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Coordinated Multivariable Calculus

Section 3.7 Triple Integrals in Cylindrical and Spherical Coordinates

We have encountered two different coordinate systems in R2 — the rectangular and polar coordinates systems — and seen how in certain situations, polar coordinates form a convenient alternative. In a similar way, there are two additional natural coordinate systems in R3. Given that we are already familiar with the Cartesian coordinate system for R3, we next investigate the cylindrical and spherical coordinate systems (each of which builds upon polar coordinates in R2). In what follows, we will see how to convert among the different coordinate systems, how to evaluate triple integrals using them, and some situations in which these other coordinate systems prove advantageous.

Preview Activity 3.7.1.

In the following questions, we investigate the two new coordinate systems that are the subject of this section: cylindrical and spherical coordinates. Our goal is to consider some examples of how to convert from rectangular coordinates to each of these systems, and vice versa. Triangles and trigonometry prove to be particularly important.
Figure 3.7.1. The cylindrical (left) and spherical (right) coordinates of a point.
  1. The cylindrical coordinates of a point in R3 are given by (r,θ,z) where r and θ are the polar coordinates of the point (x,y) and z is the same z coordinate as in Cartesian coordinates. An illustration is given at left in Figure 3.7.1.
    1. Find cylindrical coordinates for the point whose Cartesian coordinates are (1,3,3). Draw a labeled picture illustrating all of the coordinates.
    2. Find the Cartesian coordinates of the point whose cylindrical coordinates are (2,5π4,1). Draw a labeled picture illustrating all of the coordinates.
  2. The spherical coordinates of a point in R3 are ρ (rho), θ, and ϕ (phi), where ρ is the distance from the point to the origin, θ has the same interpretation it does in polar coordinates, and ϕ is the angle between the positive z axis and the vector from the origin to the point, as illustrated at right in Figure 3.7.1. You should convince yourself that any point in R3 can be represented in spherical coordinates with ρ0, 0θ<2π, and 0ϕπ.
    For the following questions, consider the point P whose Cartesian coordinates are (2,2,8).
    1. What is the distance from P to the origin? Your result is the value of ρ in the spherical coordinates of P.
    2. Determine the point that is the projection of P onto the xy-plane. Then, use this projection to find the value of θ in the polar coordinates of the projection of P that lies in the plane. Your result is also the value of θ for the spherical coordinates of the point.
    3. Based on the illustration in Figure 3.7.1, how is the angle ϕ determined by ρ and the z coordinate of P? Use a well-chosen right triangle to find the value of ϕ, which is the final component in the spherical coordinates of P. Draw a carefully labeled picture that clearly illustrates the values of ρ, θ, and ϕ in this example, along with the original rectangular coordinates of P.
    4. Based on your responses to i., ii., and iii., if we are given the Cartesian coordinates (x,y,z) of a point Q, how are the values of ρ, θ, and ϕ in the spherical coordinates of Q determined by x, y, and z?

Subsection 3.7.1 Cylindrical Coordinates

As we stated in Preview Activity 3.7.1, the cylindrical coordinates of a point are (r,θ,z), where r and θ are the polar coordinates of the point (x,y), and z is the same z coordinate as in Cartesian coordinates. The general situation is illustrated Figure 3.7.1.
Since we already know how to convert between rectangular and polar coordinates in the plane, and the z coordinate is identical in both Cartesian and cylindrical coordinates, the conversion equations between the two systems in R3 are essentially those we found for polar coordinates.

Converting between Cartesian and cylindrical coordinates.

  • Converting from Cartesian to cylindrical..
    If we are given the Cartesian coordinates (x,y,z) of a point P, then the cylindrical coordinates (r,θ,z) of P satisfy
    x=rcos(θ)     y=rsin(θ)      and      z=z.
  • Converting from cylindrical to cartesian..
    If we are given the cylindrical coordinates (r,θ,z) of a point P, then the Cartesian coordinates (x,y,z) of P satisfy
    r2=x2+y2     tan(θ)=yx      and      z=z,
    assuming x0.
Just as with rectangular coordinates, where we usually write z as a function of x and y to plot the resulting surface, in cylindrical coordinates, we often express z as a function of r and θ. In the following activity, we explore several basic equations in cylindrical coordinates and the corresponding surface each generates.

