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Calculus Volume 3

5.5 Triple Integrals in Cylindrical and Spherical Coordinates

Calculus Volume 35.5 Triple Integrals in Cylindrical and Spherical Coordinates
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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

  • 5.5.1. Evaluate a triple integral by changing to cylindrical coordinates.
  • 5.5.2. Evaluate a triple integral by changing to spherical coordinates.

Earlier in this chapter we showed how to convert a double integral in rectangular coordinates into a double integral in polar coordinates in order to deal more conveniently with problems involving circular symmetry. A similar situation occurs with triple integrals, but here we need to distinguish between cylindrical symmetry and spherical symmetry. In this section we convert triple integrals in rectangular coordinates into a triple integral in either cylindrical or spherical coordinates.

Also recall the chapter opener, which showed the opera house l’Hemisphèric in Valencia, Spain. It has four sections with one of the sections being a theater in a five-story-high sphere (ball) under an oval roof as long as a football field. Inside is an IMAX screen that changes the sphere into a planetarium with a sky full of 90009000 twinkling stars. Using triple integrals in spherical coordinates, we can find the volumes of different geometric shapes like these.

Review of Cylindrical Coordinates

As we have seen earlier, in two-dimensional space 2,2, a point with rectangular coordinates (x,y)(x,y) can be identified with (r,θ)(r,θ) in polar coordinates and vice versa, where x=rcosθ,x=rcosθ, y=rsinθ,y=rsinθ, r2=x2+y2r2=x2+y2 and tanθ=(yx)tanθ=(yx) are the relationships between the variables.

In three-dimensional space 3,3, a point with rectangular coordinates (x,y,z)(x,y,z) can be identified with cylindrical coordinates (r,θ,z)(r,θ,z) and vice versa. We can use these same conversion relationships, adding zz as the vertical distance to the point from the xyxy-plane as shown in the following figure.

In xyz space, a point is shown (x, y, z). There is also a depiction of it in polar coordinates as (r, theta, z).
Figure 5.50 Cylindrical coordinates are similar to polar coordinates with a vertical zz coordinate added.

To convert from rectangular to cylindrical coordinates, we use the conversion x=rcosθx=rcosθ and y=rsinθ.y=rsinθ. To convert from cylindrical to rectangular coordinates, we use r2=x2+y2r2=x2+y2 and θ=tan−1(yx).θ=tan−1(yx). The zz-coordinate remains the same in both cases.

In the two-dimensional plane with a rectangular coordinate system, when we say x=kx=k (constant) we mean an unbounded vertical line parallel to the yy-axis and when y=ly=l (constant) we mean an unbounded horizontal line parallel to the xx-axis. With the polar coordinate system, when we say r=cr=c (constant), we mean a circle of radius cc units and when θ=αθ=α (constant) we mean an infinite ray making an angle αα with the positive xx-axis.

Similarly, in three-dimensional space with rectangular coordinates (x,y,z),(x,y,z), the equations x=k,y=l,x=k,y=l, and z=m,z=m, where k,l,k,l, and mm are constants, represent unbounded planes parallel to the yzyz-plane, xzxz-plane and xyxy-plane, respectively. With cylindrical coordinates (r,θ,z),(r,θ,z), by r=c,θ=α,r=c,θ=α, and z=m,z=m, where c,α,c,α, and mm are constants, we mean an unbounded vertical cylinder with the zz-axis as its radial axis; a plane making a constant angle αα with the xyxy-plane; and an unbounded horizontal plane parallel to the xyxy-plane, respectively. This means that the circular cylinder x2+y2=c2x2+y2=c2 in rectangular coordinates can be represented simply as r=cr=c in cylindrical coordinates. (Refer to Cylindrical and Spherical Coordinates for more review.)

Integration in Cylindrical Coordinates

Triple integrals can often be more readily evaluated by using cylindrical coordinates instead of rectangular coordinates. Some common equations of surfaces in rectangular coordinates along with corresponding equations in cylindrical coordinates are listed in Table 5.1. These equations will become handy as we proceed with solving problems using triple integrals.

Circular cylinder Circular cone Sphere Paraboloid
Rectangular x2+y2=c2x2+y2=c2 z2=c2(x2+y2)z2=c2(x2+y2) x2+y2+z2=c2x2+y2+z2=c2 z=c(x2+y2)z=c(x2+y2)
Cylindrical r=cr=c z=crz=cr r2+z2=c2r2+z2=c2 z=cr2z=cr2
Table 5.1 Equations of Some Common Shapes

As before, we start with the simplest bounded region BB in 3,3, to describe in cylindrical coordinates, in the form of a cylindrical box, B={(r,θ,z)|arb,αθβ,czd}B={(r,θ,z)|arb,αθβ,czd} (Figure 5.51). Suppose we divide each interval into l,mandnl,mandn subdivisions such that Δr=bal,Δθ=βαm,Δr=bal,Δθ=βαm, and Δz=dcn.Δz=dcn. Then we can state the following definition for a triple integral in cylindrical coordinates.

In cylindrical box is shown with its projection onto the polar coordinate plane with inner radius a, outer radius b, and sides defined by theta = alpha and beta. The cylindrical box B starts at height c and goes to height d with the rest of the values the same as the projection onto the plane.
Figure 5.51 A cylindrical box BB described by cylindrical coordinates.

Definition

Consider the cylindrical box (expressed in cylindrical coordinates)

B={(r,θ,z)|arb,αθβ,czd}.B={(r,θ,z)|arb,αθβ,czd}.

If the function f(r,θ,z)f(r,θ,z) is continuous on BB and if (rijk*,θijk*,zijk*)(rijk*,θijk*,zijk*) is any sample point in the cylindrical subbox Bijk=[ri1,ri]×[θj1,θj]×[zk1,zk]Bijk=[ri1,ri]×[θj1,θj]×[zk1,zk] (Figure 5.51), then we can define the triple integral in cylindrical coordinates as the limit of a triple Riemann sum, provided the following limit exists:

liml,m,ni=1lj=1mk=1nf(rijk*,θijk*,zijk*)rijk*ΔrΔθΔz.liml,m,ni=1lj=1mk=1nf(rijk*,θijk*,zijk*)rijk*ΔrΔθΔz.

Note that if g(x,y,z)g(x,y,z) is the function in rectangular coordinates and the box BB is expressed in rectangular coordinates, then the triple integral Bg(x,y,z)dVBg(x,y,z)dV is equal to the triple integral Bg(rcosθ,rsinθ,z)rdrdθdzBg(rcosθ,rsinθ,z)rdrdθdz and we have

Bg(x,y,z)dV=Bg(rcosθ,rsinθ,z)rdrdθdz=Bf(r,θ,z)rdrdθdz.Bg(x,y,z)dV=Bg(rcosθ,rsinθ,z)rdrdθdz=Bf(r,θ,z)rdrdθdz.
5.11

As mentioned in the preceding section, all the properties of a double integral work well in triple integrals, whether in rectangular coordinates or cylindrical coordinates. They also hold for iterated integrals. To reiterate, in cylindrical coordinates, Fubini’s theorem takes the following form:

Theorem 5.12

Fubini’s Theorem in Cylindrical Coordinates

Suppose that g(x,y,z)g(x,y,z) is continuous on a rectangular box B,B, which when described in cylindrical coordinates looks like B={(r,θ,z)|arb,αθβ,czd}.B={(r,θ,z)|arb,αθβ,czd}.

