Triple Integrals in Cylindrical and Spherical Coordinates

Section3.7Triple Integrals in Cylindrical and Spherical Coordinates

Objectives

What are the cylindrical coordinates of a point, and how are they related to Cartesian coordinates?
What is the volume element in cylindrical coordinates? How does this inform us about evaluating a triple integral as an iterated integral in cylindrical coordinates?
What are the spherical coordinates of a point, and how are they related to Cartesian coordinates?
What is the volume element in spherical coordinates? How does this inform us about evaluating a triple integral as an iterated integral in spherical coordinates?

We have encountered two different coordinate systems in \(\R^2\) 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 \(\R^3\text{.}\) Given that we are already familiar with the Cartesian coordinate system for \(\R^3\text{,}\) we next investigate the cylindrical and spherical coordinate systems (each of which builds upon polar coordinates in \(\R^2\)). 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.

Exploration3.7.1

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.

Figure3.7.1The cylindrical (left) and spherical (right) coordinates of a point.

The cylindrical coordinates of a point in \(\R^3\) are given by \((r,\theta,z)\) where \(r\) and \(\theta\) 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 Figure3.7.1.
1. Find cylindrical coordinates for the point whose Cartesian coordinates are \((-1, \sqrt{3}, 3)\text{.}\) Draw a labeled picture illustrating all of the coordinates.
2. Find the Cartesian coordinates of the point whose cylindrical coordinates are \(\left(2, \frac{5\pi}{4}, 1\right)\text{.}\) Draw a labeled picture illustrating all of the coordinates.
The spherical coordinates of a point in \(\R^3\) are \(\rho\) (rho), \(\theta\text{,}\) and \(\phi\) (phi), where \(\rho\) is the distance from the point to the origin, \(\theta\) has the same interpretation it does in polar coordinates, and \(\phi\) is the angle between the positive \(z\) axis and the vector from the origin to the point, as illustrated at right in Figure3.7.1. You should convince yourself that any point in \(\R^3\) can be represented in spherical coordinates with \(\rho \geq 0\text{,}\) \(0 \leq \theta \lt 2 \pi\text{,}\) and \(0 \leq \phi \leq \pi\text{.}\)

For the following questions, consider the point \(P\) whose Cartesian coordinates are \((-2,2,\sqrt{8})\text{.}\)
1. What is the distance from \(P\) to the origin? Your result is the value of \(\rho\) in the spherical coordinates of \(P\text{.}\)
2. Determine the point that is the projection of \(P\) onto the \(xy\)-plane. Then, use this projection to find the value of \(\theta\) in the polar coordinates of the projection of \(P\) that lies in the plane. Your result is also the value of \(\theta\) for the spherical coordinates of the point.
3. Based on the illustration in Figure3.7.1, how is the angle \(\phi\) determined by \(\rho\) and the \(z\) coordinate of \(P\text{?}\) Use a well-chosen right triangle to find the value of \(\phi\text{,}\) which is the final component in the spherical coordinates of \(P\text{.}\) Draw a carefully labeled picture that clearly illustrates the values of \(\rho\text{,}\) \(\theta\text{,}\) and \(\phi\) in this example, along with the original rectangular coordinates of \(P\text{.}\)
4. Based on your responses to i., ii., and iii., if we are given the Cartesian coordinates \((x,y,z)\) of a point \(Q\text{,}\) how are the values of \(\rho\text{,}\) \(\theta\text{,}\) and \(\phi\) in the spherical coordinates of \(Q\) determined by \(x\text{,}\) \(y\text{,}\) and \(z\text{?}\)

Subsection3.7.1Cylindrical Coordinates

As we stated in Preview Activity3.7.1, the cylindrical coordinates of a point are \((r,\theta,z)\text{,}\) where \(r\) and \(\theta\) are the polar coordinates of the point \((x, y)\text{,}\) and \(z\) is the same \(z\) coordinate as in Cartesian coordinates. The general situation is illustrated Figure3.7.1.

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 \(\R^3\) are essentially those we found for polar coordinates.

Converting between Cartesian and cylindrical coordinates

If we are given the Cartesian coordinates \((x,y,z)\) of a point \(P\text{,}\) then the cylindrical coordinates \((r,\theta,z)\) of \(P\) satisfy

\begin{equation*} x = r \cos(\theta) \ \ \ \ \ y = r \sin(\theta) \ \ \ \ \ \text{ and } \ \ \ \ \ z = z. \end{equation*}
If we are given the cylindrical coordinates \((r,\theta,z)\) of a point \(P\text{,}\) then the Cartesian coordinates \((x,y,z)\) of \(P\) satisfy

\begin{equation*} r^2 = x^2 + y^2 \ \ \ \ \ \tan(\theta) = \frac{y}{x} \ \ \ \ \ \text{ and } \ \ \ \ \ z = z, \end{equation*}

assuming \(x \neq 0\text{.}\)

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 \(\theta\text{.}\) In the following activity, we explore several basic equations in cylindrical coordinates and the corresponding surface each generates.