Activity 3.7.2.

In this activity, we graph some surfaces using cylindrical coordinates. To improve your intuition and test your understanding, you should first think about what each graph should look like before you plot it using appropriate technology.
  1. What familiar surface is described by the points in cylindrical coordinates with r=2, 0θ2π, and 0z2? How does this example suggest that we call these coordinates cylindrical coordinates? How does your answer change if we restrict θ to 0θπ?
  2. What familiar surface is described by the points in cylindrical coordinates with θ=2, 0r2, and 0z2?
  3. What familiar surface is described by the points in cylindrical coordinates with z=2, 0θ2π, and 0r2?
  4. Plot the graph of the cylindrical equation z=r, where 0θ2π and 0r2. What familiar surface results?
  5. Plot the graph of the cylindrical equation z=θ for 0θ4π and 0r2. What does this surface look like?
Answer.
  1. The points in cylindrical coordinates with r=2, 0θ2π, and 0z2 give a cylinder with radius and height 2 (just the curved face, neither of the disks). We call these coordinates cylindrical coordinates because it’s easy to make cylinders with them. If we restrict θ to 0θπ, then we get half of the cylinder.
  2. The points in cylindrical coordinates with θ=2, 0r2, and 0z2 form a 2×2 form a rectangle.
  3. The points in cylindrical coordinates with z=2, 0θ2π, and 0r2 give a disk with radius 2.
  4. The cylindrical equation z=r, where 0θ2π and 0r2 gives a cone (just the curved face, with no disk) situated with vertex at the origin.
  5. The graph of the cylindrical equation z=θ for 0θ4π and 0r2 forms a spiral.
As the name and Activity 3.7.2 suggests, cylindrical coordinates are useful for describing surfaces that are cylindrical in nature.

Subsection 3.7.2 Triple Integrals in Cylindrical Coordinates

To evaluate a triple integral Sf(x,y,z)dV as an iterated integral in Cartesian coordinates, we use the fact that the volume element dV is equal to dzdydx (which corresponds to the volume of a small box). To evaluate a triple integral in cylindrical coordinates, we similarly must understand the volume element dV in cylindrical coordinates.

Activity 3.7.3.

A picture of a cylindrical box, B={(r,θ,z):r1rr2,θ1θθ2,z1zz2}, is shown in Figure 3.7.2. Let Δr=r2r1, Δθ=θ2θ1, and Δz=z2z1. We want to determine the volume ΔV of B in terms of Δr, Δθ, Δz, r, θ, and z.
Figure 3.7.2. A cylindrical box.
  1. Appropriately label Δr, Δθ, and Δz in Figure 3.7.2.
  2. Let ΔA be the area of the projection of the box, B, onto the xy-plane, which is shaded blue in Figure 3.7.2. Recall that we previously determined the area ΔA in polar coordinates in terms of r, Δr, and Δθ. In light of the fact that we know ΔA and that z is the standard z coordinate from Cartesian coordinates, what is the volume ΔV in cylindrical coordinates?
Answer.
  1. The base of B is the same as its projection onto the xy-plane, which has an area ΔA=rΔrΔθ. The height of B is Δz. Then we can mutltiply base and height to get that the volume of B is ΔV=rΔrΔθΔz.
Activity 3.7.3 demonstrates that the volume element dV in cylindrical coordinates is given by dV=rdzdrdθ, and hence the following rule holds in general.

Triple integrals in cylindrical coordinates.

Given a continuous function f=f(x,y,z) over a region S in R3,
Sf(x,y,z)dV=Sf(rcos(θ),rsin(θ),z)rdzdrdθ.
The latter expression is an iterated integral in cylindrical coordinates.
Of course, to complete the task of writing an iterated integral in cylindrical coordinates, we need to determine the limits on the three integrals: θ, r, and z. In the following activity, we explore how to do this in several situations where cylindrical coordinates are natural and advantageous.

Activity 3.7.4.