Then g(x,y,z)=g(rcosθ,rsinθ,z)=f(r,θ,z)g(x,y,z)=g(rcosθ,rsinθ,z)=f(r,θ,z) and

Bg(x,y,z)dV=cdαβabf(r,θ,z)rdrdθdz.Bg(x,y,z)dV=cdαβabf(r,θ,z)rdrdθdz.

The iterated integral may be replaced equivalently by any one of the other five iterated integrals obtained by integrating with respect to the three variables in other orders.

Cylindrical coordinate systems work well for solids that are symmetric around an axis, such as cylinders and cones. Let us look at some examples before we define the triple integral in cylindrical coordinates on general cylindrical regions.

Example 5.43

Evaluating a Triple Integral over a Cylindrical Box

Evaluate the triple integral B(zrsinθ)rdrdθdzB(zrsinθ)rdrdθdz where the cylindrical box BB is B={(r,θ,z)|0r2,0θπ/2,0z4}.B={(r,θ,z)|0r2,0θπ/2,0z4}.

Solution

As stated in Fubini’s theorem, we can write the triple integral as the iterated integral

B(zrsinθ)rdrdθdz=θ=0θ=π/2r=0r=2z=0z=4(zrsinθ)rdzdrdθ.B(zrsinθ)rdrdθdz=θ=0θ=π/2r=0r=2z=0z=4(zrsinθ)rdzdrdθ.

The evaluation of the iterated integral is straightforward. Each variable in the integral is independent of the others, so we can integrate each variable separately and multiply the results together. This makes the computation much easier:

θ=0θ=π/2r=0r=2z=0z=4(zrsinθ)rdzdrdθ=(0π/2sinθdθ)(02r2dr)(04zdz)=(cosθ|0π/2)(r33|02)(z22|04)=643.θ=0θ=π/2r=0r=2z=0z=4(zrsinθ)rdzdrdθ=(0π/2sinθdθ)(02r2dr)(04zdz)=(cosθ|0π/2)(r33|02)(z22|04)=643.
Checkpoint 5.27

Evaluate the triple integral θ=0θ=πr=0r=1z=0z=4rzsinθrdzdrdθ.θ=0θ=πr=0r=1z=0z=4rzsinθrdzdrdθ.

If the cylindrical region over which we have to integrate is a general solid, we look at the projections onto the coordinate planes. Hence the triple integral of a continuous function f(r,θ,z)f(r,θ,z) over a general solid region E={(r,θ,z)|(r,θ)D,u1(r,θ)zu2(r,θ)}E={(r,θ,z)|(r,θ)D,u1(r,θ)zu2(r,θ)} in 3,3, where DD is the projection of EE onto the rθrθ-plane, is

Ef(r,θ,z)rdrdθdz=D[u1(r,θ)u2(r,θ)f(r,θ,z)dz]rdrdθ.Ef(r,θ,z)rdrdθdz=D[u1(r,θ)u2(r,θ)f(r,θ,z)dz]rdrdθ.

In particular, if D={(r,θ)|g1(θ)rg2(θ),αθβ},D={(r,θ)|g1(θ)rg2(θ),αθβ}, then we have

Ef(r,θ,z)rdrdθ=θ=αθ=βr=g1(θ)r=g2(θ)z=u1(r,θ)z=u2(r,θ)f(r,θ,z)rdzdrdθ.Ef(r,θ,z)rdrdθ=θ=αθ=βr=g1(θ)r=g2(θ)z=u1(r,θ)z=u2(r,θ)f(r,θ,z)rdzdrdθ.

Similar formulas exist for projections onto the other coordinate planes. We can use polar coordinates in those planes if necessary.

Example 5.44

Setting up a Triple Integral in Cylindrical Coordinates over a General Region

Consider the region EE inside the right circular cylinder with equation r=2sinθ,r=2sinθ, bounded below by the rθrθ-plane and bounded above by the sphere with radius 44 centered at the origin (Figure 5.52). Set up a triple integral over this region with a function f(r,θ,z)f(r,θ,z) in cylindrical coordinates.

In polar coordinate space, a sphere of radius 4 is shown with equation r squared + z squared = 16 and center being the origin. There is also a cylinder described by r = 2 sin theta inside the sphere.
Figure 5.52 Setting up a triple integral in cylindrical coordinates over a cylindrical region.

Solution

First, identify that the equation for the sphere is r2+z2=16.r2+z2=16. We can see that the limits for zz are from 00 to z=16r2.z=16r2. Then the limits for rr are from 00 to r=2sinθ.r=2sinθ. Finally, the limits for θθ are from 00 to π.π. Hence the region is

E={(r,θ,z)|0θπ,0r2sinθ,0z16r2}.E={(r,θ,z)|0θπ,0r2sinθ,0z16r2}.

Therefore, the triple integral is

Ef(r,θ,z)rdzdrdθ=θ=0θ=πr=0r=2sinθz=0z=16r2f(r,θ,z)rdzdrdθ.Ef(r,θ,z)rdzdrdθ=θ=0θ=πr=0r=2sinθz=0z=16r2f(r,θ,z)rdzdrdθ.
Checkpoint 5.28

Consider the region EE inside the right circular cylinder with equation r=2sinθ,r=2sinθ, bounded below by the rθrθ-plane and bounded above by z=4y.z=4y. Set up a triple integral with a function f(r,θ,z)f(r,θ,z) in cylindrical coordinates.

Example 5.45

Setting up a Triple Integral in Two Ways

Let EE be the region bounded below by the cone z=x2+y2z=x2+y2 and above by the paraboloid z=2x2y2.z=2x2y2. (Figure 5.53). Set up a triple integral in cylindrical coordinates to find the volume of the region, using the following orders of integration:

  1. dzdrdθdzdrdθ
  2. drdzdθ.drdzdθ.
    A paraboloid with equation z = 2 minus x squared minus y squared opening down, and within it, a cone with equation z = the square root of (x squared + y squared) pointing down.
    Figure 5.53 Setting up a triple integral in cylindrical coordinates over a conical region.

Solution

  1. The cone is of radius 1 where it meets the paraboloid. Since z=2x2y2=2r2z=2x2y2=2r2 and z=x2+y2=rz=x2+y2=r (assuming rr is nonnegative), we have 2r2=r.2r2=r. Solving, we have r2+r2=(r+2)(r1)=0.r2+r2=(r+2)(r1)=0. Since r0,r0, we have r=1.r=1. Therefore z=1.z=1. So the intersection of these two surfaces is a circle of radius 11 in the plane z=1.z=1. The cone is the lower bound for zz and the paraboloid is the upper bound. The projection of the region onto the xyxy-plane is the circle of radius 11 centered at the origin.
    Thus, we can describe the region as
    E={(r,θ,z)|0θ2π,0r1,rz2r2}.E={(r,θ,z)|0θ2π,0r1,rz2r2}.

    Hence the integral for the volume is
    V=θ=0θ=2πr=0r=1z=rz=2r2rdzdrdθ.V=θ=0θ=2πr=0r=1z=rz=2r2rdzdrdθ.
  2. We can also write the cone surface as r=zr=z and the paraboloid as r2=2z.r2=2z. The lower bound for rr is zero, but the upper bound is sometimes the cone and the other times it is the paraboloid. The plane z=1z=1 divides the region into two regions. Then the region can be described as
    E={(r,θ,z)|0θ2π,0z1,0rz}{(r,θ,z)|0θ2π,1z2,0r2z}.E={(r,θ,z)|0θ2π,0z1,0rz}{(r,θ,z)|0θ2π,1z2,0r2z}.

