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Calculus Volume 3

5.4 Triple Integrals

Calculus Volume 35.4 Triple Integrals
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  1. Preface
  2. 1 Parametric Equations and Polar Coordinates
    1. Introduction
    2. 1.1 Parametric Equations
    3. 1.2 Calculus of Parametric Curves
    4. 1.3 Polar Coordinates
    5. 1.4 Area and Arc Length in Polar Coordinates
    6. 1.5 Conic Sections
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  3. 2 Vectors in Space
    1. Introduction
    2. 2.1 Vectors in the Plane
    3. 2.2 Vectors in Three Dimensions
    4. 2.3 The Dot Product
    5. 2.4 The Cross Product
    6. 2.5 Equations of Lines and Planes in Space
    7. 2.6 Quadric Surfaces
    8. 2.7 Cylindrical and Spherical Coordinates
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  4. 3 Vector-Valued Functions
    1. Introduction
    2. 3.1 Vector-Valued Functions and Space Curves
    3. 3.2 Calculus of Vector-Valued Functions
    4. 3.3 Arc Length and Curvature
    5. 3.4 Motion in Space
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  5. 4 Differentiation of Functions of Several Variables
    1. Introduction
    2. 4.1 Functions of Several Variables
    3. 4.2 Limits and Continuity
    4. 4.3 Partial Derivatives
    5. 4.4 Tangent Planes and Linear Approximations
    6. 4.5 The Chain Rule
    7. 4.6 Directional Derivatives and the Gradient
    8. 4.7 Maxima/Minima Problems
    9. 4.8 Lagrange Multipliers
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Chapter Review Exercises
  6. 5 Multiple Integration
    1. Introduction
    2. 5.1 Double Integrals over Rectangular Regions
    3. 5.2 Double Integrals over General Regions
    4. 5.3 Double Integrals in Polar Coordinates
    5. 5.4 Triple Integrals
    6. 5.5 Triple Integrals in Cylindrical and Spherical Coordinates
    7. 5.6 Calculating Centers of Mass and Moments of Inertia
    8. 5.7 Change of Variables in Multiple Integrals
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  7. 6 Vector Calculus
    1. Introduction
    2. 6.1 Vector Fields
    3. 6.2 Line Integrals
    4. 6.3 Conservative Vector Fields
    5. 6.4 Green’s Theorem
    6. 6.5 Divergence and Curl
    7. 6.6 Surface Integrals
    8. 6.7 Stokes’ Theorem
    9. 6.8 The Divergence Theorem
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Chapter Review Exercises
  8. 7 Second-Order Differential Equations
    1. Introduction
    2. 7.1 Second-Order Linear Equations
    3. 7.2 Nonhomogeneous Linear Equations
    4. 7.3 Applications
    5. 7.4 Series Solutions of Differential Equations
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  9. A | Table of Integrals
  10. B | Table of Derivatives
  11. C | Review of Pre-Calculus
  12. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
  13. Index

Learning Objectives

  • 5.4.1. Recognize when a function of three variables is integrable over a rectangular box.
  • 5.4.2. Evaluate a triple integral by expressing it as an iterated integral.
  • 5.4.3. Recognize when a function of three variables is integrable over a closed and bounded region.
  • 5.4.4. Simplify a calculation by changing the order of integration of a triple integral.
  • 5.4.5. Calculate the average value of a function of three variables.

In Double Integrals over Rectangular Regions, we discussed the double integral of a function f(x,y)f(x,y) of two variables over a rectangular region in the plane. In this section we define the triple integral of a function f(x,y,z)f(x,y,z) of three variables over a rectangular solid box in space, 3.3. Later in this section we extend the definition to more general regions in 3.3.

Integrable Functions of Three Variables

We can define a rectangular box BB in 33 as B={(x,y,z)|axb,cyd,ezf}.B={(x,y,z)|axb,cyd,ezf}. We follow a similar procedure to what we did in Double Integrals over Rectangular Regions. We divide the interval [a,b][a,b] into ll subintervals [xi1,xi][xi1,xi] of equal length Δx=xixi1l,Δx=xixi1l, divide the interval [c,d][c,d] into mm subintervals [yi1,yi][yi1,yi] of equal length Δy=yjyj1m,Δy=yjyj1m, and divide the interval [e,f][e,f] into nn subintervals [zi1,zi][zi1,zi] of equal length Δz=zkzk1n.Δz=zkzk1n. Then the rectangular box BB is subdivided into lmnlmn subboxes Bijk=[xi1,xi]×[yi1,yi]×[zi1,zi],Bijk=[xi1,xi]×[yi1,yi]×[zi1,zi], as shown in Figure 5.40.

In x y z space, there is a box B with a subbox Bijk with sides of length Delta x, Delta y, and Delta z.
Figure 5.40 A rectangular box in 33 divided into subboxes by planes parallel to the coordinate planes.

For each i,j,andk,i,j,andk, consider a sample point (xijk*,yijk*,zijk*)(xijk*,yijk*,zijk*) in each sub-box Bijk.Bijk. We see that its volume is ΔV=ΔxΔyΔz.ΔV=ΔxΔyΔz. Form the triple Riemann sum

i=1lj=1mk=1nf(xijk*,yijk*,zijk*)ΔxΔyΔz.i=1lj=1mk=1nf(xijk*,yijk*,zijk*)ΔxΔyΔz.

We define the triple integral in terms of the limit of a triple Riemann sum, as we did for the double integral in terms of a double Riemann sum.

Definition

The triple integral of a function f(x,y,z)f(x,y,z) over a rectangular box BB is defined as

liml,m,ni=1lj=1mk=1nf(xijk*,yijk*,zijk*)ΔxΔyΔz=Bf(x,y,z)dVliml,m,ni=1lj=1mk=1nf(xijk*,yijk*,zijk*)ΔxΔyΔz=Bf(x,y,z)dV
5.10

if this limit exists.

