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APEX Calculus

Section 14.7 Triple Integration with Cylindrical and Spherical Coordinates

Just as polar coordinates gave us a new way of describing curves in the plane, in this section we will see how cylindrical and spherical coordinates give us new ways of desribing surfaces and regions in space.

Subsection 14.7.1 Cylindrical Coordinates

In short, cylindrical coordinates can be thought of as a combination of the polar and rectangular coordinate systems. One can identify a point (x0,y0,z0), given in rectangular coordinates, with the point (r0,θ0,z0), given in cylindrical coordinates, where the z-value in both systems is the same, and the point (x0,y0) in the xy-plane is identified with the polar point P(r0,θ0); see Figure 14.7.1. So that each point in space that does not lie on the z-axis is defined uniquely, we will restrict r0 and 0θ2π.
Figure 14.7.1. Illustrating the principles behind cylindrical coordinates
We use the identity z=z along with the identities found in Key Idea 10.4.6 to convert between the rectangular coordinate (x,y,z) and the cylindrical coordinate (r,θ,z), namely:
 From rectangular to cylindrical: r=x2+y2,tan(θ)=y/x and z=z; From cylindrical to rectangular: x=rcos(θ),y=rsin(θ) and z=z.
These identities, along with conversions related to spherical coordinates, are given later in Key Idea 14.7.15.
Figure 14.7.2. Explaining the cylindrical coordinate system

Example 14.7.3. Converting between rectangular and cylindrical coordinates.

Convert the rectangular point (2,2,1) to cylindrical coordinates, and convert the cylindrical point (4,3π/4,5) to rectangular.
Solution.
Following the identities given above (and, later in Key Idea 14.7.15), we have r=22+(2)2=22. Using tan(θ)=y/x, we find θ=tan1(2/2)=π/4. As we restrict θ to being between 0 and 2π, we set θ=7π/4. Finally, z=1, giving the cylindrical point (22,7π/4,1).
In converting the cylindrical point (4,3π/4,5) to rectangular, we have x=4cos(3π/4)=22, y=4sin(3π/4)=22 and z=5, giving the rectangular point (22,22,5).
Setting each of r, θ and z equal to a constant defines a surface in space, as illustrated in the following example.

Example 14.7.4. Canonical surfaces in cylindrical coordinates.

Describe the surfaces r=1, θ=π/3 and z=2, given in cylindrical coordinates.
Solution.
The equation r=1 describes all points in space that are 1 unit away from the z-axis. This surface is a “tube” or “cylinder” of radius 1, centered on the z-axis, as graphed in Figure 11.1.17 (which describes the cylinder x2+y2=1 in space).
The equation θ=π/3 describes the plane formed by extending the line θ=π/3, as given by polar coordinates in the xy-plane, parallel to the z-axis.
The equation z=2 describes the plane of all points in space that are 2 units above the xy-plane. This plane is the same as the plane described by z=2 in rectangular coordinates.
Figure 14.7.5. Graphing the canonical surfaces in cylindrical coordinates from Example 14.7.4
All three surfaces are graphed in Figure 14.7.5. Note how their intersection uniquely defines the point P=(1,π/3,2).
Cylindrical coordinates are useful when describing certain domains in space, allowing us to evaluate triple integrals over these domains more easily than if we used rectangular coordinates.
Theorem 14.6.29 shows how to evaluate Dh(x,y,z)dV using rectangular coordinates. In that evaluation, we use dV=dzdydx (or one of the other five orders of integration). Recall how, in this order of integration, the bounds on y are “curve to curve” and the bounds on x are “point to point”: these bounds describe a region R in the xy-plane. We could describe R using polar coordinates as done in Section 14.3. In that section, we saw how we used dA=rdrdθ instead of dA=dydx.
Considering the above thoughts, we have dV=dz(rdrdθ)=rdzdrdθ. We set bounds on z as “surface to surface” as done in the previous section, and then use “curve to curve” and “point to point” bounds on r and θ, respectively. Finally, using the identities given above, we change the integrand h(x,y,z) to h(r,θ,z).
This process should sound plausible; the following theorem states it is truly a way of evaluating a triple integral.
Figure 14.7.7. Using cylindrical coordinates to find the volume of a cone

Example 14.7.8. Evaluating a triple integral with cylindrical coordinates.