    Now the integral for the volume becomes
    V=θ=0θ=2πz=0z=1r=0r=zrdrdzdθ+θ=0θ=2πz=1z=2r=0r=2zrdrdzdθ.V=θ=0θ=2πz=0z=1r=0r=zrdrdzdθ+θ=0θ=2πz=1z=2r=0r=2zrdrdzdθ.
Checkpoint 5.29

Redo the previous example with the order of integration dθdzdr.dθdzdr.

Example 5.46

Finding a Volume with Triple Integrals in Two Ways

Let E be the region bounded below by the rθrθ-plane, above by the sphere x2+y2+z2=4,x2+y2+z2=4, and on the sides by the cylinder x2+y2=1x2+y2=1 (Figure 5.54). Set up a triple integral in cylindrical coordinates to find the volume of the region using the following orders of integration, and in each case find the volume and check that the answers are the same:

  1. dzdrdθdzdrdθ
  2. drdzdθ.drdzdθ.
    A hemisphere with equation x squared + y squared + z squared = 4 in the upper half plane, and within it, a cylinder with equation x squared + y squared = 1.
    Figure 5.54 Finding a cylindrical volume with a triple integral in cylindrical coordinates.

Solution

  1. Note that the equation for the sphere is
    x2+y2+z2=4orr2+z2=4x2+y2+z2=4orr2+z2=4

    and the equation for the cylinder is
    x2+y2=1orr2=1.x2+y2=1orr2=1.

    Thus, we have for the region EE
    E={(r,θ,z)|0z4r2,0r1,0θ2π}E={(r,θ,z)|0z4r2,0r1,0θ2π}

    Hence the integral for the volume is
    V(E)=θ=0θ=2πr=0r=1z=0z=4r2rdzdrdθ=θ=0θ=2πr=0r=1[rz|z=0z=4r2]drdθ=θ=0θ=2πr=0r=1(r4r2)drdθ=02π(833)dθ=2π(833)cubic units.V(E)=θ=0θ=2πr=0r=1z=0z=4r2rdzdrdθ=θ=0θ=2πr=0r=1[rz|z=0z=4r2]drdθ=θ=0θ=2πr=0r=1(r4r2)drdθ=02π(833)dθ=2π(833)cubic units.
  2. Since the sphere is x2+y2+z2=4,x2+y2+z2=4, which is r2+z2=4,r2+z2=4, and the cylinder is x2+y2=1,x2+y2=1, which is r2=1,r2=1, we have 1+z2=4,1+z2=4, that is, z2=3.z2=3. Thus we have two regions, since the sphere and the cylinder intersect at (1,3)(1,3) in the rzrz-plane
    E1={(r,θ,z)|0r4r2,3z2,0θ2π}E1={(r,θ,z)|0r4r2,3z2,0θ2π}

    and
    E2={(r,θ,z)|0r1,0z3,0θ2π}.E2={(r,θ,z)|0r1,0z3,0θ2π}.

    Hence the integral for the volume is
    V(E)=θ=0θ=2πz=3z=2r=0r=4r2rdrdzdθ+θ=0θ=2πz=0z=3r=0r=1rdrdzdθ=3π+(16333)π=2π(833)cubic units.V(E)=θ=0θ=2πz=3z=2r=0r=4r2rdrdzdθ+θ=0θ=2πz=0z=3r=0r=1rdrdzdθ=3π+(16333)π=2π(833)cubic units.
Checkpoint 5.30

Redo the previous example with the order of integration dθdzdr.dθdzdr.

Review of Spherical Coordinates

In three-dimensional space 33 in the spherical coordinate system, we specify a point PP by its distance ρρ from the origin, the polar angle θθ from the positive x-axisx-axis (same as in the cylindrical coordinate system), and the angle φφ from the positive z-axisz-axis and the line OPOP (Figure 5.55). Note that ρ0ρ0 and 0φπ.0φπ. (Refer to Cylindrical and Spherical Coordinates for a review.) Spherical coordinates are useful for triple integrals over regions that are symmetric with respect to the origin.

A depiction of the spherical coordinate system: a point (x, y, z) is shown, which is equal to (rho, theta, phi) in spherical coordinates. Rho serves as the spherical radius, theta serves as the angle from the x axis in the xy plane, and phi serves as the angle from the z axis.
Figure 5.55 The spherical coordinate system locates points with two angles and a distance from the origin.

Recall the relationships that connect rectangular coordinates with spherical coordinates.

From spherical coordinates to rectangular coordinates:

x=ρsinφcosθ,y=ρsinφsinθ,andz=ρcosφ.x=ρsinφcosθ,y=ρsinφsinθ,andz=ρcosφ.

From rectangular coordinates to spherical coordinates:

ρ2=x2+y2+z2,tanθ=yx,φ=arccos(zx2+y2+z2).ρ2=x2+y2+z2,tanθ=yx,φ=arccos(zx2+y2+z2).

Other relationships that are important to know for conversions are

r=ρsinφθ=θThese equations are used to convert fromspherical coordinates to cylindrical coordinatesz=ρcosφr=ρsinφθ=θThese equations are used to convert fromspherical coordinates to cylindrical coordinatesz=ρcosφ

and

ρ=r2+z2θ=θThese equations are used to convert fromcylindrical coordinates to sphericalcoordinates.φ=arccos(zr2+z2)ρ=r2+z2θ=θThese equations are used to convert fromcylindrical coordinates to sphericalcoordinates.φ=arccos(zr2+z2)

The following figure shows a few solid regions that are convenient to express in spherical coordinates.

This figure consists of four figures. In the first, a sphere is shown with the note Sphere rho = c (constant). In the second, a half plane is drawn from the z axis with the note Half plane theta = c (constant). In the last two figures, a half cone is drawn in each with the note Half cone phi = c (constant). In the first of these, the cone opens up and it is marked 0 < c < pi/2. In the second of these, the cone opens down and it is marked pi/2 < c < pi.
Figure 5.56 Spherical coordinates are especially convenient for working with solids bounded by these types of surfaces. (The letter cc indicates a constant.)

Integration in Spherical Coordinates

We now establish a triple integral in the spherical coordinate system, as we did before in the cylindrical coordinate system. Let the function f(ρ,θ,φ)f(ρ,θ,φ) be continuous in a bounded spherical box, B={(ρ,θ,φ)|aρb,αθβ,γφψ}.B={(ρ,θ,φ)|aρb,αθβ,γφψ}. We then divide each interval into l,mandnl,mandn subdivisions such that Δρ=bal,Δθ=βαm,Δφ=ψγn.Δρ=bal,Δθ=βαm,Δφ=ψγn.

Now we can illustrate the following theorem for triple integrals in spherical coordinates with (ρijk*,θijk*,φijk*)(ρijk*,θijk*,φijk*) being any sample point in the spherical subbox Bijk.Bijk. For the volume element of the subbox ΔVΔV in spherical coordinates, we have ΔV=(Δρ)(ρΔφ)(ρsinφΔθ),,ΔV=(Δρ)(ρΔφ)(ρsinφΔθ),, as shown in the following figure.

In the spherical coordinate space, a box is projected onto the polar coordinate plane. On the polar coordinate plane, the projection has area rho sin phi Delta theta. On the z axis, a distance Delta rho is indicated, and from these boundaries, angles are made that project through the edges of the box. There is also a blown up version of the box that shows it has sides Delta rho, rho Delta phi, and rho sin phi Delta theta, with overall volume Delta V = rho squared sin phi Delta rho Delta phi Delta theta.
Figure 5.57 The volume element of a box in spherical coordinates.