When the triple integral exists on B,B, the function f(x,y,z)f(x,y,z) is said to be integrable on B.B. Also, the triple integral exists if f(x,y,z)f(x,y,z) is continuous on B.B. Therefore, we will use continuous functions for our examples. However, continuity is sufficient but not necessary; in other words, ff is bounded on BB and continuous except possibly on the boundary of B.B. The sample point (xijk*,yijk*,zijk*)(xijk*,yijk*,zijk*) can be any point in the rectangular sub-box BijkBijk and all the properties of a double integral apply to a triple integral. Just as the double integral has many practical applications, the triple integral also has many applications, which we discuss in later sections.

Now that we have developed the concept of the triple integral, we need to know how to compute it. Just as in the case of the double integral, we can have an iterated triple integral, and consequently, a version of Fubini’s thereom for triple integrals exists.

Theorem 5.9

Fubini’s Theorem for Triple Integrals

If f(x,y,z)f(x,y,z) is continuous on a rectangular box B=[a,b]×[c,d]×[e,f],B=[a,b]×[c,d]×[e,f], then

Bf(x,y,z)dV=efcdabf(x,y,z)dxdydz.Bf(x,y,z)dV=efcdabf(x,y,z)dxdydz.

This integral is also equal to any of the other five possible orderings for the iterated triple integral.

For a,b,c,d,e,a,b,c,d,e, and ff real numbers, the iterated triple integral can be expressed in six different orderings:

efcdabf(x,y,z)dxdydz=ef(cd(abf(x,y,z)dx)dy)dz=cd(ef(abf(x,y,z)dx)dz)dy=ab(ef(cdf(x,y,z)dy)dz)dx=ef(ab(cdf(x,y,z)dy)dx)dz=ce(ab(eff(x,y,z)dz)dx)dy=ab(ce(eff(x,y,z)dz)dy)dx.efcdabf(x,y,z)dxdydz=ef(cd(abf(x,y,z)dx)dy)dz=cd(ef(abf(x,y,z)dx)dz)dy=ab(ef(cdf(x,y,z)dy)dz)dx=ef(ab(cdf(x,y,z)dy)dx)dz=ce(ab(eff(x,y,z)dz)dx)dy=ab(ce(eff(x,y,z)dz)dy)dx.

For a rectangular box, the order of integration does not make any significant difference in the level of difficulty in computation. We compute triple integrals using Fubini’s Theorem rather than using the Riemann sum definition. We follow the order of integration in the same way as we did for double integrals (that is, from inside to outside).

Example 5.36

Evaluating a Triple Integral

Evaluate the triple integral z=0z=1y=2y=4x=−1x=5(x+yz2)dxdydz.z=0z=1y=2y=4x=−1x=5(x+yz2)dxdydz.

Solution

The order of integration is specified in the problem, so integrate with respect to xx first, then y, and then z.z.

z=0z=1y=2y=4x=−1x=5(x+yz2)dxdydz=z=0z=1y=2y=4[x22+xyz2|x=−1x=5]dydzIntegrate with respect tox.=z=0z=1y=2y=4[12+6yz2]dydzEvaluate.=z=0z=1[12y+6y22z2|y=2y=4]dzIntegrate with respect toy.=z=0z=1[24+36z2]dzEvaluate.=[24z+36z33]z=0z=1=36.Integrate with respect toz.z=0z=1y=2y=4x=−1x=5(x+yz2)dxdydz=z=0z=1y=2y=4[x22+xyz2|x=−1x=5]dydzIntegrate with respect tox.=z=0z=1y=2y=4[12+6yz2]dydzEvaluate.=z=0z=1[12y+6y22z2|y=2y=4]dzIntegrate with respect toy.=z=0z=1[24+36z2]dzEvaluate.=[24z+36z33]z=0z=1=36.Integrate with respect toz.

Example 5.37

Evaluating a Triple Integral

Evaluate the triple integral Bx2yzdVBx2yzdV where B={(x,y,z)|2x1,0y3,1z5}B={(x,y,z)|2x1,0y3,1z5} as shown in the following figure.

In x y z space, there is a box given with corners (1, 0, 5), (1, 0, 1), (1, 3, 1), (1, 3, 5), (negative 2, 0, 5), (negative 2, 0, 1), (negative 2, 3, 1), and (negative 2, 3, 5).
Figure 5.41 Evaluating a triple integral over a given rectangular box.

Solution

The order is not specified, but we can use the iterated integral in any order without changing the level of difficulty. Choose, say, to integrate y first, then x, and then z.

Bx2yzdV=15−2103[x2yz]dydxdz=15−21[x2y22z|03]dxdz=15−2192x2zdxdz=15[92x33z|−21]dz=15272zdz=272z22|15=162.Bx2yzdV=15−2103[x2yz]dydxdz=15−21[x2y22z|03]dxdz=15−2192x2zdxdz=15[92x33z|−21]dz=15272zdz=272z22|15=162.

Now try to integrate in a different order just to see that we get the same answer. Choose to integrate with respect to xx first, then z,z, and then y.y.

Bx2yzdV=0315−21[x2yz]dxdzdy=0315[x33yz|−21]dzdy=03153yzdzdy=03[3yz22|15]dy=0336ydy=36y22|03=18(90)=162.Bx2yzdV=0315−21[x2yz]dxdzdy=0315[x33yz|−21]dzdy=03153yzdzdy=03[3yz22|15]dy=0336ydy=36y22|03=18(90)=162.
Checkpoint 5.23

Evaluate the triple integral BzsinxcosydVBzsinxcosydV where B={(x,y,z)|0xπ,3π2y2π,1z3}.B={(x,y,z)|0xπ,3π2y2π,1z3}.