Find the mass of the solid represented by the region in space bounded by z=0, z=4x2y2+3 and the cylinder x2+y2=4 (as shown in Figure 14.7.9), with density function δ(x,y,z)=x2+y2+z+1, using a triple integral in cylindrical coordinates. Distances are measured in centimeters and density is measured in grams per cubic centimeter.
Figure 14.7.9. Visualizing the solid used in Example 14.7.8
Solution.
We begin by describing this region of space with cylindrical coordinates. The plane z=0 is left unchanged; with the identity r=x2+y2, we convert the hemisphere of radius 2 to the equation z=4r2; the cylinder x2+y2=4 is converted to r2=4, or, more simply, r=2. We also convert the density function: δ(r,θ,z)=r2+z+1.
To describe this solid with the bounds of a triple integral, we bound z with 0z4r2+3; we bound r with 0r2; we bound θ with 0θ2π.
Using Definition 14.6.30 and Theorem 14.7.6, we have the mass of the solid is
M=Dδ(x,y,z)dV=02π0204r2+3(r2+z+1)rdzdrdθ=02π02((r3+4r)4r2+52r3+192r)drdθ=1318π15276.04g,
where we leave the details of the remaining double integral to the reader.

Example 14.7.10. Finding the center of mass using cylindrical coordinates.

Find the center of mass of the solid with constant density whose base can be described by the polar curve r=cos(3θ) and whose top is defined by the plane z=1x+0.1y, where distances are measured in feet, as seen in Figure 14.7.11. (The volume of this solid was found in Example 14.3.13.)
Figure 14.7.11. Visualizing the solid used in Example 14.7.10
Solution.
We convert the equation of the plane to use cylindrical coordinates: z=1rcos(θ)+0.1rsin(θ). Thus the region is space is bounded by 0z1rcos(θ)+0.1rsin(θ), 0rcos(3θ), 0θπ (recall that the rose curve r=cos(3θ) is traced out once on [0,π].
Since density is constant, we set δ=1 and finding the mass is equivalent to finding the volume of the solid. We set up the triple integral to compute this but do not evaluate it; we leave it to the reader to confirm it evaluates to the same result found in Example 14.3.13.
M=DδdV=0π0cos(3θ)01rcos(θ)+0.1rsin(θ)rdzdrdθ=π4.
From Definition 14.6.30 we set up the triple integrals to compute the moments about the three coordinate planes. The computation of each is left to the reader (using technology is recommended):
Myz=DxdV=0π0cos(3θ)01rcos(θ)+0.1rsin(θ)(rcos(θ))rdzdrdθ=3π640.147.
Mxz=DydV=0π0cos(3θ)01rcos(θ)+0.1rsin(θ)(rsin(θ))rdzdrdθ=3π6400.015.Mxy=DzdV=0π0cos(3θ)01rcos(θ)+0.1rsin(θ)(z)rdzdrdθ=1903π128000.467.
The center of mass in rectangular coordinates, found by dividing the respective moments by the mass, is approximately located at (0.188,0.019,0.595), which lies outside the bounds of the solid.

Subsection 14.7.2 Spherical Coordinates

In short, spherical coordinates can be thought of as a “double application” of the polar coordinate system. In spherical coordinates, a point P is identified with (ρ,θ,φ), where ρ is the distance from the origin to P, θ is the same angle as would be used to describe P in the cylindrical coordinate system, and φ is the angle between the xy-plane and the ray from the origin to P; see Figure 14.7.13. So that each point in space that does not lie on the z-axis is defined uniquely, we will restrict ρ0, 0θ2π and π/2φπ/2.

Convention 14.7.12.