Definition

The triple integral in spherical coordinates is the limit of a triple Riemann sum,

liml,m,ni=1lj=1mk=1nf(ρijk*,θijk*,φijk*)(ρijk*)2sinφΔρΔθΔφliml,m,ni=1lj=1mk=1nf(ρijk*,θijk*,φijk*)(ρijk*)2sinφΔρΔθΔφ

provided the limit exists.

As with the other multiple integrals we have examined, all the properties work similarly for a triple integral in the spherical coordinate system, and so do the iterated integrals. Fubini’s theorem takes the following form.

Theorem 5.13

Fubini’s Theorem for Spherical Coordinates

If f(ρ,θ,φ)f(ρ,θ,φ) is continuous on a spherical solid box B=[a,b]×[α,β]×[γ,ψ],B=[a,b]×[α,β]×[γ,ψ], then

Bf(ρ,θ,φ)ρ2sinφdρdφdθ=φ=γφ=ψθ=αθ=βρ=aρ=bf(ρ,θ,φ)ρ2sinφdρdφdθ.Bf(ρ,θ,φ)ρ2sinφdρdφdθ=φ=γφ=ψθ=αθ=βρ=aρ=bf(ρ,θ,φ)ρ2sinφdρdφdθ.
5.12

This iterated integral may be replaced by other iterated integrals by integrating with respect to the three variables in other orders.

As stated before, spherical coordinate systems work well for solids that are symmetric around a point, such as spheres and cones. Let us look at some examples before we consider triple integrals in spherical coordinates on general spherical regions.

Example 5.47

Evaluating a Triple Integral in Spherical Coordinates

Evaluate the iterated triple integral θ=0θ=2πφ=0φ=π/2p=0ρ=1ρ2sinφdρdφdθ.θ=0θ=2πφ=0φ=π/2p=0ρ=1ρ2sinφdρdφdθ.

Solution

As before, in this case the variables in the iterated integral are actually independent of each other and hence we can integrate each piece and multiply:

02π0π/201ρ2sinφdρdφdθ=02πdθ0π/2sinφdφ01ρ2dρ=(2π)(1)(13)=2π3.02π0π/201ρ2sinφdρdφdθ=02πdθ0π/2sinφdφ01ρ2dρ=(2π)(1)(13)=2π3.

The concept of triple integration in spherical coordinates can be extended to integration over a general solid, using the projections onto the coordinate planes. Note that dVdV and dAdA mean the increments in volume and area, respectively. The variables VV and AA are used as the variables for integration to express the integrals.

The triple integral of a continuous function f(ρ,θ,φ)f(ρ,θ,φ) over a general solid region

E={(ρ,θ,φ)|(ρ,θ)D,u1(ρ,θ)φu2(ρ,θ)}E={(ρ,θ,φ)|(ρ,θ)D,u1(ρ,θ)φu2(ρ,θ)}

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

Ef(ρ,θ,φ)dV=D[u1(ρ,θ)u2(ρ,θ)f(ρ,θ,φ)dφ]dA.Ef(ρ,θ,φ)dV=D[u1(ρ,θ)u2(ρ,θ)f(ρ,θ,φ)dφ]dA.

In particular, if D={(ρ,θ)|g1(θ)ρg2(θ),αθβ},D={(ρ,θ)|g1(θ)ρg2(θ),αθβ}, then we have

Ef(ρ,θ,φ)dV=αβg1(θ)g2(θ)u1(ρ,θ)u2(ρ,θ)f(ρ,θ,φ)ρ2sinφdφdρdθ.Ef(ρ,θ,φ)dV=αβg1(θ)g2(θ)u1(ρ,θ)u2(ρ,θ)f(ρ,θ,φ)ρ2sinφdφdρdθ.

Similar formulas occur for projections onto the other coordinate planes.

Example 5.48

Setting up a Triple Integral in Spherical Coordinates

Set up an integral for the volume of the region bounded by the cone z=3(x2+y2)z=3(x2+y2) and the hemisphere z=4x2y2z=4x2y2 (see the figure below).

A hemisphere with equation z = the square root of (4 minus x squared minus y squared) in the upper half plane, and within it, a cone with equation z = the square root of (3 times (x squared + y squared)) that is pointing down, with vertex at the origin.
Figure 5.58 A region bounded below by a cone and above by a hemisphere.

Solution

Using the conversion formulas from rectangular coordinates to spherical coordinates, we have:

For the cone: z=3(x2+y2)z=3(x2+y2) or ρcosφ=3ρsinφρcosφ=3ρsinφ or tanφ=13tanφ=13 or φ=π6.φ=π6.

For the sphere: z=4x2y2z=4x2y2 or z2+x2+y2=4z2+x2+y2=4 or ρ2=4ρ2=4 or ρ=2.ρ=2.

Thus, the triple integral for the volume is V(E)=θ=0θ=2πϕ=0φ=π/6ρ=0ρ=2ρ2sinφdρdφdθ.V(E)=θ=0θ=2πϕ=0φ=π/6ρ=0ρ=2ρ2sinφdρdφdθ.

Checkpoint 5.31

Set up a triple integral for the volume of the solid region bounded above by the sphere ρ=2ρ=2 and bounded below by the cone φ=π/3.φ=π/3.

Example 5.49

Interchanging Order of Integration in Spherical Coordinates

Let EE be the region bounded below by the cone z=x2+y2z=x2+y2 and above by the sphere z=x2+y2+z2z=x2+y2+z2 (Figure 5.59). Set up a triple integral in spherical coordinates and find the volume of the region using the following orders of integration:

  1. dρdϕdθ,dρdϕdθ,
  2. dφdρdθ.dφdρdθ.
    A sphere with equation z = x squared + y squared + z squared, and within it, a cone with equation z = the square root of (x squared + y squared) that is pointing down, with vertex at the origin.
    Figure 5.59 A region bounded below by a cone and above by a sphere.

Solution

  1. Use the conversion formulas to write the equations of the sphere and cone in spherical coordinates.
    For the sphere:
    x2+y2+z2=zρ2=ρcosφρ=cosφ.x2+y2+z2=zρ2=ρcosφρ=cosφ.

    For the cone:
    z=x2+y2ρcosφ=ρ2sin2φcos2ϕ+ρ2sin2φsin2ϕρcosφ=ρ2sin2φ(cos2ϕ+sin2ϕ)ρcosφ=ρsinφcosφ=sinφφ=π/4.z=x2+y2ρcosφ=ρ2sin2φcos2ϕ+ρ2sin2φsin2ϕρcosφ=ρ2sin2φ(cos2ϕ+sin2ϕ)ρcosφ=ρsinφcosφ=sinφφ=π/4.

    Hence the integral for the volume of the solid region EE becomes
    V(E)=θ=0θ=2πφ=0φ=π/4ρ=0ρ=cosφρ2sinφdρdφdθ.V(E)=θ=0θ=2πφ=0φ=π/4ρ=0ρ=cosφρ2sinφdρdφdθ.
  2. Consider the φρφρ-plane. Note that the ranges for φφ and ρρ (from part a.) are
    0φπ/40ρcosφ.0φπ/40ρcosφ.

