12.4

Find the center of mass of a lamina in the shape of an isosceles right triangle with equal sides of length if the density at any point is proportional to the square of the dis- tance from the vertex opposite the hypotenuse.

14. A lamina occupies the region inside the circle

but outside the circle . Find the center of mass if the density at any point is inversely proportional to its distance from the origin.

15. Find the moments of inertia , , for the lamina of Exercise 7.

16. Find the moments of inertia , , for the lamina of Exercise 12.

17. Find the moments of inertia , , for the lamina of Exercise 9.

18. Consider a square fan blade with sides of length 2 and the lower left corner placed at the origin. If the density of the blade is , is it more difficult to rotate the blade about the -axis or the -axis?

19–20 ^■ Use a computer algebra system to find the mass, center of mass, and moments of inertia of the lamina that occupies the region and has the given density function.

19. ;

20. is enclosed by the cardioid ;

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

21. A lamina with constant density occupies a square with vertices , , , and . Find the moments of inertia and and the radii of gyration and . 22. A lamina with constant density occupies the

region under the curve from to . Find the moments of inertia and and the radii of gyration and .y

x Iy

Ix

x x 0

y sin xx, y  

y x Iy

Ix

0, a

a, a

a, 0

0, 0 x, y  

x, y  sx² y²

r 1  cos D

x, y  xy D x, y



D

CAS

y x

x, y  1  0.1x I0

Iy

Ix

I0

Iy

Ix

I0

Iy

Ix

x² y² 1

x² y² 2y a

Electric charge is distributed over the rectangle 13.

, so that the charge density at

is (measured in coulombs per

square meter). Find the total charge on the rectangle.

2. Electric charge is distributed over the disk so that the charge density at is

(measured in coulombs per square meter). Find the total charge on the disk.

3–10 ^■ Find the mass and center of mass of the lamina that occupies the region and has the given density function .

3. ;

4. ;

is the triangular region with vertices , , ;

6. is the triangular region with vertices , , ;

7. is bounded by , , , and ;

8. is bounded by , , and ;

9. is bounded by the parabola and the line

;

10. ;

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

11. A lamina occupies the part of the disk in the first quadrant. Find its center of mass if the density at any point is proportional to its distance from the -axis.

12. Find the center of mass of the lamina in Exercise 11 if the density at any point is proportional to the square of its dis- tance from the origin.

x x² y² 1

x, y  x D x, y



x, y  3 y x  2

x y² D

x, y  x x 1

y 0 y sx D

x, y  y

x 1 x 0

y 0 y e^x D

x, y  x 0, 0 1, 1 4, 0

D

x, y  x  y 0, 0 2, 1 0, 3

5. D

x, y  cxy D x, y



x, y  xy² D x, y



 D

!x, y  x  y  xx² y² 4 ² y² x, y

!x, y  2xy  y²

x, y

0 y 2 1 x 3

1.

EXERCISES

12.4

TRIPLE INTEGRALS

Just as we defined single integrals for functions of one variable and double integrals for functions of two variables, so we can define triple integrals for functions of three variables. Let’s first deal with the simplest case where is defined on a rectangular box:

The first step is to divide B into sub-boxes. We do this by dividing the interval into l subintervals with lengths , dividing into m sub- intervals with lengths , and dividing into n subintervals with lengthsz^k z^k zk1. The planes through the endpoints of these subintervals paral-

r, s

y^j y^j yj1

c, d

xⁱ xⁱ xi1

xi1, xi a, b

B



1

f

12.5

lel to the coordinate planes divide the box into sub-boxes

which are shown in Figure 1. The sub-box has volume . Then we form the triple Riemann sum

where the sample point is in . By analogy with the definition of a double integral (12.1.5), we define the triple integral as the limit of the triple Riemann sums in (2) as the sub-boxes shrink.

DEFINITION The triple integral of over the box is

if this limit exists.