In this activity we work with triple integrals in cylindrical coordinates.
  1. Let S be the solid bounded above by the graph of z=x2+y2 and below by z=0 on the unit disk in the xy-plane.
    1. The projection of the solid S onto the xy-plane is a disk. Describe this disk using polar coordinates.
    2. Now describe the surfaces bounding the solid S using cylindrical coordinates.
    3. Determine an iterated triple integral expression in cylindrical coordinates that gives the volume of S. You do not need to evaluate this integral.
  2. Suppose the density of the cone defined by r=1z, with z0, is given by δ(r,θ,z)=z. A picture of the cone is shown at left in Figure 3.7.3, and the projection of the cone onto the xy-plane in given at right in Figure 3.7.3. Set up an iterated integral in cylindrical coordinates that gives the mass of the cone. You do not need to evaluate this integral.
    Figure 3.7.3. The cylindrical cone r=1z and its projection onto the xy-plane.
  3. Determine an iterated integral expression in cylindrical coordinates whose value is the volume of the solid bounded below by the cone z=x2+y2 and above by the cone z=4x2+y2. A picture is shown in Figure 3.7.4. You do not need to evaluate this integral.
    Figure 3.7.4. A solid bounded by the cones z=x2+y2 and z=4x2+y2.
Answer.
    1. Since S is bounded below by the unit disk in the xy-plane, the projection of S onto the xy-plane is just the unit disk. We can describe the unit disk in polar coordinates with 0r1 and 0θ2π.
    2. We can think of the solid S as being bounded by three surfaces: above by the parabolic surface z=x2+y2, below by the unit disk in the xy-plane, and on the sides by a cylinder.
      • Parabolic surface: z=r2 with 0r1 and 0θ2π.
      • Unit disk in the xy-plane: z=0 with 0r1 and 0θ2π.
      • Cylinder: r=1 with 0z1 and 0θ2π.
    3. To find the volume of a solid, we integrate 1 over the solid:
      S1dV=02π010r1rdzdrdθ.
  1. To find the mass of the cone, we integrate the density over the solid:
    Sδ(r,θ,z)dV=02π0101zzrdrdzdθ.
  2. To find the volume of a solid, we integrate 1 over the solid:
    S1dV=02π02r4r1rdzdrdθ.

Subsection 3.7.3 Spherical Coordinates

As we saw in Preview Activity 3.7.1, the spherical coordinates of a point in 3-space have the form (ρ,θ,ϕ), where ρ is the distance from the point to the origin, θ has the same meaning as in polar coordinates, and ϕ is the angle between the positive z axis and the vector from the origin to the point. The overall situation is illustrated at right in Figure 3.7.1.
Figure 3.7.5. Converting from spherical to Cartesian coordinates.
The example in Preview Activity 3.7.1 and Figure 3.7.5 suggest how to convert between Cartesian and spherical coordinates.

Coverting between Cartesian and spherical coordinates.

  • Converting from Cartesian to spherical..
    If we are given the Cartesian coordinates (x,y,z) of a point P, then the spherical coordinates (ρ,θ,ϕ) of P satisfy
    ρ=x2+y2+z2     tan(θ)=yx      and      cos(ϕ)=zρ,
    where in the latter two equations, we require x0 and ρ0.
  • Converting from spherical to Cartesian..
    If we are given the spherical coordinates (ρ,θ,ϕ) of a point P, then the Cartesian coordinates (x,y,z) of P satisfy
    x=ρsin(ϕ)cos(θ)     y=ρsin(ϕ)sin(θ)      and      z=ρcos(ϕ).
When it comes to thinking about particular surfaces in spherical coordinates, similar to our work with cylindrical and Cartesian coordinates, we usually write ρ as a function of θ and ϕ; this is a natural analog to polar coordinates, where we often think of our distance from the origin in the plane as being a function of θ. In spherical coordinates, we likewise often view ρ as a function of θ and ϕ, thus viewing distance from the origin as a function of two key angles.
In the following activity, we explore several basic equations in spherical coordinates and the surfaces they generate.

Activity 3.7.5.