Note that most mathematics textbooks define φ to be measured from the positive z-axis, with values in [0,π], rather than from the xy-plane.
We have chosen our convention with a number of considerations in mind:
  • The coordinates (ρ,θ,φ) form a right-handed coordinate system: one in which the orientation matches that of our usual (x,y,z) coordinates, where the “right-hand rule” applies. If φ is measured from the z-axis, the order (ρ,φ,θ) is needed to get a right-handed system.
  • Points of the form (a,α,0) are the same in both cylindrical and spherical coordinates.
  • Some integration problems become slightly easier: we will see soon that the volume element in spherical coordinates involves cos(φ), which integrates to sin(φ). In the usual convention, the volume element involves sin(φ), which integrates to cos(φ) – a source of many common sign errors.
Students of Physics will encounter yet another convention. In Physics, the variable r is preferred as the radial coordinate, and spherical coordinates are given as (r,θ,φ); however, in Physics, φ becomes the angle in the xy-plane, while θ is the angle measured from the positive z-axis.
Note that the angle in the xy-plane (θ, in our case) is known as the azimuthal angle. Our angle φ is known as the elevation angle. The angle used in other conventions that is measured from the positive z-axis (often identified with the north pole) is known as the polar angle. For further discussion, the Wikipedia article
 1 
en.wikipedia.org/wiki/Spherical_coordinate_system
is quite useful.
Figure 14.7.13. Illustrating the principles behind spherical coordinates
Figure 14.7.14. Introducing spherical coordinates
The following Key Idea gives conversions to/from our three spatial coordinate systems.

Key Idea 14.7.15. Converting Between Rectangular, Cylindrical and Spherical Coordinates.

  • Rectangular and Cylindrical.
    r2=x2+y2,tan(θ)=y/x,z=zx=rcos(θ),y=rsin(θ),z=z
  • Rectangular and Spherical.
    ρ=x2+y2+z2,tan(θ)=y/x,sin(φ)=z/x2+y2+z2x=ρcos(φ)cos(θ),y=ρcos(φ)sin(θ),z=ρsin(φ)
  • Cylindrical and Spherical.
    ρ=r2+z2,θ=θ,tan(φ)=z/rr=ρcos(φ),θ=θ,z=ρsin(φ)

Example 14.7.16. Converting between rectangular and spherical coordinates.

Convert the rectangular point (2,2,1) to spherical coordinates, and convert the spherical point (6,π/3,0) to rectangular and cylindrical coordinates.
Solution.
This rectangular point is the same as used in Example 14.7.3. Using Key Idea 14.7.15, we find ρ=22+(1)2+12=3. Using the same logic as in Example 14.7.3, we find θ=7π/4. Finally, sin(φ)=1/3, giving φ=sin1(1/3)0.34, or about 19.47. Thus the spherical coordinates are approximately (3,7π/4,0.34).
Converting the spherical point (6,π/3,0) to rectangular, we have x=6cos(0)cos(π/3)=3, y=6cos(0)sin(π/3)=33 and z=6sin(0)=0. Thus the rectangular coordinates are (3,33,0).
To convert this spherical point to cylindrical, we have r=6cos(0)=6, θ=π/3 and z=6sin(0)=0, giving the cylindrical point (6,π/3,0).

Example 14.7.17. Canonical surfaces in spherical coordinates.