Again, the triple integral always exists if is continuous. We can choose the sample point to be any point in the sub-box, but if we choose it to be the point , and if we choose sub-boxes with the same dimensions. so that , we get a simpler- looking expression for the triple integral:

Just as for double integrals, the practical method for evaluating triple integrals is to express them as iterated integrals as follows.

FUBINI’S THEOREM FOR TRIPLE INTEGRALS If is continuous on the

rectangular box , then

The iterated integral on the right side of Fubini’s Theorem means that we integrate first with respect to (keeping and fixed), then we integrate with respect to (keep- ing fixed), and finally we integrate with respect to . There are five other possible orders in which we can integrate, all of which give the same value. For instance, if we integrate with respect to , then , and then , we have

yyy

B

fx, y, z dV 

y

y

y

y

z

z x y z y

yyy

B

fx, y, z dV 

y

y

y

f

4

yyy

B

fx, y, z dV  lim_{l, m, nl i}







V^ijk V xⁱ, yj,z^k f

yyy

B

fx, y, z dV  lim

maxxi,yj,zkl 0 _i







B f

3

Bi jk

x^{i jk}* , yi jk* ,z^{i jk}*



i1_j





2

V^ijk xⁱy^jz^k Bi jk

Bi jk xⁱ1, xi y^j1, yj z^k1,z^k lmn

B

FIGURE 1 B

Bijk

z

x y

z

x y

Îy_j Î x_i Îz_k

EXAMPLE 1 Evaluate the triple integral , where is the rectangular box given by

SOLUTION We could use any of the six possible orders of integration. If we choose to integrate with respect to , then , and then , we obtain

■

Now we define the triple integral over a general bounded region E in three- dimensional space (a solid) by much the same procedure that we used for double inte- grals (12.2.2). We enclose in a box of the type given by Equation 1. Then we define a function so that it agrees with on but is 0 for points in that are out- side . By definition,

This integral exists if is continuous and the boundary of is “reasonably smooth.”

The triple integral has essentially the same properties as the double integral (Proper- ties 6 – 9 in Section 12.2).

We restrict our attention to continuous functions and to certain simple types of regions. A solid region is said to be of type 1 if it lies between the graphs of two continuous functions of and , that is,

where is the projection of onto the -plane as shown in Figure 2. Notice that the upper boundary of the solid is the surface with equation , while the lower boundary is the surface .

By the same sort of argument that led to (12.2.3), it can be shown that if is a type 1 region given by Equation 5, then

The meaning of the inner integral on the right side of Equation 6 is that and are held fixed, and therefore and are regarded as constants, while

is integrated with respect to .

In particular, if the projection of onto the -plane is a type I plane region (as in Figure 3), then

E



xy E

D

z u2x, y fx, y, z

u1x, y x y

yyy

E

fx, y, z dV 

yy

D

 ^y



6

E z  u¹x, y z  u²x, y

E

xy E

D

E



5

y x E

f E f

yyy

E

fx, y, z dV 

yyy

B

Fx, y, z dV E

B E

f F

B E



y

3z²

4 dz  z³ 4



3

 27



y

y

y





y2

dz

yyy

B

xyz²dV

y

y

y

y

y





dy dz z

y x

B



 xxx

V

FIGURE 2

A type 1 solid region z

0

x D y

E

z=u™ (x, y)

z=u¡ (x, y)

FIGURE 3

A type 1 solid region

z=u™(x, y)

0

D E

y=g™(x) y=g¡(x)

z

x y a b

z=u¡(x, y)

and Equation 6 becomes

If, on the other hand, is a type II plane region (as in Figure 4), then

and Equation 6 becomes

EXAMPLE 2 Evaluate , where is the solid tetrahedron bounded by the

four planes , , , and .