In this activity, we graph some surfaces using spherical coordinates. To improve your intuition and test your understanding, you should first think about what each graph should look like before you plot it using appropriate technology.
  1. What familiar surface is described by the points in spherical coordinates with ρ=1, 0θ2π, and 0ϕπ? How does this particular example demonstrate the reason for the name of this coordinate system? What if we restrict ϕ to 0ϕπ2?
  2. What familiar surface is described by the points in spherical coordinates with ϕ=π3, 0ρ1, and 0θ2π?
  3. What familiar shape is described by the points in spherical coordinates with θ=π6, 0ρ1, and 0ϕπ?
  4. Plot the graph of ρ=θ, for 0ϕπ and 0θ2π. How does the resulting surface appear?
Answer.
  1. The points in spherical coordinates with ρ=1, 0θ2π, and 0ϕπ give (the surface of) a sphere with radius 1. The name of this coordinate system is "sphereical" because it’s particularly nice for describing spheres. If we restrict ϕ to 0ϕπ2 we get the upper hemisphere.
  2. The points in spherical coordinates with ϕ=π3, 0ρ1, and 0θ2π give a cone (the curved face, not the disk).
  3. The points in spherical coordinates with θ=π6, 0ρ1, and 0ϕπ give a semicircle.
  4. The graph of ρ=θ, for 0ϕπ and 0θ2π looks like a seashell.
As the name and Activity 3.7.5 indicate, spherical coordinates are particularly useful for describing surfaces that are spherical in nature; they are also convenient for working with certain conical surfaces.

Subsection 3.7.4 Triple Integrals in Spherical Coordinates

As with rectangular and cylindrical coordinates, a triple integral Sf(x,y,z)dV in spherical coordinates can be evaluated as an iterated integral once we understand the volume element dV.

Activity 3.7.6.

To find the volume element dV in spherical coordinates, we need to understand how to determine the volume of a spherical box of the form ρ1ρρ2 (with Δρ=ρ2ρ1), ϕ1ϕϕ2 (with Δϕ=ϕ2ϕ1), and θ1θθ2 (with Δθ=θ2θ1). An illustration of such a box is given at left in Figure 3.7.6. This spherical box is a bit more complicated than the cylindrical box we encountered earlier. In this situation, it is easier to approximate the volume ΔV than to compute it directly. Here we can approximate the volume ΔV of this spherical box with the volume of a Cartesian box whose sides have the lengths of the sides of this spherical box. In other words,
ΔV|PS| |PR| |PQ|,
where |PR| denotes the length of the circular arc from P to R.
Figure 3.7.6. Left: A spherical box. Right: A spherical volume element.
  1. What is the length |PS| in terms of ρ?
  2. What is the length of the arc PR? (Hint: The arc PR is an arc of a circle of radius ρ2, and arc length along a circle is the product of the angle measure (in radians) and the circle’s radius.)
  3. What is the length of the arc PQ? (Hint: The arc PQ lies on a horizontal circle as illustrated at right in Figure 3.7.6. What is the radius of this circle?)
  4. Use your work in (a), (b), and (c) to determine an approximation for ΔV in spherical coordinates.
Answer.
  1. The length of PS is the change in ρ, so |PS|=Δρ.
  2. We can find the length of an arc by multiplying the angle by the distance between the arc and the origin, so the length of the arc PR is |PR|=ρ2Δϕ.
  3. What is the radius of the circle in Figure 3.7.6 is ρ2sin(ϕ2), so we can do another arc length calculation to get |PQ|=ρ2sin(ϕ2)Δθ.
  4. We can then approximate
    ΔV|PS| |PR| |PQ|=(Δρ)(ρ2Δϕ)(ρ2sin(ϕ2)Δθ)=ρ22sin(ϕ2)ΔρΔθΔϕ.
Letting Δρ, Δθ, and Δϕ go to 0, it follows from the final result in Activity 3.7.6 that dV=ρ2sin(ϕ)dρdθdϕ in spherical coordinates, and thus allows us to state the following general rule.

Triple integrals in spherical coordinates.

Given a continuous function f=f(x,y,z) over a region S in R3, the triple integral Sf(x,y,z)dV is converted to the integral
Sf(ρsin(ϕ)cos(θ),ρsin(ϕ)sin(θ),ρcos(ϕ))ρ2sin(ϕ)dρdθdϕ
in spherical coordinates.
The latter expression is an iterated integral in spherical coordinates.
Finally, in order to actually evaluate an iterated integral in spherical coordinates, we must of course determine the limits of integration in ϕ, θ, and ρ. The process is similar to our earlier work in the other two coordinate systems.

Activity 3.7.7.