Triple Integrals over a General Bounded Region

We now expand the definition of the triple integral to compute a triple integral over a more general bounded region EE in 3.3. The general bounded regions we will consider are of three types. First, let DD be the bounded region that is a projection of EE onto the xyxy-plane. Suppose the region EE in 33 has the form

E={(x,y,z)|(x,y)D,u1(x,y)zu2(x,y)}.E={(x,y,z)|(x,y)D,u1(x,y)zu2(x,y)}.

For two functions z=u1(x,y)z=u1(x,y) and z=u2(x,y),z=u2(x,y), such that u1(x,y)u2(x,y)u1(x,y)u2(x,y) for all (x,y)(x,y) in DD as shown in the following figure.

In x y z space, there is a shape E with top surface z = u2(x, y) and bottom surface z = u1(x, y). The bottom projects onto the x y plane as region D.
Figure 5.42 We can describe region EE as the space between u1(x,y)u1(x,y) and u2(x,y)u2(x,y) above the projection DD of EE onto the xyxy-plane.
Theorem 5.10

Triple Integral over a General Region

The triple integral of a continuous function f(x,y,z)f(x,y,z) over a general three-dimensional region

E={(x,y,z)|(x,y)D,u1(x,y)zu2(x,y)}E={(x,y,z)|(x,y)D,u1(x,y)zu2(x,y)}

in 3,3, where DD is the projection of EE onto the xyxy-plane, is

Ef(x,y,z)dV=D[u1(x,y)u2(x,y)f(x,y,z)dz]dA.Ef(x,y,z)dV=D[u1(x,y)u2(x,y)f(x,y,z)dz]dA.

Similarly, we can consider a general bounded region DD in the xyxy-plane and two functions y=u1(x,z)y=u1(x,z) and y=u2(x,z)y=u2(x,z) such that u1(x,z)u2(x,z)u1(x,z)u2(x,z) for all (x,z)(x,z) in D.D. Then we can describe the solid region EE in 33 as

E={(x,y,z)|(x,z)D,u1(x,z)yu2(x,z)}E={(x,y,z)|(x,z)D,u1(x,z)yu2(x,z)}

where DD is the projection of EE onto the xyxy-plane and the triple integral is

Ef(x,y,z)dV=D[u1(x,z)u2(x,z)f(x,y,z)dy]dA.Ef(x,y,z)dV=D[u1(x,z)u2(x,z)f(x,y,z)dy]dA.

Finally, if DD is a general bounded region in the yzyz-plane and we have two functions x=u1(y,z)x=u1(y,z) and x=u2(y,z)x=u2(y,z) such that u1(y,z)u2(y,z)u1(y,z)u2(y,z) for all (y,z)(y,z) in D,D, then the solid region EE in 33 can be described as

E={(x,y,z)|(y,z)D,u1(y,z)xu2(y,z)}E={(x,y,z)|(y,z)D,u1(y,z)xu2(y,z)}

where DD is the projection of EE onto the yzyz-plane and the triple integral is

Ef(x,y,z)dV=D[u1(y,z)u2(y,z)f(x,y,z)dx]dA.Ef(x,y,z)dV=D[u1(y,z)u2(y,z)f(x,y,z)dx]dA.

Note that the region DD in any of the planes may be of Type I or Type II as described in Double Integrals over General Regions. If DD in the xyxy-plane is of Type I (Figure 5.43), then

E={(x,y,z)|axb,g1(x)yg2(x),u1(x,y)zu2(x,y)}.E={(x,y,z)|axb,g1(x)yg2(x),u1(x,y)zu2(x,y)}.
In x y z space, there is a complex shape E with top surface z = u2(x, y) and bottom surface z = u1(x, y). The bottom projects onto the xy plane as region D with boundaries x = a, x = b, y = g1(x), and y = g2(x).
Figure 5.43 A box EE where the projection DD in the xyxy-plane is of Type I.

Then the triple integral becomes

Ef(x,y,z)dV=abg1(x)g2(x)u1(x,y)u2(x,y)f(x,y,z)dzdydx.Ef(x,y,z)dV=abg1(x)g2(x)u1(x,y)u2(x,y)f(x,y,z)dzdydx.

If DD in the xyxy-plane is of Type II (Figure 5.44), then

E={(x,y,z)|cxd,h1(x)yh2(x),u1(x,y)zu2(x,y)}.E={(x,y,z)|cxd,h1(x)yh2(x),u1(x,y)zu2(x,y)}.
In x y z space, there is a complex shape E with top surface z = u2(x, y) and bottom surface z = u1(x, y). The bottom projects onto the xy plane as region D with boundaries y = c, y = d, x = h1(y), and x = h2(y).
Figure 5.44 A box EE where the projection DD in the xyxy-plane is of Type II.

Then the triple integral becomes

Ef(x,y,z)dV=y=cy=dx=h1(y)x=h2(y)z=u1(x,y)z=u2(x,y)f(x,y,z)dzdxdy.Ef(x,y,z)dV=y=cy=dx=h1(y)x=h2(y)z=u1(x,y)z=u2(x,y)f(x,y,z)dzdxdy.

Example 5.38

Evaluating a Triple Integral over a General Bounded Region

Evaluate the triple integral of the function f(x,y,z)=5x3yf(x,y,z)=5x3y over the solid tetrahedron bounded by the planes x=0,y=0,z=0,x=0,y=0,z=0, and x+y+z=1.x+y+z=1.

Solution

Figure 5.45 shows the solid tetrahedron EE and its projection DD on the xyxy-plane.

In x y z space, there is a solid E with boundaries being the x y, z y, and x z planes and z = 1 minus x minus y. The points are the origin, (1, 0, 0), (0, 0, 1), and (0, 1, 0). Its surface on the x y plane is shown as being a rectangle marked D with line y = 1 minus x. Additionally, there is a vertical line shown on D.
Figure 5.45 The solid EE has a projection DD on the xyxy-plane of Type I.

We can describe the solid region tetrahedron as

E={(x,y,z)|0x1,0y1x,0z1xy}.E={(x,y,z)|0x1,0y1x,0z1xy}.