SOLUTION When we set up a triple integral it’s wise to draw two diagrams: one of the solid region (see Figure 5) and one of its projection on the -plane (see Figure 6). The lower boundary of the tetrahedron is the plane and the upper

boundary is the plane (or ), so we use

and in Formula 7. Notice that the planes and

intersect in the line (or ) in the -plane. So the projec- tion of is the triangular region shown in Figure 6, and we have

This description of as a type 1 region enables us to evaluate the integral as follows:

■ A solid region is of type 2 if it is of the form

where, this time, is the projection of onto the -plane (see Figure 7). The back surface is , the front surface is , and we have

yyy

E

fx, y, z dV 

yy

D

 ^y



10

x u²y, z

x u¹y, zD E yz

E



E

¹6

y

6





¹2

y





y1x

¹2

y

y



y

y





dy dx

yyy

E

z dV 

y

y

y

E



9

E

xy y 1  x

x y  1

z  0u2x, y  1  x  yE x y  z  1 z  1  x  y D z  0x y  z  1xyu1x, y  0 x y  z  1

z  0 y 0

x 0

xxx

yyy

E

fx, y, z dV 

y

y

y

8

E



D

yyy

E

fx, y, z dV 

y

y

y

7

FIGURE 4

Another type 1 solid region z=u™(x, y)

x=h™(y) x=h¡(y) E

D

z=u¡(x, y)

x 0

z

y

c d

FIGURE 5 x

0

z

(1, 0, 0) y

(0, 1, 0) (0, 0, 1)

E

z=1-x-y

z=0

0 1

1 x y=0 y=1-x

D y

FIGURE 6

0 z

x E y

D

x=u¡( y, z)

x=u™( y, z) FIGURE 7

A type 2 region

Finally, a type 3 region is of the form

where is the projection of onto the -plane, is the left surface, and is the right surface (see Figure 8). For this type of region we have

In each of Equations 10 and 11 there may be two possible expressions for the inte- gral depending on whether is a type I or type II plane region (and corresponding to Equations 7 and 8).

EXAMPLE 3 Evaluate , where is the region bounded by the

paraboloid and the plane .

SOLUTION The solid is shown in Figure 9. If we regard it as a type 1 region, then we need to consider its projection onto the -plane, which is the parabolic region in Figure 10. (The trace of in the plane is the parabola .)

From we obtain , so the lower boundary surface of is and the upper surface is . Therefore, the description of as a type 1 region is

and so we obtain

Although this expression is correct, it is extremely difficult to evaluate. So let’s instead consider as a type 3 region. As such, its projection onto the -plane is the disk shown in Figure 11.

Then the left boundary of is the paraboloid and the right boundary is the plane y 4, so taking uE1x, z  x² z²andy xu2x, z  4² z² in Equation 11, we

x² z² 4E D3 xz

yy

E

y

y

y

y

E



E

z  sy  x²

z  sy  xy x² z^{2 2} z  sy  x² E

0

FIGURE 10

Projection on xy-plane FIGURE 9

Region of integration 0

4 y=≈+z@

E

x y

y=4

y=≈

D¡

x z

y

y x² z  0

y x²D z1 ² xy E

y 4

y x² z²

xxx

V

D

yyy

E

fx, y, z dV 

yy

D

 ^y



11

y u²Dx, z E xz y u¹x, z

E



FIGURE 8 A type 3 region

z

y=u™(x, z)

y=u¡(x, z) x

0

y D

E

FIGURE 11

Projection on xz-plane x 0

z

≈+z@=4

_2 2

D£

Visual 12.5 illustrates how solid regions (including the one in Figure 9) project onto coordinate planes.

have

Although this integral could be written as

it’s easier to convert to polar coordinates in the -plane: , . This gives

■

APPLICATIONS OF TRIPLE INTEGRALS

Recall that if , then the single integral represents the area under the curve from to , and if , then the double integral

represents the volume under the surface and above . The corresponding interpretation of a triple integral , where , is not very useful because it would be the “hypervolume” of a four-dimensional object and, of course, that is very difficult to visualize. (Remember that is just the domain of the function ; the graph of lies in four-dimensional space.) Nonetheless, the triple inte- gral can be interpreted in different ways in different physical situa- tions, depending on the physical interpretations of , , and .

