple integral will give us a volume instead of a mass.

z

xy

zy 2

0y 0x2zx

12.7 - 18Cross-section Rxy (keeping z constant)

We integrate (x,y,z)=1 over Rxy

treating z as constant to get the area of Rxy

Its horizontal traces define regions R=Rxy(z) above the xy-plane. Each of these planar regions should be Type I (or Type II) so that the areas of the cross-sections can be evaluated as double integrals

Between z and z+dz the total volume is dV=Axydz. Sum from z=0 to 1 to get the total volume: that sum converges 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

3

8)2(2

)2(

mdzzz

dzdyz

dzdydx

dzdAdV

z

z

z

z

z

z

z

zz

z

z D

xy

xy

kgdzzz

dzdyzyz

dzdydxyz

dzdAyzdVyz

z

z

z

z

z

z

z

zz

z

z D

xy

xy

5

7)2(2

)2(

1

0

3

1

0

2

0

1

0

2

0

2

0

1

0E

12.7 - 20

Challenge

Rewrite

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 5

72

5

2122

1

2

33

2

1

0

32

0

x

dxdzzdxdzz

12.7 - 22

Example: Tetrahedron(x,y, and z-simple :-) Evaluate the triple integral

where E is the solid tetrahedron with vertices (0,0,0), (1,0,0), (0,2,0), and (0,0,3).

E

dVxy

12.7 - 23

Example: Step 1

Visualize the solid. You need to get equations of the 4 planar sides of the tetrahedron.

Consider P(0,0,0), Q(1,0,0), R(0,2,0)Convince yourself -- or show using

that the equation of the plane through these 3 points is z=0.

0,,, zyx PzPyPxnPRPQn

12.7 - 24

Consider P(0,0,0), Q(1,0,0), S(0,0,3)Similarly the equation of the plane

through these 3 points is y=0. Consider P(0,0,0), R(0,2,0), S(0,0,3)

This corresponds the plan x=0. Now 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

The region is described by 0 x, 0 y, 0 z, and 6x+3y+2z 6.

This solid is x-simple, y-simple and z-simple. To describe it as z-simple, we let z1=0 and

z2=3-3x-1.5y. Equating z=z1=z2 we obtain the region Dxy

described by 3x+1.5y=3 in the xy-plane. Since 0 y 2-2x for 0 x 1 ...

12.7 - 26

Example: Step 3

We now are in the position to set up the triple 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

32222

22

0

221

0

22

0

1

0

2

1

2

3

2

3

)5.133(

)5.133(

dxyxyxxy

dxdyxyyxxy

dxdyyxxyI

xy

y

x

x

12.7 - 28

…and finis.

101

52

233

2662

1

0

5432

1

0

432

x

xxxxx

dxxxxxI