We can use spherical coordinates to help us more easily understand some natural geometric objects.
  1. Recall that the sphere of radius a has spherical equation ρ=a. Set up and evaluate an iterated integral in spherical coordinates to determine the volume of a sphere of radius a.
  2. Set up, but do not evaluate, an iterated integral expression in spherical coordinates whose value is the mass of the solid obtained by removing the cone ϕ=π4 from the sphere ρ=2 if the density δ at the point (x,y,z) is δ(x,y,z)=x2+y2+z2. An illustration of the solid is shown in Figure 3.7.7.
    Figure 3.7.7. The solid cut from the sphere ρ=2 by the cone ϕ=π4.
Answer.
  1. Recall that the sphere of radius a has spherical equation ρ=a. An iterated integral in spherical coordinates to determine the volume of a sphere of radius a is:
    S1ΔV=0π02π0a1ρ2sin(ϕ)dρdθdϕ.
  2. Since we want to work in spherical coordinates, we should change the density function to be in spherical coordinates:
    δ(ρsin(ϕ)cos(θ),ρsin(ϕ)sin(θ),ρcos(ϕ))=ρ2sin2(ϕ)cos2(θ)+ρ2sin2(ϕ)sin2(θ)+ρ2cos2(ϕ)=ρ2(sin2(ϕ)(cos2(θ)+sin2(θ))+cos2(ϕ))=ρsin2(ϕ)(1)+cos2(ϕ)=ρ1=ρ.
    We can now find the mass by integrating the density over the solid:
    Sδ(ρsin(ϕ)cos(θ),ρsin(ϕ)sin(θ),ρcos(ϕ))ΔV=π/4π02π02ρρ2sin(ϕ)dρdθdϕ=π/4π02π02ρ3sin(ϕ)dρdθdϕ.

Subsection 3.7.5 Summary

  • The cylindrical coordinates of a point P are (r,θ,z) where r is the distance from the origin to the projection of P onto the xy-plane, θ is the angle that the projection of P onto the xy-plane makes with the positive x-axis, and z is the vertical distance from P to the projection of P onto the xy-plane. When P has rectangular coordinates (x,y,z), it follows that its cylindrical coordinates are given by
    r2=x2+y2,     tan(θ)=yx,     z=z.
    When P has given cylindrical coordinates (r,θ,z), its rectangular coordinates are
    x=rcos(θ),     y=rsin(θ),     z=z.
  • The volume element dV in cylindrical coordinates is dV=rdzdrdθ. Hence, a triple integral Sf(x,y,z)dA can be evaluated as the iterated integral
    Sf(rcos(θ),rsin(θ),z)rdzdrdθ.
  • The spherical coordinates of a point P in 3-space are ρ (rho), θ, and ϕ (phi), where ρ is the distance from P to the origin, θ is the angle that the projection of P onto the xy-plane makes with the positive x-axis, and ϕ is the angle between the positive z axis and the vector from the origin to P. When P has Cartesian coordinates (x,y,z), the spherical coordinates are given by
    ρ2=x2+y2+z2,     tan(θ)=yx,     cos(ϕ)=zρ.
    Given the point P in spherical coordinates (ρ,ϕ,θ), its rectangular coordinates are
    x=ρsin(ϕ)cos(θ),     y=ρsin(ϕ)sin(θ),     z=ρcos(ϕ).
  • The volume element dV in spherical coordinates is dV=ρ2sin(ϕ)dρdθdϕ. Thus, a triple integral Sf(x,y,z)dA can be evaluated as the iterated integral
    Sf(ρsin(ϕ)cos(θ),ρsin(ϕ)sin(θ),ρcos(ϕ))ρ2sin(ϕ)dρdθdϕ.

Exercises 3.7.6 Exercises

1.

What are the rectangular coordinates of the point whose cylindrical coordinates are
(r=10, θ=3π7, z=7) ?
x=
y=
z=

2.

What are the rectangular coordinates of the point whose spherical coordinates are
(1,12π,13π) ?
x=
y=
z=

3.

What are the cylindrical coordinates of the point whose spherical coordinates are
(1, 3, 4π6) ?
r =
θ =
z=

4.

Find an equation for the paraboloid z=x2+y2 in spherical coordinates. (Enter rho, phi and theta for ρ, ϕ and θ, respectively.)
equation:

5.