Let’s begin with the special case where for all points in . Then the triple integral does represent the volume of :

For example, you can see this in the case of a type 1 region by putting in Formula 6:

and from Section 12.2 we know this represents the volume that lies between the sur- faces and .

EXAMPLE 4 Use a triple integral to find the volume of the tetrahedron bounded by the planes x 2y  z  2, x 2y, x 0, and z  0. T

z  u²x, y

z  u¹x, y

yyy

E

1 dV

yy

D

 ^y



^yy

u²x, y  u¹x, y dA

fx, y, z  1 VE 

yyy

E

dV

12

E

E fx, y, z  1x y z fx, y, z

xxx

E

fx, y, z  0

xxx

xx

a

y f xfx  0

x

 2







y

y



y

y

yyy

E

sx² z² dV

yy

D3

4  x² z²sx² z² dA

z  r sin x r cos

xz

y

y



yy

D3

4  x² z²sx² z²dA

yyy

E

sx² z² dV

yy

D3

 ^y



| The most difficult step in evaluating a triple integral is setting up an expres- sion for the region of integration (such as Equation 9 in Example 2). Remember that the limits of integration in the inner integral contain at most two variables, the limits of integration in the middle integral contain at most one variable, and the limits of integration in the outer inte- gral must be constants.

SOLUTION The tetrahedron and its projection on the -plane are shown in Figures 12 and 13. The lower boundary of is the plane and the upper boundary is the plane , that is, . Therefore, we have

by the same calculation as in Example 4 in Section 12.2.

(Notice that it is not necessary to use triple integrals to compute volumes. They simply give an alternative method for setting up the calculation.) _■

All the applications of double integrals in Section 12.4 can be immediately ex- tended to triple integrals. For example, if the density function of a solid object that occupies the region is , in units of mass per unit volume, at any given point

, then its mass is

and its moments about the three coordinate planes are

The center of mass is located at the point , where

If the density is constant, the center of mass of the solid is called the centroid of . The moments of inertia about the three coordinate axes are

As in Section 12.4, the total electric charge on a solid object occupying a region and having charge density is

Q

yyy

E

!x, y, z dV

!x, y, z

E

Iz

yyy

E

x² y²x, y, z dV Iy

yyy

E

x² z²x, y, z dV Ix

yyy

E

y² z²x, y, z dV

16

E z  Mxy

y Mxz m x Myz m

15 m

x, y, z

Mx y

yyy

E

zx, y, z dV Mxz 

yyy

E

yx, y, z dV Myz 

yyy

E

xx, y, z dV

14

m

yyy

E

x, y, z dV

13

x, y, z E x, y, z



y

y

VT 

yyy

T

dV

y

y

y

x 2y  z  2 T z  0

xy D

T

”or y=1-    ’ FIGURE 12

(0, 1, 0) (0, 0, 2)

y

x 0

z

x+2y+z=2 x=2y

”1,  , 0’¹₂ T

FIGURE 13

y=^x₂

”1,   ’₂¹ D

y

0 1

1 x x 2 x+2y=2

EXAMPLE 5 Find the center of mass of a solid of constant density that is bounded by the parabolic cylinder and the planes , , and . SOLUTION The solid and its projection onto the -plane are shown in Figure 14.

The lower and upper surfaces of are the planes and , so we describe as a type 1 region:

Then, if the density is , the mass is

Because of the symmetry of and about the -plane, we can immediately say that and, therefore, . The other moments are

Therefore, the center of mass is

x, y, z 





⁽

⁾

 

3

y

  7

2

y

y



y

y





dx dy Mxy

yyy

E

z dV 

y

y

y

 2

3

y

3







y





dy



y

y

yyy

E

x dV 

y

y

y

Mxz 0 E  xz









y

 

2

y



y





dy



y

y

yyy

E

 dV 

y

y

y

x, y, z  

E



z  x E z  0

E

xy E

x 1 z  0

x z x y²

V

0 y

x x=1 x=¥

D

0

1 E

z=x

x

z

y

FIGURE 14

3–6 ^■ Evaluate the iterated integral.