    The curve ρ=cosφρ=cosφ meets the line φ=π/4φ=π/4 at the point (π/4,2/2).(π/4,2/2). Thus, to change the order of integration, we need to use two pieces:
    0ρ2/20φπ/4and2/2ρ10φcos−1ρ.0ρ2/20φπ/4and2/2ρ10φcos−1ρ.

    Hence the integral for the volume of the solid region EE becomes
    V(E)=θ=0θ=2πρ=0ρ=2/2φ=0φ=π/4ρ2sinφdφdρdθ+θ=0θ=2πρ=2/2ρ=1φ=0φ=cos−1ρρ2sinφdφdρdθ.V(E)=θ=0θ=2πρ=0ρ=2/2φ=0φ=π/4ρ2sinφdφdρdθ+θ=0θ=2πρ=2/2ρ=1φ=0φ=cos−1ρρ2sinφdφdρdθ.

    In each case, the integration results in V(E)=π8.V(E)=π8.

Before we end this section, we present a couple of examples that can illustrate the conversion from rectangular coordinates to cylindrical coordinates and from rectangular coordinates to spherical coordinates.

Example 5.50

Converting from Rectangular Coordinates to Cylindrical Coordinates

Convert the following integral into cylindrical coordinates:

y=−1y=1x=0x=1y2z=x2+y2z=x2+y2xyzdzdxdy.y=−1y=1x=0x=1y2z=x2+y2z=x2+y2xyzdzdxdy.

Solution

The ranges of the variables are

1y10x1y2x2+y2zx2+y2.1y10x1y2x2+y2zx2+y2.

The first two inequalities describe the right half of a circle of radius 1.1. Therefore, the ranges for θθ and rr are

π2θπ2and0r1.π2θπ2and0r1.

The limits of zz are r2zr,r2zr, hence

y=−1y=1x=0x=1y2z=x2+y2z=x2+y2xyzdzdxdy=θ=π/2θ=π/2r=0r=1z=r2z=rr(rcosθ)(rsinθ)zdzdrdθ.y=−1y=1x=0x=1y2z=x2+y2z=x2+y2xyzdzdxdy=θ=π/2θ=π/2r=0r=1z=r2z=rr(rcosθ)(rsinθ)zdzdrdθ.

Example 5.51

Converting from Rectangular Coordinates to Spherical Coordinates

Convert the following integral into spherical coordinates:

y=0y=3x=0x=9y2z=x2+y2z=18x2y2(x2+y2+z2)dzdxdy.y=0y=3x=0x=9y2z=x2+y2z=18x2y2(x2+y2+z2)dzdxdy.

Solution

The ranges of the variables are

0y30x9y2x2+y2z18x2y2.0y30x9y2x2+y2z18x2y2.

The first two ranges of variables describe a quarter disk in the first quadrant of the xyxy-plane. Hence the range for θθ is 0θπ2.0θπ2.

The lower bound z=x2+y2z=x2+y2 is the upper half of a cone and the upper bound z=18x2y2z=18x2y2 is the upper half of a sphere. Therefore, we have 0ρ18,0ρ18, which is 0ρ32.0ρ32.

For the ranges of φ,φ, we need to find where the cone and the sphere intersect, so solve the equation

r2+z2=18(x2+y2)2+z2=18z2+z2=182z2=18z2=9z=3.r2+z2=18(x2+y2)2+z2=18z2+z2=182z2=18z2=9z=3.

This gives

32cosφ=3cosφ=12φ=π4.32cosφ=3cosφ=12φ=π4.

Putting this together, we obtain

y=0y=3x=0x=9y2z=x2+y2z=18x2y2(x2+y2+z2)dzdxdy=φ=0φ=π/4θ=0θ=π/2ρ=0ρ=32ρ4sinφdρdθdφ.y=0y=3x=0x=9y2z=x2+y2z=18x2y2(x2+y2+z2)dzdxdy=φ=0φ=π/4θ=0θ=π/2ρ=0ρ=32ρ4sinφdρdθdφ.
Checkpoint 5.32

Use rectangular, cylindrical, and spherical coordinates to set up triple integrals for finding the volume of the region inside the sphere x2+y2+z2=4x2+y2+z2=4 but outside the cylinder x2+y2=1.x2+y2=1.

A sphere with equation x squared + y squared + z squared = 4, and within it, a cylinder with equation x squared + y squared = 1.

Now that we are familiar with the spherical coordinate system, let’s find the volume of some known geometric figures, such as spheres and ellipsoids.

Example 5.52

Chapter Opener: Finding the Volume of l’Hemisphèric

Find the volume of the spherical planetarium in l’Hemisphèric in Valencia, Spain, which is five stories tall and has a radius of approximately 5050 ft, using the equation x2+y2+z2=r2.x2+y2+z2=r2.

A picture of l’Hemisphèric, which is a giant glass structure that is in the shape of an ellipsoid.
Figure 5.60 (credit: modification of work by Javier Yaya Tur, Wikimedia Commons)

Solution

We calculate the volume of the ball in the first octant, where x0,y0,x0,y0, and z0,z0, using spherical coordinates, and then multiply the result by 88 for symmetry. Since we consider the region DD as the first octant in the integral, the ranges of the variables are

0φπ2,0ρr,0θπ2.0φπ2,0ρr,0θπ2.

Therefore,

V=Ddxdydz=8θ=0θ=π/2ρ=0ρ=πφ=0φ=π/2ρ2sinθdφdρdθ=8φ=0φ=π/2dφρ=0ρ=rρ2dρθ=0θ=π/2sinθdθ=8(π2)(r33)(1)=43πr3.V=Ddxdydz=8θ=0θ=π/2ρ=0ρ=πφ=0φ=π/2ρ2sinθdφdρdθ=8φ=0φ=π/2dφρ=0ρ=rρ2dρθ=0θ=π/2sinθdθ=8(π2)(r33)(1)=43πr3.

This exactly matches with what we knew. So for a sphere with a radius of approximately 5050 ft, the volume is 43π(50)3523,600ft3.43π(50)3523,600ft3.

For the next example we find the volume of an ellipsoid.

Example 5.53

Finding the Volume of an Ellipsoid

Find the volume of the ellipsoid x2a2+y2b2+z2c2=1.x2a2+y2b2+z2c2=1.

Solution

We again use symmetry and evaluate the volume of the ellipsoid using spherical coordinates. As before, we use the first octant x0,y0,x0,y0, and z0z0 and then multiply the result by 8.8.

In this case the ranges of the variables are

0φπ2,0ρπ2,0ρ1,and0θπ2.0φπ2,0ρπ2,0ρ1,and0θπ2.

Also, we need to change the rectangular to spherical coordinates in this way:

x=aρcosφsinθ,y=bρsinφsinθ,andz=cρcosθ.x=aρcosφsinθ,y=bρsinφsinθ,andz=cρcosθ.

Then the volume of the ellipsoid becomes

V=Ddxdydz=8θ=0θ=π/2ρ=0ρ=1φ=0φ=π/2abcρ2sinθdφdρdθ=8abcφ=0φ=π/2dφρ=0ρ=1ρ2dρθ=0θ=π/2sinθdθ=8abc(π2)(13)(1)=43πabc.V=Ddxdydz=8θ=0θ=π/2ρ=0ρ=1φ=0φ=π/2abcρ2sinθdφdρdθ=8abcφ=0φ=π/2dφρ=0ρ=1ρ2dρθ=0θ=π/2sinθdθ=8abc(π2)(13)(1)=43πabc.