Hence, the triple integral is

Ef(x,y,z)dV=x=0x=1y=0y=1xz=0z=1xy(5x3y)dzdydx.Ef(x,y,z)dV=x=0x=1y=0y=1xz=0z=1xy(5x3y)dzdydx.

To simplify the calculation, first evaluate the integral z=0z=1xy(5x3y)dz.z=0z=1xy(5x3y)dz. We have

z=0z=1xy(5x3y)dz=(5x3y)(1xy).z=0z=1xy(5x3y)dz=(5x3y)(1xy).

Now evaluate the integral y=0y=1x(5x3y)(1xy)dy,y=0y=1x(5x3y)(1xy)dy, obtaining

y=0y=1x(5x3y)(1xy)dy=12(x1)2(6x1).y=0y=1x(5x3y)(1xy)dy=12(x1)2(6x1).

Finally, evaluate

x=0x=112(x1)2(6x1)dx=112.x=0x=112(x1)2(6x1)dx=112.

Putting it all together, we have

Ef(x,y,z)dV=x=0x=1y=0y=1xz=0z=1xy(5x3y)dzdydx=112.Ef(x,y,z)dV=x=0x=1y=0y=1xz=0z=1xy(5x3y)dzdydx=112.

Just as we used the double integral D1dAD1dA to find the area of a general bounded region D,D, we can use E1dVE1dV to find the volume of a general solid bounded region E.E. The next example illustrates the method.

Example 5.39

Finding a Volume by Evaluating a Triple Integral

Find the volume of a right pyramid that has the square base in the xyxy-plane [−1,1]×[−1,1][−1,1]×[−1,1] and vertex at the point (0,0,1)(0,0,1) as shown in the following figure.

In x y z space, there is a pyramid with a square base centered at the origin. The apex of the pyramid is (0, 0, 1).
Figure 5.46 Finding the volume of a pyramid with a square base.

Solution

In this pyramid the value of zz changes from 0to1,0to1, and at each height z,z, the cross section of the pyramid for any value of zz is the square [−1+z,1z]×[−1+z,1z].[−1+z,1z]×[−1+z,1z]. Hence, the volume of the pyramid is E1dVE1dV where

E={(x,y,z)|0z1,−1+zy1z,−1+zx1z}.E={(x,y,z)|0z1,−1+zy1z,−1+zx1z}.

Thus, we have

E1dV=z=0z=1y=1+zy=1zx=1+zx=1z1dxdydz=z=0z=1y=1+zy=1z(22z)dydz=z=0z=1(22z)2dz=43.E1dV=z=0z=1y=1+zy=1zx=1+zx=1z1dxdydz=z=0z=1y=1+zy=1z(22z)dydz=z=0z=1(22z)2dz=43.

Hence, the volume of the pyramid is 4343 cubic units.

Checkpoint 5.24

Consider the solid sphere E={(x,y,z)|x2+y2+z2=9}.E={(x,y,z)|x2+y2+z2=9}. Write the triple integral Ef(x,y,z)dVEf(x,y,z)dV for an arbitrary function ff as an iterated integral. Then evaluate this triple integral with f(x,y,z)=1.f(x,y,z)=1. Notice that this gives the volume of a sphere using a triple integral.

Changing the Order of Integration

As we have already seen in double integrals over general bounded regions, changing the order of the integration is done quite often to simplify the computation. With a triple integral over a rectangular box, the order of integration does not change the level of difficulty of the calculation. However, with a triple integral over a general bounded region, choosing an appropriate order of integration can simplify the computation quite a bit. Sometimes making the change to polar coordinates can also be very helpful. We demonstrate two examples here.

Example 5.40

Changing the Order of Integration

Consider the iterated integral

x=0x=1y=0y=x2z=0z=y2f(x,y,z)dzdydx.x=0x=1y=0y=x2z=0z=y2f(x,y,z)dzdydx.

The order of integration here is first with respect to z, then y, and then x. Express this integral by changing the order of integration to be first with respect to x, then z, and then y.y. Verify that the value of the integral is the same if we let f(x,y,z)=xyz.f(x,y,z)=xyz.

Solution

The best way to do this is to sketch the region EE and its projections onto each of the three coordinate planes. Thus, let

E={(x,y,z)|0x1,0yx2,0zy2}.E={(x,y,z)|0x1,0yx2,0zy2}.

and

x=0x=1y=0y=x2z=0z=y2f(x,y,z)dzdydx=Ef(x,y,z)dV.x=0x=1y=0y=x2z=0z=y2f(x,y,z)dzdydx=Ef(x,y,z)dV.

We need to express this triple integral as

y=cy=dz=v1(y)z=v2(y)x=u1(y,z)x=u2(y,z)f(x,y,z)dxdzdy.y=cy=dz=v1(y)z=v2(y)x=u1(y,z)x=u2(y,z)f(x,y,z)dxdzdy.

Knowing the region EE we can draw the following projections (Figure 5.47):

on the xyxy-plane is D1={(x,y)|0x1,0yx2}={(x,y)|0y1,yx1},D1={(x,y)|0x1,0yx2}={(x,y)|0y1,yx1},

on the yzyz-plane is D2={(y,z)|0y1,0zy2},D2={(y,z)|0y1,0zy2}, and

on the xzxz-plane is D3={(x,z)|0x1,0zx2}.D3={(x,z)|0x1,0zx2}.

Three similar versions of the following graph are shown: In the x y plane, a region D1 is bounded by the x axis, the line x = 1, and the curve y = x squared. In the second version, region D2 on the z y plane is shown with equation z = y squared. And in the third version, region D3 on the x z plane is shown with equation z = x squared.
Figure 5.47 The three cross sections of EE on the three coordinate planes.