3. 4.

5. 6.

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

y

y

y

y

y

y

y

y

y

y

y

y

respect to , then , and then .

2. Evaluate the integral , where

using three different orders of integration.

E



y z x

EXERCISES

12.5

(b) Use a computer algebra system to approximate the inte- gral in part (a) correct to the nearest integer. Compare with the answer to part (a).

23–24 ^■ Use the Midpoint Rule for triple integrals (Exer- cise 22) to estimate the value of the integral. Divide into eight sub-boxes of equal size.

23. , where

24. , where

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

25–26 ^■ Sketch the solid whose volume is given by the iterated integral.

26.

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

27–30 ^■ Express the integral as an iterated integral in six different ways, where is the solid bounded by the given surfaces.

27. , ,

28. , , ,

29. , ,

30.

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

31. The figure shows the region of integration for the integral

Rewrite this integral as an equivalent iterated integral in the five other orders.

0 z 1

x

1 y

z=1-y

y=œ„x

y

y

y

x² 1  y z  y

z  0

z  y  2x y 2

x 0 z  0

y 6 y 0 x² z² 4

E

xxxEfx, y, z dV

y

y

y

y

y

y

B x, y, z



xxxBsinxy²z³ dV

B x, y, z



xxxB

1

ln1  x  y  zdV

B 7–16 ^■ Evaluate the triple integral. CAS

7. , where

8. , where

, where lies under the plane

and above the region in the -plane bounded by the curves

, , and

10. , where is bounded by the planes , , , and

11. , where is the solid tetrahedron with vertices

, , , and

12. , where is the solid tetrahedron with vertices

, , , and

13. , where is bounded by the parabolic cylinder and the planes , , and

14. , where is bounded by the parabolic cylinder and the planes , , and 15. , where is bounded by the paraboloid

and the plane

16. , where is bounded by the cylinder

and the planes , , and in the first octant

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

17–20 ^■ Use a triple integral to find the volume of the given solid.

The tetrahedron enclosed by the coordinate planes and the plane

18. The solid bounded by the cylinder and the planes , and

19. The solid enclosed by the cylinder and the planes and

20. The solid enclosed by the paraboloid and the plane

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

(a) Express the volume of the wedge in the first octant that is cut from the cylinder by the planes

and as a triple integral.

(b) Use either the Table of Integrals (on Reference Pages 6 –10) or a computer algebra system to find the exact value of the triple integral in part (a).

22. (a) In the Midpoint Rule for triple integrals we use a triple Riemann sum to approximate a triple integral over a box , where is evaluated at the center

of the box . Use the Midpoint Rule to esti-

mate , where is the cube

defined by , , . Divide

into eight cubes of equal size.

B

0 z 4 0 y 4

0 x 4

xxxBsx² y² z²dV B Bi j k

xⁱ, yj,z^k

fx, y, z

B

CAS

x 1 y x

y² z² 1 21.

x 16

x y² z² z  1

y z  5

x² y² 9 y 9

z  4 z  0,

y x² 2x y  z  4

17.

z  0 y 3x

x 0

y² z² 9

xxxEz dV E

x 4 x 4y² 4z²

xxxEx dV E

z  0 x y

x z y x²

xxxEx  2y dV E

x 1 x 1

z  0 z  1  y²

xxxEx²e^ydV E

0, 1, 1

1, 1, 0

0, 1, 0

0, 0, 0

xxxExz dV E

0, 0, 3

0, 2, 0

1, 0, 0

0, 0, 0

xxxExy dV E

2x 2y  z  4 z  0

y 0 x 0

xxxEy dV E

x 1 y 0

y sx

xy

z  1  x  y

xxxE 6xy dV E 9.

E x, y, z



xxxEyz cosx⁵ dV

E

{



}