Example 5.54

Finding the Volume of the Space Inside an Ellipsoid and Outside a Sphere

Find the volume of the space inside the ellipsoid x2752+y2802+z2902=1x2752+y2802+z2902=1 and outside the sphere x2+y2+z2=502.x2+y2+z2=502.

Solution

This problem is directly related to the l’Hemisphèric structure. The volume of space inside the ellipsoid and outside the sphere might be useful to find the expense of heating or cooling that space. We can use the preceding two examples for the volume of the sphere and ellipsoid and then substract.

First we find the volume of the ellipsoid using a=75ft,a=75ft, b=80ft,b=80ft, and c=90ftc=90ft in the result from Example 5.53. Hence the volume of the ellipsoid is

Vellipsoid=43π(75)(80)(90)2,262,000ft3.Vellipsoid=43π(75)(80)(90)2,262,000ft3.

From Example 5.52, the volume of the sphere is

Vsphere523,600ft3.Vsphere523,600ft3.

Therefore, the volume of the space inside the ellipsoid x2752+y2802+z2902=1x2752+y2802+z2902=1 and outside the sphere x2+y2+z2=502x2+y2+z2=502 is approximately

VHemisferic=VellipsoidVsphere=1,738,400ft3.VHemisferic=VellipsoidVsphere=1,738,400ft3.

Student Project

Hot air balloons

Hot air ballooning is a relaxing, peaceful pastime that many people enjoy. Many balloonist gatherings take place around the world, such as the Albuquerque International Balloon Fiesta. The Albuquerque event is the largest hot air balloon festival in the world, with over 500500 balloons participating each year.

A picture of many hot air balloons.
Figure 5.61 Balloons lift off at the 20012001 Albuquerque International Balloon Fiesta. (credit: David Herrera, Flickr)

As the name implies, hot air balloons use hot air to generate lift. (Hot air is less dense than cooler air, so the balloon floats as long as the hot air stays hot.) The heat is generated by a propane burner suspended below the opening of the basket. Once the balloon takes off, the pilot controls the altitude of the balloon, either by using the burner to heat the air and ascend or by using a vent near the top of the balloon to release heated air and descend. The pilot has very little control over where the balloon goes, however—balloons are at the mercy of the winds. The uncertainty over where we will end up is one of the reasons balloonists are attracted to the sport.

In this project we use triple integrals to learn more about hot air balloons. We model the balloon in two pieces. The top of the balloon is modeled by a half sphere of radius 2828 feet. The bottom of the balloon is modeled by a frustum of a cone (think of an ice cream cone with the pointy end cut off). The radius of the large end of the frustum is 2828 feet and the radius of the small end of the frustum is 66 feet. A graph of our balloon model and a cross-sectional diagram showing the dimensions are shown in the following figure.

This figure consists of two parts, a and b. Figure a shows a representation of a hot air balloon in xyz space as a half sphere on top of a frustrum of a cone. Figure b shows the dimensions, namely, the radius of the half sphere is 28 ft, the distance from the bottom to the top of the frustrum is 44 ft, and the diameter of the circle at the top of the frustrum is 12 ft.
Figure 5.62 (a) Use a half sphere to model the top part of the balloon and a frustum of a cone to model the bottom part of the balloon. (b) A cross section of the balloon showing its dimensions.

We first want to find the volume of the balloon. If we look at the top part and the bottom part of the balloon separately, we see that they are geometric solids with known volume formulas. However, it is still worthwhile to set up and evaluate the integrals we would need to find the volume. If we calculate the volume using integration, we can use the known volume formulas to check our answers. This will help ensure that we have the integrals set up correctly for the later, more complicated stages of the project.

  1. Find the volume of the balloon in two ways.
    1. Use triple integrals to calculate the volume. Consider each part of the balloon separately. (Consider using spherical coordinates for the top part and cylindrical coordinates for the bottom part.)
    2. Verify the answer using the formulas for the volume of a sphere, V=43πr3,V=43πr3, and for the volume of a cone, V=13πr2h.V=13πr2h.
    In reality, calculating the temperature at a point inside the balloon is a tremendously complicated endeavor. In fact, an entire branch of physics (thermodynamics) is devoted to studying heat and temperature. For the purposes of this project, however, we are going to make some simplifying assumptions about how temperature varies from point to point within the balloon. Assume that just prior to liftoff, the temperature (in degrees Fahrenheit) of the air inside the balloon varies according to the function
    T0(r,θ,z)=zr10+210.T0(r,θ,z)=zr10+210.
  2. What is the average temperature of the air in the balloon just prior to liftoff? (Again, look at each part of the balloon separately, and do not forget to convert the function into spherical coordinates when looking at the top part of the balloon.)
    Now the pilot activates the burner for 1010 seconds. This action affects the temperature in a 1212-foot-wide column 2020 feet high, directly above the burner. A cross section of the balloon depicting this column in shown in the following figure.
    This figure shows the dimensions of the balloon and the hot air, namely, the radius of the half sphere is 28 ft, the distance from the bottom to the top of the frustrum is 44 ft, the diameter of the circle at the top of the frustrum is 12 ft, and the inner column of hot air has height 20 ft and diameter 12 ft.
    Figure 5.63 Activating the burner heats the air in a 2020-foot-high, 1212-foot-wide column directly above the burner.

    Assume that after the pilot activates the burner for 1010 seconds, the temperature of the air in the column described above increases according to the formula
    H(r,θ,z)=−2z48.H(r,θ,z)=−2z48.

    Then the temperature of the air in the column is given by
    T1(r,θ,z)=zr10+210+(−2z48),T1(r,θ,z)=zr10+210+(−2z48),

    while the temperature in the remainder of the balloon is still given by
    T0(r,θ,z)=zr10+210.T0(r,θ,z)=zr10+210.
  3. Find the average temperature of the air in the balloon after the pilot has activated the burner for 1010 seconds.

Section 5.5 Exercises

In the following exercises, evaluate the triple integrals Ef(x,y,z)dVEf(x,y,z)dV over the solid E.E.

241.

f(x,y,z)=z,f(x,y,z)=z, B={(x,y,z)|x2+y29,x0,y0,0z1}B={(x,y,z)|x2+y29,x0,y0,0z1}

A quarter section of a cylinder with height 1 and radius 3.
242.

f(x,y,z)=xz2,f(x,y,z)=xz2, B={(x,y,z)|x2+y216,x0,y0,−1z1}B={(x,y,z)|x2+y216,x0,y0,−1z1}

243.

f(x,y,z)=xy,f(x,y,z)=xy, B={(x,y,z)|x2+y21,x0,xy,−1z1}B={(x,y,z)|x2+y21,x0,xy,−1z1}

A wedge with radius 1, height 1, and angle pi/4.
244.

f(x,y,z)=x2+y2,f(x,y,z)=x2+y2, B={(x,y,z)|x2+y24,x0,xy,0z3}B={(x,y,z)|x2+y24,x0,xy,0z3}

245.

f(x,y,z)=ex2+y2,f(x,y,z)=ex2+y2, B={(x,y,z)|1x2+y24,y0,xy3,2z3}B={(x,y,z)|1x2+y24,y0,xy3,2z3}

246.

f(x,y,z)=x2+y2,f(x,y,z)=x2+y2, B={(x,y,z)|1x2+y29,y0,0z1}B={(x,y,z)|1x2+y29,y0,0z1}