Now we can describe the same region EE as {(x,y,z)|0y1,0zy2,yx1},{(x,y,z)|0y1,0zy2,yx1}, and consequently, the triple integral becomes

y=cy=dz=v1(y)z=v2(y)x=u1(y,z)x=u2(y,z)f(x,y,z)dxdzdy=y=0y=1z=0z=x2x=yx=1f(x,y,z)dxdzdy.y=cy=dz=v1(y)z=v2(y)x=u1(y,z)x=u2(y,z)f(x,y,z)dxdzdy=y=0y=1z=0z=x2x=yx=1f(x,y,z)dxdzdy.

Now assume that f(x,y,z)=xyzf(x,y,z)=xyz in each of the integrals. Then we have

x=0x=1y=0y=x2z=0z=y2xyzdzdydx=x=0x=1y=0y=x2[xyz22|z=0z=y2]dydx=x=0x=1y=0y=x2(xy52)dydx=x=0x=1[xy612|y=0y=x2]dx=x=0x=1x1312dx=1168,y=0y=1z=0z=y2x=yx=1xyzdxdzdy=y=0y=1z=0z=y2[yzx22|y1]dzdy=y=0y=1z=0z=y2(yz2y2z2)dzdy=y=0y=1[yz24y2z24|z=0z=y2]dy=y=0y=1(y54y64)dy=1168.x=0x=1y=0y=x2z=0z=y2xyzdzdydx=x=0x=1y=0y=x2[xyz22|z=0z=y2]dydx=x=0x=1y=0y=x2(xy52)dydx=x=0x=1[xy612|y=0y=x2]dx=x=0x=1x1312dx=1168,y=0y=1z=0z=y2x=yx=1xyzdxdzdy=y=0y=1z=0z=y2[yzx22|y1]dzdy=y=0y=1z=0z=y2(yz2y2z2)dzdy=y=0y=1[yz24y2z24|z=0z=y2]dy=y=0y=1(y54y64)dy=1168.

The answers match.

Checkpoint 5.25

Write five different iterated integrals equal to the given integral

z=0z=4y=0y=4zx=0x=yf(x,y,z)dxdydz.z=0z=4y=0y=4zx=0x=yf(x,y,z)dxdydz.

Example 5.41

Changing Integration Order and Coordinate Systems

Evaluate the triple integral Ex2+z2dV,Ex2+z2dV, where EE is the region bounded by the paraboloid y=x2+z2y=x2+z2 (Figure 5.48) and the plane y=4.y=4.

The paraboloid y = x squared + z squared is shown opening up along the y axis to y = 4.
Figure 5.48 Integrating a triple integral over a paraboloid.

Solution

The projection of the solid region EE onto the xyxy-plane is the region bounded above by y=4y=4 and below by the parabola y=x2y=x2 as shown.

In the x y plane, the graph of y = x squared is shown with the line y = 4 intersecting the graph at (negative 2, 4) and (2, 4).
Figure 5.49 Cross section in the xyxy-plane of the paraboloid in Figure 5.48.

Thus, we have

E={(x,y,z)|2x2,x2y4,yx2zyx2}.E={(x,y,z)|2x2,x2y4,yx2zyx2}.

The triple integral becomes

Ex2+z2dV=x=−2x=2y=x2y=4z=yx2z=yx2x2+z2dzdydx.Ex2+z2dV=x=−2x=2y=x2y=4z=yx2z=yx2x2+z2dzdydx.

This expression is difficult to compute, so consider the projection of EE onto the xzxz-plane. This is a circular disc x2+z24.x2+z24. So we obtain

Ex2+z2dV=x=−2x=2y=x2y=4z=yx2z=yx2x2+z2dzdydx=x=−2x=2z=4x2z=4x2y=x2+z2y=4x2+z2dydzdx.Ex2+z2dV=x=−2x=2y=x2y=4z=yx2z=yx2x2+z2dzdydx=x=−2x=2z=4x2z=4x2y=x2+z2y=4x2+z2dydzdx.

Here the order of integration changes from being first with respect to z,z, then y,y, and then xx to being first with respect to y,y, then to z,z, and then to x.x. It will soon be clear how this change can be beneficial for computation. We have

x=−2x=2z=4x2z=4x2y=x2+z2y=4x2+z2dydzdx=x=−2x=2z=4x2z=4x2(4x2z2)x2+z2dzdx.x=−2x=2z=4x2z=4x2y=x2+z2y=4x2+z2dydzdx=x=−2x=2z=4x2z=4x2(4x2z2)x2+z2dzdx.

Now use the polar substitution x=rcosθ,z=rsinθ,x=rcosθ,z=rsinθ, and dzdx=rdrdθdzdx=rdrdθ in the xzxz-plane. This is essentially the same thing as when we used polar coordinates in the xyxy-plane, except we are replacing yy by z.z. Consequently the limits of integration change and we have, by using r2=x2+z2,r2=x2+z2,

x=−2x=2z=4x2z=4x2(4x2z2)x2+z2dzdx=θ=0θ=2πr=0r=2(4r2)rrdrdθ=02π[4r33r55|02]dθ=02π6415dθ=128π15.x=−2x=2z=4x2z=4x2(4x2z2)x2+z2dzdx=θ=0θ=2πr=0r=2(4r2)rrdrdθ=02π[4r33r55|02]dθ=02π6415dθ=128π15.

Average Value of a Function of Three Variables

Recall that we found the average value of a function of two variables by evaluating the double integral over a region on the plane and then dividing by the area of the region. Similarly, we can find the average value of a function in three variables by evaluating the triple integral over a solid region and then dividing by the volume of the solid.

Theorem 5.11

Average Value of a Function of Three Variables

If f(x,y,z)f(x,y,z) is integrable over a solid bounded region EE with positive volume V(E),V(E), then the average value of the function is

fave=1V(E)Ef(x,y,z)dV.fave=1V(E)Ef(x,y,z)dV.

Note that the volume is V(E)=E1dV.V(E)=E1dV.