Describe the surfaces ρ=1, θ=π/3 and φ=π/3, given in spherical coordinates.
Solution.
The equation ρ=1 describes all points in space that are 1 unit away from the origin: this is the sphere of radius 1, centered at the origin.
The equation θ=π/3 describes the same surface in spherical coordinates as it does in cylindrical coordinates: beginning with the line θ=π/3 in the xy-plane as given by polar coordinates, extend the line parallel to the z-axis, forming a plane.
The equation φ=π/3 describes all points P in space where the ray from the origin to P makes an angle of π/3 with the xy-plane. This describes a cone, with the positive z-axis its axis of symmetry, with point at the origin.
Figure 14.7.18. Graphing the canonical surfaces in spherical coordinates from Example 14.7.17
All three surfaces are graphed in Figure 14.7.18. Note how their intersection uniquely defines the point P=(1,π/3,π/6).
Spherical coordinates are useful when describing certain domains in space, allowing us to evaluate triple integrals over these domains more easily than if we used rectangular coordinates or cylindrical coordinates. The crux of setting up a triple integral in spherical coordinates is appropriately describing the “small amount of volume,” dV, used in the integral.
Considering Figure 14.7.19, we can make a small “spherical wedge” by varying ρ, θ and φ each a small amount, Δρ, Δθ and Δφ, respectively. This wedge is approximately a rectangular solid when the change in each coordinate is small, giving a volume of about
ΔVΔρ × ρΔφ × ρcos(φ)Δθ.
Figure 14.7.19. Approximating the volume of a standard region in space using spherical coordinates
Given a region D in space, we can approximate the volume of D with many such wedges. As the size of each of Δρ, Δθ and Δφ goes to zero, the number of wedges increases to infinity and the volume of D is more accurately approximated, giving
dV=dρ × ρdφ × ρcos(φ)dθ=ρ2cos(φ)dρdθdφ.
Again, this development of dV should sound reasonable, and the following theorem states it is the appropriate manner by which triple integrals are to be evaluated in spherical coordinates.

Example 14.7.21. Establishing the volume of a sphere.

Let D be the region in space bounded by the sphere, centered at the origin, of radius r. Use a triple integral in spherical coordinates to find the volume V of D.
Solution.
The sphere of radius r, centered at the origin, has equation ρ=r. To obtain the full sphere, the bounds on θ and φ are 0θ2π and π/2φπ/2. This leads us to:
V=DdV=π/2π/202π0r(ρ2cos(φ))dρdθdφ=π/2π/202π(13ρ3cos(φ)|0r)dθdφ=π/2π/202π(13r3cos(φ))dθdφ=π/2π/2(2π3r3cos(φ))dφ=(2π3r3sin(φ))|π/2π/2=4π3r3,
the familiar formula for the volume of a sphere. Note how the integration steps were easy, not using square roots nor integration steps such as Substitution.
Figure 14.7.22. Setting up an integral in spherical coorindates

Example 14.7.23. Finding the center of mass using spherical coordinates.