247.
  1. Let BB be a cylindrical shell with inner radius a,a, outer radius b,b, and height c,c, where 0<a<b0<a<b and c>0.c>0. Assume that a function FF defined on BB can be expressed in cylindrical coordinates as F(x,y,z)=f(r)+h(z),F(x,y,z)=f(r)+h(z), where ff and hh are differentiable functions. If abf˜(r)dr=0abf˜(r)dr=0 and h˜(0)=0,h˜(0)=0, where f˜f˜ and h˜h˜ are antiderivatives of ff and h,h, respectively, show that
    BF(x,y,z)dV=2πc(bf˜(b)af˜(a))+π(b2a2)h˜(c).BF(x,y,z)dV=2πc(bf˜(b)af˜(a))+π(b2a2)h˜(c).
  2. Use the previous result to show that B(z+sinx2+y2)dxdydz=6π2(π2),B(z+sinx2+y2)dxdydz=6π2(π2), where BB is a cylindrical shell with inner radius π,π, outer radius 2π,2π, and height 2.2.
248.
  1. Let BB be a cylindrical shell with inner radius a,a, outer radius b,b, and height c,c, where 0<a<b0<a<b and c>0.c>0. Assume that a function FF defined on BB can be expressed in cylindrical coordinates as F(x,y,z)=f(r)g(θ)h(z),F(x,y,z)=f(r)g(θ)h(z), where f,g,andhf,g,andh are differentiable functions. If abf˜(r)dr=0,abf˜(r)dr=0, where f˜f˜ is an antiderivative of f,f, show that
    BF(x,y,z)dV=[bf˜(b)af˜(a)][g˜(2π)g˜(0)][h˜(c)h˜(0)],BF(x,y,z)dV=[bf˜(b)af˜(a)][g˜(2π)g˜(0)][h˜(c)h˜(0)],

    where g˜g˜ and h˜h˜ are antiderivatives of gg and h,h, respectively.
  2. Use the previous result to show that Bzsinx2+y2dxdydz=−12π2,Bzsinx2+y2dxdydz=−12π2, where BB is a cylindrical shell with inner radius π,π, outer radius 2π,2π, and height 2.2.

In the following exercises, the boundaries of the solid EE are given in cylindrical coordinates.

  1. Express the region EE in cylindrical coordinates.
  2. Convert the integral Ef(x,y,z)dVEf(x,y,z)dV to cylindrical coordinates.
249.

E is bounded by the right circular cylinder r=4sinθ,r=4sinθ, the rθrθ-plane, and the sphere r2+z2=16.r2+z2=16.

250.

EE is bounded by the right circular cylinder r=cosθ,r=cosθ, the rθrθ-plane, and the sphere r2+z2=9.r2+z2=9.

251.

EE is located in the first octant and is bounded by the circular paraboloid z=93r2,z=93r2, the cylinder r=3,r=3, and the plane r(cosθ+sinθ)=20z.r(cosθ+sinθ)=20z.

252.

EE is located in the first octant outside the circular paraboloid z=102r2z=102r2 and inside the cylinder r=5r=5 and is bounded also by the planes z=20z=20 and θ=π4.θ=π4.

In the following exercises, the function ff and region EE are given.

  1. Express the region EE and the function ff in cylindrical coordinates.
  2. Convert the integral Bf(x,y,z)dVBf(x,y,z)dV into cylindrical coordinates and evaluate it.
253.

f(x,y,z)=1x+3,f(x,y,z)=1x+3, E={(x,y,z)|0x2+y29,x0,y0,0zx+3}E={(x,y,z)|0x2+y29,x0,y0,0zx+3}

254.

f(x,y,z)=x2+y2,f(x,y,z)=x2+y2, E={(x,y,z)|0x2+y24,y0,0z3x}E={(x,y,z)|0x2+y24,y0,0z3x}

255.

f(x,y,z)=x,f(x,y,z)=x, E={(x,y,z)|1y2+z29,0x1y2z2}E={(x,y,z)|1y2+z29,0x1y2z2}

256.

f(x,y,z)=y,f(x,y,z)=y, E={(x,y,z)|1x2+z29,0y1x2z2}E={(x,y,z)|1x2+z29,0y1x2z2}

In the following exercises, find the volume of the solid EE whose boundaries are given in rectangular coordinates.

257.

EE is above the xyxy-plane, inside the cylinder x2+y2=1,x2+y2=1, and below the plane z=1.z=1.

258.

EE is below the plane z=1z=1 and inside the paraboloid z=x2+y2.z=x2+y2.

259.

EE is bounded by the circular cone z=x2+y2z=x2+y2 and z=1.z=1.

260.

EE is located above the xyxy-plane, below z=1,z=1, outside the one-sheeted hyperboloid x2+y2z2=1,x2+y2z2=1, and inside the cylinder x2+y2=2.x2+y2=2.

261.

EE is located inside the cylinder x2+y2=1x2+y2=1 and between the circular paraboloids z=1x2y2z=1x2y2 and z=x2+y2.z=x2+y2.

262.

EE is located inside the sphere x2+y2+z2=1,x2+y2+z2=1, above the xyxy-plane, and inside the circular cone z=x2+y2.z=x2+y2.

263.

EE is located outside the circular cone x2+y2=(z1)2x2+y2=(z1)2 and between the planes z=0z=0 and z=2.z=2.

264.

EE is located outside the circular cone z=1x2+y2,z=1x2+y2, above the xyxy-plane, below the circular paraboloid, and between the planes z=0andz=2.z=0andz=2.

265.

[T] Use a computer algebra system (CAS) to graph the solid whose volume is given by the iterated integral in cylindrical coordinates π/2π/201r2rrdzdrdθ.π/2π/201r2rrdzdrdθ. Find the volume VV of the solid. Round your answer to four decimal places.

266.

[T] Use a CAS to graph the solid whose volume is given by the iterated integral in cylindrical coordinates 0π/201r4rrdzdrdθ.0π/201r4rrdzdrdθ. Find the volume VV of the solid Round your answer to four decimal places.

267.

Convert the integral 011y21y2x2+y2x2+y2xzdzdxdy011y21y2x2+y2x2+y2xzdzdxdy into an integral in cylindrical coordinates.

268.

Convert the integral 020x01(xy+z)dzdxdy020x01(xy+z)dzdxdy into an integral in cylindrical coordinates.

In the following exercises, evaluate the triple integral Bf(x,y,z)dVBf(x,y,z)dV over the solid B.B.

269.

f(x,y,z)=1,f(x,y,z)=1, B={(x,y,z)|x2+y2+z290,z0}B={(x,y,z)|x2+y2+z290,z0}

A filled-in half-sphere with radius 3 times the square root of 10.
270.

f(x,y,z)=1x2+y2+z2,f(x,y,z)=1x2+y2+z2, B={(x,y,z)|x2+y2+z29,y0,z0}B={(x,y,z)|x2+y2+z29,y0,z0}

A quarter section of an ovoid with height 8, width 8 and length 18.
271.

f(x,y,z)=x2+y2,f(x,y,z)=x2+y2, BB is bounded above by the half-sphere x2+y2+z2=9x2+y2+z2=9 with z0z0 and below by the cone 2z2=x2+y2.2z2=x2+y2.

272.

f(x,y,z)=z,f(x,y,z)=z, BB is bounded above by the half-sphere x2+y2+z2=16x2+y2+z2=16 with z0z0 and below by the cone 2z2=x2+y2.2z2=x2+y2.

273.