Match the given equation with the verbal description of the surface:
  1. Half plane
  2. Cone
  3. Sphere
  4. Plane
  5. Elliptic or Circular Paraboloid
  6. Circular Cylinder
  1. ρcos(ϕ)=4
  2. ϕ=π3
  3. r=2cos(θ)
  4. r2+z2=16
  5. θ=π3
  6. z=r2
  7. ρ=4
  8. ρ=2cos(ϕ)
  9. r=4

6.

Match the integrals with the type of coordinates which make them the easiest to do. Put the letter of the coordinate system to the left of the number of the integral.
  1. E dV where E is: x2+y2+z24,  x0,  y0,  z0
  2. E z dV where E is: 1x2,  3y4,  5z6
  3. E z2 dV where E is: 2z2,  1x2+y22
  4. 010y2 1x dx dy
  5. D 1x2+y2 dA where D is: x2+y24
  1. cartesian coordinates
  2. cylindrical coordinates
  3. polar coordinates
  4. spherical coordinates

7.

Evaluate the integral.
0525x225x225x2z225x2z21(x2+y2+z2)1/2dydzdx =

8.

Use cylindrical coordinates to evaluate the triple integral Ex2+y2dV, where E is the solid bounded by the circular paraboloid z=19(x2+y2) and the xy -plane.

9.

Use spherical coordinates to evaluate the triple integral Ex2+y2+z2dV, where E is the ball: x2+y2+z264.

10.

Find the volume of the solid enclosed by the paraboloids z=16(x2+y2) and z=816(x2+y2).

11.

FInd the volume of the ellipsoid x2+y2+4z2=16.

12.

The density, δ, of the cylinder x2+y21, 0z4 varies with the distance, r, from the z-axis:
δ=3+r g/cm3.
Find the mass of the cylinder, assuming x,y,z are in cm.
mass =
(Include units
 1 
/webwork2_files/helpFiles/Units.html
.)

13.

Suppose f(x,y,z)=1x2+y2+z2 and W is the bottom half of a sphere of radius 3. Enter ρ as rho, ϕ as phi, and θ as theta.
(a) As an iterated integral,
WfdV=ABCDEF dρdϕdθ
with limits of integration
A =
B =
C =
D =
E =
F =
(b) Evaluate the integral.

14.

In each of the following questions, set up an iterated integral expression whose value determines the desired result. Then, evaluate the integral first by hand, and then using appropriate technology.
  1. Find the volume of the “cap” cut from the solid sphere x2+y2+z2=4 by the plane z=1, as well as the z-coordinate of its centroid.
  2. Find the x-coordinate of the center of mass of the portion of the unit sphere that lies in the first octant (i.e., where x, y, and z are all nonnegative). Assume that the density of the solid given by δ(x,y,z)=11+x2+y2+z2.
  3. Find the volume of the solid bounded below by the xy-plane, on the sides by the sphere ρ=2, and above by the cone ϕ=π/3.
  4. Find the z coordinate of the center of mass of the region that is bounded above by the surface z=x2+y2, on the sides by the cylinder x2+y2=4, and below by the xy-plane. Assume that the density of the solid is uniform and constant.
  5. Find the volume of the solid that lies outside the sphere x2+y2+z2=1 and inside the sphere x2+y2+z2=2z.

15.

For each of the following questions,
  • sketch the region of integration,
  • change the coordinate system in which the iterated integral is written to one of the remaining two,
  • evaluate the iterated integral you deem easiest to evaluate by hand.
  1. 0101x2x2+y22x2y2xydzdydx
  2. 0π/20π01ρ2sin(ϕ)dρdϕdθ
  3. 02π01r1r2cos(θ)dzdrdθ

16.

Consider the solid region S bounded above by the paraboloid z=16x2y2 and below by the paraboloid z=3x2+3y2.
  1. Describe parametrically the curve in R3 in which these two surfaces intersect.
  2. In terms of x and y, write an equation to describe the projection of the curve onto the xy-plane.
  3. What coordinate system do you think is most natural for an iterated integral that gives the volume of the solid?
  4. Set up, but do not evaluate, an iterated integral expression whose value is average z-value of points in the solid region S.
  5. Use technology to plot the two surfaces and evaluate the integral in (c). Write at least one sentence to discuss how your computations align with your intuition about where the average z-value of the solid should fall.