Find the center of mass of the solid with constant density enclosed above by ρ=4 and below by φ=π/3, as illustrated in Figure 14.7.24.
Figure 14.7.24. Graphing the solid, and its center of mass, from Example 14.7.23
Solution.
We will set up the four triple integrals needed to find the center of mass (i.e., to compute M, Myz, Mxz and Mxy) and leave it to the reader to evaluate each integral. Because of symmetry, we expect the x- and y- coordinates of the center of mass to be 0.
While the surfaces describing the solid are given in the statement of the problem, to describe the full solid D, we use the following bounds: 0ρ4, 0θ2π and 0φπ/3. Since density δ is constant, we assume δ=1.
The mass of the solid:
M=Ddm=DdV=0π/302π04(ρ2cos(φ))dρdθdφ=643(23)π17.958.
To compute Myz, the integrand is x; using Key Idea 14.7.15, we have x=ρcos(φ)cos(θ). This gives:
Myz=Dxdm=0π/302π04((ρcos(φ)cos(θ))ρ2cos(φ))dρdθdφ=0π/302π04(ρ3cos2(φ)cos(θ))dρdθdφ=0,
which we expected as we expect x=0.
To compute Mxz, the integrand is y; using Key Idea 14.7.15, we have y=ρcos(φ)sin(θ). This gives:
Mxz=Dydm=0π/302π04((ρcos(φ)sin(θ))ρ2cos(φ))dρdθdφ=0π/302π04(ρ3cos2(φ)sin(θ))dρdθdφ=0,
which we also expected as we expect y=0.
To compute Mxy, the integrand is z; using Key Idea 14.7.15, we have z=ρsin(φ). This gives:
Mxy=Dzdm=0π/302π04((ρsin(φ))ρ2cos(φ))dρdθdφ=0π/302π04(ρ3sin(φ)cos(φ))dρdθdφ=16π50.266.
Thus the center of mass is (0,0,Mxy/M)(0,0,2.799), as indicated in Figure 14.7.24.
Figure 14.7.25. Spherical coordinates example: volume of a spherical wedge
Figure 14.7.26. Spherical coordinates example: volume of a conical cup
This section has provided a brief introduction into two new coordinate systems useful for identifying points in space. Each can be used to define a variety of surfaces in space beyond the canonical surfaces graphed as each system was introduced.
However, the usefulness of these coordinate systems does not lie in the variety of surfaces that they can describe nor the regions in space these surfaces may enclose. Rather, cylindrical coordinates are mostly used to describe cylinders and spherical coordinates are mostly used to describe spheres. These shapes are of special interest in the sciences, especially in physics, and computations on/inside these shapes is difficult using rectangular coordinates. For instance, in the study of electricity and magnetism, one often studies the effects of an electrical current passing through a wire; that wire is essentially a cylinder, described well by cylindrical coordinates.
This chapter investigated the natural follow-on to partial derivatives: iterated integration. We learned how to use the bounds of a double integral to describe a region in the plane using both rectangular and polar coordinates, then later expanded to use the bounds of a triple integral to describe a region in space. We used double integrals to find volumes under surfaces, surface area, and the center of mass of lamina; we used triple integrals as an alternate method of finding volumes of space regions and also to find the center of mass of a region in space.
Integration does not stop here. We could continue to iterate our integrals, next investigating “quadruple integrals” whose bounds describe a region in 4-dimensional space (which are very hard to visualize). We can also look back to “regular” integration where we found the area under a curve in the plane. A natural analogue to this is finding the “area under a curve,” where the curve is in space, not in a plane. These are just two of many avenues to explore under the heading of “integration.”

Exercises 14.7.3 Exercises

Terms and Concepts

1.
Explain the difference between the roles r, in cylindrical coordinates, and ρ, in spherical coordinates, play in determining the location of a point.
2.
Why are points on the z-axis not determined uniquely when using cylindrical and spherical coordinates?
3.
What surfaces are naturally defined using cylindrical coordinates?
4.
What surfaces are naturally defined using spherical coordinates?