Show that if F(ρ,θ,φ)=f(ρ)g(θ)h(φ)F(ρ,θ,φ)=f(ρ)g(θ)h(φ) is a continuous function on the spherical box B={(ρ,θ,φ)|aρb,αθβ,γφψ},B={(ρ,θ,φ)|aρb,αθβ,γφψ}, then

BFdV=(abρ2f(ρ)dr)(αβg(θ)dθ)(γψh(φ)sinφdφ).BFdV=(abρ2f(ρ)dr)(αβg(θ)dθ)(γψh(φ)sinφdφ).
274.
  1. A function FF is said to have spherical symmetry if it depends on the distance to the origin only, that is, it can be expressed in spherical coordinates as F(x,y,z)=f(ρ),F(x,y,z)=f(ρ), where ρ=x2+y2+z2.ρ=x2+y2+z2. Show that
    BF(x,y,z)dV=2πabρ2f(ρ)dρ,BF(x,y,z)dV=2πabρ2f(ρ)dρ,

    where BB is the region between the upper concentric hemispheres of radii aa and bb centered at the origin, with 0<a<b0<a<b and FF a spherical function defined on B.B.
  2. Use the previous result to show that B(x2+y2+z2)x2+y2+z2dV=21π,B(x2+y2+z2)x2+y2+z2dV=21π, where
    B={(x,y,z)|1x2+y2+z22,z0}.B={(x,y,z)|1x2+y2+z22,z0}.
275.
  1. Let BB be the region between the upper concentric hemispheres of radii a and b centered at the origin and situated in the first octant, where 0<a<b.0<a<b. Consider F a function defined on B whose form in spherical coordinates (ρ,θ,φ)(ρ,θ,φ) is F(x,y,z)=f(ρ)cosφ.F(x,y,z)=f(ρ)cosφ. Show that if g(a)=g(b)=0g(a)=g(b)=0 and abh(ρ)dρ=0,abh(ρ)dρ=0, then
    BF(x,y,z)dV=π24[ah(a)bh(b)],BF(x,y,z)dV=π24[ah(a)bh(b)],

    where gg is an antiderivative of ff and hh is an antiderivative of g.g.
  2. Use the previous result to show that Bzcosx2+y2+z2x2+y2+z2dV=3π22,Bzcosx2+y2+z2x2+y2+z2dV=3π22, where BB is the region between the upper concentric hemispheres of radii ππ and 2π2π centered at the origin and situated in the first octant.

In the following exercises, the function ff and region EE are given.

  1. Express the region EE and function ff in cylindrical coordinates.
  2. Convert the integral Bf(x,y,z)dVBf(x,y,z)dV into cylindrical coordinates and evaluate it.
276.

f(x,y,z)=z;f(x,y,z)=z; E={(x,y,z)|0x2+y2+z21,z0}E={(x,y,z)|0x2+y2+z21,z0}

277.

f(x,y,z)=x+y;f(x,y,z)=x+y; E={(x,y,z)|1x2+y2+z22,z0,y0}E={(x,y,z)|1x2+y2+z22,z0,y0}

278.

f(x,y,z)=2xy;f(x,y,z)=2xy; E={(x,y,z)|x2+y2z1x2y2,x0,y0}E={(x,y,z)|x2+y2z1x2y2,x0,y0}

279.

f(x,y,z)=z;f(x,y,z)=z; E={(x,y,z)|x2+y2+z22z0,x2+y2z}E={(x,y,z)|x2+y2+z22z0,x2+y2z}

In the following exercises, find the volume of the solid EE whose boundaries are given in rectangular coordinates.

280.

E={(x,y,z)|x2+y2z16x2y2,x0,y0}E={(x,y,z)|x2+y2z16x2y2,x0,y0}

281.

E={(x,y,z)|x2+y2+z22z0,x2+y2z}E={(x,y,z)|x2+y2+z22z0,x2+y2z}

282.

Use spherical coordinates to find the volume of the solid situated outside the sphere ρ=1ρ=1 and inside the sphere ρ=cosφ,ρ=cosφ, with φ[0,π2].φ[0,π2].

283.

Use spherical coordinates to find the volume of the ball ρ3ρ3 that is situated between the cones φ=π4andφ=π3.φ=π4andφ=π3.

284.

Convert the integral −4416y216y216x2y216x2y2(x2+y2+z2)dzdxdy−4416y216y216x2y216x2y2(x2+y2+z2)dzdxdy into an integral in spherical coordinates.

285.

Convert the integral 04016x216x2y216x2y2(x2+y2+z2)2dzdydx04016x216x2y216x2y2(x2+y2+z2)2dzdydx into an integral in spherical coordinates.

286.

Convert the integral −224x24x2x2+y216x2y2dzdydx−224x24x2x2+y216x2y2dzdydx into an integral in spherical coordinates and evaluate it.

287.

[T] Use a CAS to graph the solid whose volume is given by the iterated integral in spherical coordinates π/2π5π/6π/602ρ2sinφdρdφdθ.π/2π5π/6π/602ρ2sinφdρdφdθ. Find the volume VV of the solid. Round your answer to three decimal places.

288.

[T] Use a CAS to graph the solid whose volume is given by the iterated integral in spherical coordinates as 02π3π/4π/401ρ2sinφdρdφdθ.02π3π/4π/401ρ2sinφdρdφdθ. Find the volume VV of the solid. Round your answer to three decimal places.

289.

[T] Use a CAS to evaluate the integral E(x2+y2)dVE(x2+y2)dV where EE lies above the paraboloid z=x2+y2z=x2+y2 and below the plane z=3y.z=3y.

290.

[T]

  1. Evaluate the integral Eex2+y2+z2dV,Eex2+y2+z2dV, where EE is bounded by the spheres 4x2+4y2+4z2=14x2+4y2+4z2=1 and x2+y2+z2=1.x2+y2+z2=1.
  2. Use a CAS to find an approximation of the previous integral. Round your answer to two decimal places.
291.

Express the volume of the solid inside the sphere x2+y2+z2=16x2+y2+z2=16 and outside the cylinder x2+y2=4x2+y2=4 as triple integrals in cylindrical coordinates and spherical coordinates, respectively.

292.

Express the volume of the solid inside the sphere x2+y2+z2=16x2+y2+z2=16 and outside the cylinder x2+y2=4x2+y2=4 that is located in the first octant as triple integrals in cylindrical coordinates and spherical coordinates, respectively.

293.

The power emitted by an antenna has a power density per unit volume given in spherical coordinates by

p(ρ,θ,φ)=P0ρ2cos2θsin4φ,p(ρ,θ,φ)=P0ρ2cos2θsin4φ,where P0P0 is a constant with units in watts. The total power within a sphere BB of radius rr meters is defined as P=Bp(ρ,θ,φ)dV.P=Bp(ρ,θ,φ)dV. Find the total power P.P.

294.

Use the preceding exercise to find the total power within a sphere BB of radius 5 meters when the power density per unit volume is given by p(ρ,θ,φ)=30ρ2cos2θsin4φ.p(ρ,θ,φ)=30ρ2cos2θsin4φ.

295.

A charge cloud contained in a sphere BB of radius r centimeters centered at the origin has its charge density given by q(x,y,z)=kx2+y2+z2μCcm3,q(x,y,z)=kx2+y2+z2μCcm3, where k>0.k>0. The total charge contained in BB is given by Q=Bq(x,y,z)dV.Q=Bq(x,y,z)dV. Find the total charge Q.Q.

296.

Use the preceding exercise to find the total charge cloud contained in the unit sphere if the charge density is q(x,y,z)=20x2+y2+z2μCcm3.q(x,y,z)=20x2+y2+z2μCcm3.

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