Example 5.42

Finding an Average Temperature

The temperature at a point (x,y,z)(x,y,z) of a solid EE bounded by the coordinate planes and the plane x+y+z=1x+y+z=1 is T(x,y,z)=(xy+8z+20)°C.T(x,y,z)=(xy+8z+20)°C. Find the average temperature over the solid.

Solution

Use the theorem given above and the triple integral to find the numerator and the denominator. Then do the division. Notice that the plane x+y+z=1x+y+z=1 has intercepts (1,0,0),(0,1,0),(1,0,0),(0,1,0), and (0,0,1).(0,0,1). The region EE looks like

E={(x,y,z)|0x1,0y1x,0z1xy}.E={(x,y,z)|0x1,0y1x,0z1xy}.

Hence the triple integral of the temperature is

Ef(x,y,z)dV=x=0x=1y=0y=1xz=0z=1xy(xy+8z+20)dzdydx=14740.Ef(x,y,z)dV=x=0x=1y=0y=1xz=0z=1xy(xy+8z+20)dzdydx=14740.

The volume evaluation is V(E)=E1dV=x=0x=1y=0y=1xz=0z=1xy1dzdydx=16.V(E)=E1dV=x=0x=1y=0y=1xz=0z=1xy1dzdydx=16.

Hence the average value is Tave=147/401/6=6(147)40=44120Tave=147/401/6=6(147)40=44120 degrees Celsius.

Checkpoint 5.26

Find the average value of the function f(x,y,z)=xyzf(x,y,z)=xyz over the cube with sides of length 44 units in the first octant with one vertex at the origin and edges parallel to the coordinate axes.

Section 5.4 Exercises

In the following exercises, evaluate the triple integrals over the rectangular solid box B.B.

181.

B(2x+3y2+4z3)dV,B(2x+3y2+4z3)dV, where B={(x,y,z)|0x1,0y2,0z3}B={(x,y,z)|0x1,0y2,0z3}

182.

B(xy+yz+xz)dV,B(xy+yz+xz)dV, where B={(x,y,z)|1x2,0y2,1z3}B={(x,y,z)|1x2,0y2,1z3}

183.

B(xcosy+z)dV,B(xcosy+z)dV, where B={(x,y,z)|0x1,0yπ,−1z1}B={(x,y,z)|0x1,0yπ,−1z1}

184.

B(zsinx+y2)dV,B(zsinx+y2)dV, where B={(x,y,z)|0xπ,0y1,−1z2}B={(x,y,z)|0xπ,0y1,−1z2}

In the following exercises, change the order of integration by integrating first with respect to z,z, then x,x, then y.y.

185.

011223(x2+lny+z)dxdydz011223(x2+lny+z)dxdydz

186.

01−1103(zex+2y)dxdydz01−1103(zex+2y)dxdydz

187.

−121304(x2z+1y)dxdydz−121304(x2z+1y)dxdydz

188.

12−2−101x+yzdxdydz12−2−101x+yzdxdydz

189.

Let F,G,andHF,G,andH be continuous functions on [a,b],[c,d],[a,b],[c,d], and [e,f],[e,f], respectively, where a,b,c,d,e,andfa,b,c,d,e,andf are real numbers such that a<b,c<d,ande<f.a<b,c<d,ande<f. Show that

abcdefF(x)G(y)H(z)dzdydx=(abF(x)dx)(cdG(y)dy)(efH(z)dz).abcdefF(x)G(y)H(z)dzdydx=(abF(x)dx)(cdG(y)dy)(efH(z)dz).
190.

Let F,G,andHF,G,andH be differential functions on [a,b],[c,d],[a,b],[c,d], and [e,f],[e,f], respectively, where a,b,c,d,e,andfa,b,c,d,e,andf are real numbers such that a<b,c<d,ande<f.a<b,c<d,ande<f. Show that

abcdefF(x)G(y)H(z)dzdydx=[F(b)F(a)][G(d)G(c)][H(f)H(e)].abcdefF(x)G(y)H(z)dzdydx=[F(b)F(a)][G(d)G(c)][H(f)H(e)].

In the following exercises, evaluate the triple integrals over the bounded region E={(x,y,z)|axb,h1(x)yh2(x),ezf}.E={(x,y,z)|axb,h1(x)yh2(x),ezf}.

191.

E(2x+5y+7z)dV,E(2x+5y+7z)dV, where E={(x,y,z)|0x1,0yx+1,1z2}E={(x,y,z)|0x1,0yx+1,1z2}

192.

E(ylnx+z)dV,E(ylnx+z)dV, where E={(x,y,z)|1xe,0ylnx,0z1}E={(x,y,z)|1xe,0ylnx,0z1}

193.

E(sinx+siny)dV,E(sinx+siny)dV, where E={(x,y,z)|0xπ2,cosxycosx,−1z1}E={(x,y,z)|0xπ2,cosxycosx,−1z1}

194.

E(xy+yz+xz)dV,E(xy+yz+xz)dV, where E={(x,y,z)|0x1,x2yx2,0z1}E={(x,y,z)|0x1,x2yx2,0z1}

In the following exercises, evaluate the triple integrals over the indicated bounded region E.E.

195.

E(x+2yz)dV,E(x+2yz)dV, where E={(x,y,z)|0x1,0yx,0z5xy}E={(x,y,z)|0x1,0yx,0z5xy}

196.

E(x3+y3+z3)dV,E(x3+y3+z3)dV, where E={(x,y,z)|0x2,0y2x,0z4xy}E={(x,y,z)|0x2,0y2x,0z4xy}

197.

EydV,EydV, where E={(x,y,z)|1x1,1x2y1x2,0z1x2y2}E={(x,y,z)|1x1,1x2y1x2,0z1x2y2}

198.