Problems

Exercise Group.
In the following exercises, points are given in either the rectangular, cylindrical or spherical coordinate systems. Find the coordinates of the points in the other systems.
5.
(a)
Points in rectangular coordinates: (2,2,1) and (3,1,0)
(b)
Points in cylindrical coordinates: (2,π/4,2) and (3,3π/2,4)
(c)
Points in spherical coordinates: (2,π/4,π/4) and (1,0,0)
6.
(a)
Points in rectangular coordinates: (0,1,1) and (1,0,1)
(b)
Points in cylindrical coordinates: (0,π,1) and (2,4π/3,0)
(c)
Points in spherical coordinates: (2,π/6,0) and (3,π,π/2)
Exercise Group.
In the following exercises, describe the curve, surface or region in space determined by the given bounds in cylindrical coordinates.
Exercise Group.
In the following exercises, describe the curve, surface or region in space determined by the given bounds in spherical coordinates.
9.
(a)
ρ=3, 0θ2π, 0φπ/2
(b)
2ρ3, 0θ2π, π/2φπ/2
10.
(a)
0ρ2, 0θπ, φ=π/4
(b)
ρ=2, 0θ2π, φ=π/3
(c)
This is a curve, a circle of radius 1 centered at (0,0,3), lying parallel to the xy-plane.
Exercise Group.
In the following exercises, standard regions in space, as defined by cylindrical and spherical coordinates, are shown. Set up the triple integral that integrates the given function over the graphed region.
Exercise Group.
In the following exercises, a triple integral in cylindrical coordinates is given. Describe the region in space defined by the bounds of the integral.
15.
02π0101rrdzdrdθ
16.
0π0102rrdzdrdθ
17.
0π0309r2rdzdrdθ
18.
02π0a0a2r2+brdzdrdθ
Exercise Group.
In the following exercises, a triple integral in spherical coordinates is given. Describe the region in space defined by the bounds of the integral.
19.
0π/20π/201ρ2cos(φ)dρdθdφ
20.
π/2π/20π11.1ρ2cos(φ)dρdθdφ
21.
π/4π/202π02ρ2cos(φ)dρdθdφ
22.
π/4π/302π02ρ2cos(φ)dρdθdφ
23.
π/3π/202π0csc(φ)ρ2cos(φ)dρdθdφ
24.
π/3π/202π0acsc(φ)ρ2cos(φ)dρdθdφ
Exercise Group.
In the following exercises, a solid is described along with its density function. Find the mass of the solid using cylindrical coordinates.
25.
Bounded by the cylinder x2+y2=4 and the planes z=0 and z=4 with density function δ(x,y,z)=x2+y2+1.
26.
Bounded by the cylinders x2+y2=4 and x2+y2=9, between the planes z=0 and z=10 with density function δ(x,y,z)=z.
27.
Bounded by y0, the cylinder x2+y2=1, and between the planes z=0 and z=4y with density function δ(x,y,z)=1.
28.
The upper half of the unit ball, bounded between z=0 and z=1x2y2, with density function δ(x,y,z)=1.
Exercise Group.
In the following exercises, a solid is described along with its density function. Find the center of mass of the solid using cylindrical coordinates. (Note: these are the same solids and density functions as found in Exercises 25–28.)
29.
Bounded by the cylinder x2+y2=4 and the planes z=0 and z=4 with density function δ(x,y,z)=x2+y2+1.
30.
Bounded by the cylinders x2+y2=4 and x2+y2=9, between the planes z=0 and z=10 with density function δ(x,y,z)=z.
31.
Bounded by y0, the cylinder x2+y2=1, and between the planes z=0 and z=4y with density function δ(x,y,z)=1.
32.
The upper half of the unit ball, bounded between z=0 and z=1x2y2, with density function δ(x,y,z)=1.
Exercise Group.
In the following exercises, a solid is described along with its density function. Find the mass of the solid using spherical coordinates.
33.
The upper half of the unit ball, bounded between z=0 and z=1x2y2, with density function δ(x,y,z)=1.
34.
The spherical shell bounded between x2+y2+z2=16 and x2+y2+z2=25 with density function δ(x,y,z)=x2+y2+z2.
35.
The conical region bounded below by z=x2+y2 and above by the sphere x2+y2+z2=1 with density function δ(x,y,z)=z.
36.
The cone that lies above the cone z=x2+y2 and below the plane z=1 with density function δ(x,y,z)=z.
Exercise Group.
In the following exercises, a solid is described along with its density function. Find the center of mass of the solid using spherical coordinates. (Note: these are the same solids and density functions as found in Exercises 33–36.)
37.
The upper half of the unit ball, bounded between z=0 and z=1x2y2, with density function δ(x,y,z)=1.
38.
The spherical shell bounded between x2+y2+z2=16 and x2+y2+z2=25 with density function δ(x,y,z)=x2+y2+z2.
39.
The conical region bounded above z=x2+y2 and below the sphere x2+y2+z2=1 with density function δ(x,y,z)=z.
40.
The cone bounded above z=x2+y2 and below the plane z=1 with density function δ(x,y,z)=z.
Exercise Group.
In the following exercises, a region is space is described. Set up the triple integrals that find the volume of this region using rectangular, cylindrical and spherical coordinates, then comment on which of the three appears easiest to evaluate.
41.
The region enclosed by the unit sphere, x2+y2+z2=1.
42.
The region enclosed by the cylinder x2+y2=1 and planes z=0 and z=1.
43.
The region enclosed by the cone z=x2+y2 and plane z=1.
44.
The cube enclosed by the planes x=0, x=1, y=0, y=1, z=0 and z=1. (Hint: in spherical, use order of integration dρdφdθ.)
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