ExdV,ExdV, where E={(x,y,z)|2x2,−41x2y4x2,0z4x2y2}E={(x,y,z)|2x2,−41x2y4x2,0z4x2y2}

In the following exercises, evaluate the triple integrals over the bounded region EE of the form E={(x,y,z)|g1(y)xg2(y),cyd,ezf}.E={(x,y,z)|g1(y)xg2(y),cyd,ezf}.

199.

Ex2dV,Ex2dV, where E={(x,y,z)|1y2xy21,−1y1,1z2}E={(x,y,z)|1y2xy21,−1y1,1z2}

200.

E(sinx+y)dV,E(sinx+y)dV, where E={(x,y,z)|y4xy4,0y2,0z4}E={(x,y,z)|y4xy4,0y2,0z4}

201.

E(xyz)dV,E(xyz)dV, where E={(x,y,z)|y6xy,0y1x,−1z1}E={(x,y,z)|y6xy,0y1x,−1z1}

202.

EzdV,EzdV, where E={(x,y,z)|22yx2+y,0y1x,2z3}E={(x,y,z)|22yx2+y,0y1x,2z3}

In the following exercises, evaluate the triple integrals over the bounded region

E={(x,y,z)|g1(y)xg2(y),cyd,u1(x,y)zu2(x,y)}.E={(x,y,z)|g1(y)xg2(y),cyd,u1(x,y)zu2(x,y)}.
203.

EzdV,EzdV, where E={(x,y,z)|yxy,0y1,0z1x4y4}E={(x,y,z)|yxy,0y1,0z1x4y4}

204.

E(xz+1)dV,E(xz+1)dV, where E={(x,y,z)|0xy,0y2,0z1x2y2}E={(x,y,z)|0xy,0y2,0z1x2y2}

205.

E(xz)dV,E(xz)dV, where E={(x,y,z)|1y2xy,0y12x,0z1x2y2}E={(x,y,z)|1y2xy,0y12x,0z1x2y2}

206.

E(x+y)dV,E(x+y)dV, where E={(x,y,z)|0x1y2,0y1x,0z1x}E={(x,y,z)|0x1y2,0y1x,0z1x}

In the following exercises, evaluate the triple integrals over the bounded region

E={(x,y,z)|(x,y)D,u1(x,y)xzu2(x,y)},E={(x,y,z)|(x,y)D,u1(x,y)xzu2(x,y)}, where DD is the projection of EE onto the xyxy-plane.

207.

D(12(x+z)dz)dA,D(12(x+z)dz)dA, where D={(x,y)|x2+y21}D={(x,y)|x2+y21}

208.

D(13x(z+1)dz)dA,D(13x(z+1)dz)dA, where D={(x,y)|x2y21,x5}D={(x,y)|x2y21,x5}

209.

D(010xy(x+2z)dz)dA,D(010xy(x+2z)dz)dA, where D={(x,y)|y0,x0,x+y10}D={(x,y)|y0,x0,x+y10}

210.

D(04x2+4y2ydz)dA,D(04x2+4y2ydz)dA, where D={(x,y)|x2+y24,y1,x0}D={(x,y)|x2+y24,y1,x0}

211.

The solid EE bounded by y2+z2=9,z=0,x=0,y2+z2=9,z=0,x=0, and x=5x=5 is shown in the following figure. Evaluate the integral EzdVEzdV by integrating first with respect to z,z, then y,and thenx.y,and thenx.

A solid arching shape that reaches its maximum along the y axis with z = 3. The shape reach zero at y = plus or minus 3, and the graph is truncated at x = 0 and 5.
212.

The solid EE bounded by y=x,y=x, x=4,x=4, y=0,y=0, and z=1z=1 is given in the following figure. Evaluate the integral ExyzdVExyzdV by integrating first with respect to x,x, then y,y, and then z.z.

A quarter section of an oval cylinder with z from negative 2 to positive 1. The solid is bounded by y = 0 and x = 4, and the top of the shape runs from (0, 0, 1) to (4, 2, 1) in a gentle arc.
213.

[T] The volume of a solid EE is given by the integral −20x00x2+y2dzdydx.−20x00x2+y2dzdydx. Use a computer algebra system (CAS) to graph EE and find its volume. Round your answer to two decimal places.

214.

[T] The volume of a solid EE is given by the integral −10x2001+x2+y2dzdydx.−10x2001+x2+y2dzdydx. Use a CAS to graph EE and find its volume V.V. Round your answer to two decimal places.

In the following exercises, use two circular permutations of the variables x,y,andzx,y,andz to write new integrals whose values equal the value of the original integral. A circular permutation of x,y,andzx,y,andz is the arrangement of the numbers in one of the following orders: y,z,andxorz,x,andy.y,z,andxorz,x,andy.

215.

011324(x2z2+1)dxdydz011324(x2z2+1)dxdydz

216.

13010x+1(2x+5y+7z)dydxdz13010x+1(2x+5y+7z)dydxdz

217.

01yy01x4y4lnxdzdxdy01yy01x4y4lnxdzdxdy

218.

−1101y6y(x+yz)dxdydz−1101y6y(x+yz)dxdydz

219.

Set up the integral that gives the volume of the solid EE bounded by y2=x2+z2y2=x2+z2 and y=a2,y=a2, where a>0.a>0.

220.

Set up the integral that gives the volume of the solid EE bounded by x=y2+z2x=y2+z2 and x=a2,x=a2, where a>0.a>0.

221.

Find the average value of the function f(x,y,z)=x+y+zf(x,y,z)=x+y+z over the parallelepiped determined by x=0,x=1,y=0,y=3,z=0,x=0,x=1,y=0,y=3,z=0, and z=5.z=5.

222.

Find the average value of the function f(x,y,z)=xyzf(x,y,z)=xyz over the solid E=[0,1]×[0,1]×[0,1]E=[0,1]×[0,1]×[0,1] situated in the first octant.

223.

Find the volume of the solid EE that lies under the plane x+y+z=9x+y+z=9 and whose projection onto the xyxy-plane is bounded by x=y1,x=0,x=y1,x=0, and x+y=7.x+y=7.

224.

Find the volume of the solid E that lies under the plane 2x+y+z=82x+y+z=8 and whose projection onto the xyxy-plane is bounded by x=sin−1y,y=0,x=sin−1y,y=0, and x=π2.x=π2.

225.

Consider the pyramid with the base in the xyxy-plane of [−2,2]×[−2,2][−2,2]×[−2,2] and the vertex at the point (0,0,8).(0,0,8).

  1. Show that the equations of the planes of the lateral faces of the pyramid are 4y+z=8,4y+z=8, 4yz=−8,4yz=−8, 4x+z=8,4x+z=8, and −4x+z=8.−4x+z=8.
  2. Find the volume of the pyramid.
226.

Consider the pyramid with the base in the xyxy-plane of [−3,3]×[−3,3][−3,3]×[−3,3] and the vertex at the point (0,0,9).(0,0,9).

  1. Show that the equations of the planes of the side faces of the pyramid are 3y+z=9,3y+z=9, 3y+z=9,3y+z=9, y=0y=0 and x=0.x=0.
  2. Find the volume of the pyramid.
227.

The solid EE bounded by the sphere of equation x2+y2+z2=r2x2+y2+z2=r2 with r>0r>0 and located in the first octant is represented in the following figure.

The eighth of a sphere of radius 2 with center at the origin for positive x, y, and z.
  1. Write the triple integral that gives the volume of EE by integrating first with respect to z,z, then with y,y, and then with x.x.
  2. Rewrite the integral in part a. as an equivalent integral in five other orders.
228.

The solid EE bounded by the equation 9x2+4y2+z2=19x2+4y2+z2=1 and located in the first octant is represented in the following figure.

In the first octant, a complex shape is shown that is roughly a solid ovoid with center the origin, height 1, width 0.5, and length 0.35.
  1. Write the triple integral that gives the volume of EE by integrating first with respect to z,z, then with y,y, and then with x.x.
  2. Rewrite the integral in part a. as an equivalent integral in five other orders.
229.

Find the volume of the prism with vertices (0,0,0),(2,0,0),(2,3,0),(0,0,0),(2,0,0),(2,3,0), (0,3,0),(0,0,1),and(2,0,1).(0,3,0),(0,0,1),and(2,0,1).

230.

Find the volume of the prism with vertices (0,0,0),(4,0,0),(4,6,0),(0,0,0),(4,0,0),(4,6,0), (0,6,0),(0,0,1),and(4,0,1).(0,6,0),(0,0,1),and(4,0,1).

231.

The solid EE bounded by z=102xyz=102xy and situated in the first octant is given in the following figure. Find the volume of the solid.

A tetrahedron bounded by the x y, y z, and x z planes and a triangle with vertices (0, 0, 10), (5, 0, 0), and (0, 10, 0).
232.

The solid EE bounded by z=1x2z=1x2 and situated in the first octant is given in the following figure. Find the volume of the solid.

A complex shape in the first octant with height 1, width 5, and length 1. The shape appears to be a slightly deformed quarter of a cylinder of radius 1 and width 5.
233.

The midpoint rule for the triple integral Bf(x,y,z)dVBf(x,y,z)dV over the rectangular solid box BB is a generalization of the midpoint rule for double integrals. The region BB is divided into subboxes of equal sizes and the integral is approximated by the triple Riemann sum i=1lj=1mk=1nf(xi,yj,zk)ΔV,i=1lj=1mk=1nf(xi,yj,zk)ΔV, where (xi,yj,zk)(xi,yj,zk) is the center of the box BijkBijk and ΔVΔV is the volume of each subbox. Apply the midpoint rule to approximate Bx2dVBx2dV over the solid B={(x,y,z)|0x1,0y1,0z1}B={(x,y,z)|0x1,0y1,0z1} by using a partition of eight cubes of equal size. Round your answer to three decimal places.

234.

[T]

  1. Apply the midpoint rule to approximate Bex2dVBex2dV over the solid B={(x,y,z)|0x1,0y1,0z1}B={(x,y,z)|0x1,0y1,0z1} by using a partition of eight cubes of equal size. Round your answer to three decimal places.
  2. Use a CAS to improve the above integral approximation in the case of a partition of n3n3 cubes of equal size, where n=3,4,…,10.n=3,4,…,10.
235.

Suppose that the temperature in degrees Celsius at a point (x,y,z)(x,y,z) of a solid EE bounded by the coordinate planes and x+y+z=5x+y+z=5 is T(x,y,z)=xz+5z+10.T(x,y,z)=xz+5z+10. Find the average temperature over the solid.

236.

Suppose that the temperature in degrees Fahrenheit at a point (x,y,z)(x,y,z) of a solid EE bounded by the coordinate planes and x+y+z=5x+y+z=5 is T(x,y,z)=x+y+xy.T(x,y,z)=x+y+xy. Find the average temperature over the solid.

237.

Show that the volume of a right square pyramid of height hh and side length aa is v=ha23v=ha23 by using triple integrals.

238.

Show that the volume of a regular right hexagonal prism of edge length aa is 3a3323a332 by using triple integrals.

239.

Show that the volume of a regular right hexagonal pyramid of edge length aa is a332a332 by using triple integrals.

240.

If the charge density at an arbitrary point (x,y,z)(x,y,z) of a solid EE is given by the function ρ(x,y,z),ρ(x,y,z), then the total charge inside the solid is defined as the triple integral Eρ(x,y,z)dV.Eρ(x,y,z)dV. Assume that the charge density of the solid EE enclosed by the paraboloids x=5y2z2x=5y2z2 and x=y2+z25x=y2+z25 is equal to the distance from an arbitrary point of EE to the origin. Set up the integral that gives the total charge inside the solid E.E.

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