Activity3.7.2

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.

What familiar surface is described by the points in cylindrical coordinates with \(r=2\text{,}\) \(0 \leq \theta \leq 2\pi\text{,}\) and \(0 \leq z \leq 2\text{?}\) How does this example suggest that we call these coordinates cylindrical coordinates? How does your answer change if we restrict \(\theta\) to \(0 \leq \theta \leq \pi\text{?}\)
What familiar surface is described by the points in cylindrical coordinates with \(\theta=2\text{,}\) \(0 \leq r \leq 2\text{,}\) and \(0 \leq z \leq 2\text{?}\)
What familiar surface is described by the points in cylindrical coordinates with \(z=2\text{,}\) \(0 \leq \theta \leq 2\pi\text{,}\) and \(0 \leq r \leq 2\text{?}\)
Plot the graph of the cylindrical equation \(z=r\text{,}\) where \(0 \leq \theta \leq 2\pi\) and \(0 \leq r \leq 2\text{.}\) What familiar surface results?
Plot the graph of the cylindrical equation \(z= \theta\) for \(0 \leq \theta \leq 4 \pi\) and \(0 \leq r \leq 2\text{.}\) What does this surface look like?

Answer

The points in cylindrical coordinates with \(r=2\text{,}\) \(0 \leq \theta \leq 2\pi\text{,}\) and \(0 \leq z \leq 2\) 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 \(\theta\) to \(0 \leq \theta \leq \pi\text{,}\) then we get half of the cylinder.
The points in cylindrical coordinates with \(\theta=2\text{,}\) \(0 \leq r \leq 2\text{,}\) and \(0 \leq z \leq 2\) form a \(2\times 2\) form a rectangle.
The points in cylindrical coordinates with \(z=2\text{,}\) \(0 \leq \theta \leq 2\pi\text{,}\) and \(0 \leq r \leq 2\) give a disk with radius 2.
The cylindrical equation \(z=r\text{,}\) where \(0 \leq \theta \leq 2\pi\) and \(0 \leq r \leq 2\) gives a cone (just the curved face, with no disk) situated with vertex at the origin.
The graph of the cylindrical equation \(z= \theta\) for \(0 \leq \theta \leq 4 \pi\) and \(0 \leq r \leq 2\) forms a spiral.

As the name and Activity3.7.2 suggests, cylindrical coordinates are useful for describing surfaces that are cylindrical in nature.

Subsection3.7.2Triple Integrals in Cylindrical Coordinates

To evaluate a triple integral \(\iiint_S f(x,y,z) \, dV\) as an iterated integral in Cartesian coordinates, we use the fact that the volume element \(dV\) is equal to \(dz \, dy \, dx\) (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.

Activity3.7.3

A picture of a cylindrical box, \(B = \{(r,\theta,z) : r_1 \leq r \leq r_2, \theta_1 \leq \theta \leq \theta_2, z_1 \leq z \leq z_2\},\) is shown in Figure3.7.2. Let \(\Delta r = r_2-r_1\text{,}\) \(\Delta \theta = \theta_2 - \theta_1\text{,}\) and \(\Delta z = z_2-z_1\text{.}\) We want to determine the volume \(\Delta V\) of \(B\) in terms of \(\Delta r\text{,}\) \(\Delta \theta\text{,}\) \(\Delta z\text{,}\) \(r\text{,}\) \(\theta\text{,}\) and \(z\text{.}\)

Figure3.7.2A cylindrical box.

Appropriately label \(\Delta r\text{,}\) \(\Delta \theta\text{,}\) and \(\Delta z\) in Figure3.7.2.
Let \(\Delta A\) be the area of the projection of the box, \(B\text{,}\) onto the \(xy\)-plane, which is shaded blue in Figure3.7.2. Recall that we previously determined the area \(\Delta A\) in polar coordinates in terms of \(r\text{,}\) \(\Delta r\text{,}\) and \(\Delta \theta\text{.}\) In light of the fact that we know \(\Delta A\) and that \(z\) is the standard \(z\) coordinate from Cartesian coordinates, what is the volume \(\Delta V\) in cylindrical coordinates?

Answer

The base of \(B\) is the same as its projection onto the \(xy\)-plane, which has an area \(\Delta A = r\, \Delta r\, \Delta \theta\text{.}\) The height of \(B\) is \(\Delta z\text{.}\) Then we can mutltiply base and height to get that the volume of \(B\) is \(\Delta V = r\, \Delta r\, \Delta \theta\,\Delta z\text{.}\)

Activity3.7.3 demonstrates that the volume element \(dV\) in cylindrical coordinates is given by \(dV = r \, dz \, dr \, d\theta\text{,}\) 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 \(\R^3\text{,}\)

\begin{equation*} \iiint_S f(x,y,z) \, dV = \iiint_S f(r\cos(\theta), r\sin(\theta), z) \, r \, dz \, dr \, d\theta. \end{equation*}

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: \(\theta\text{,}\) \(r\text{,}\) and \(z\text{.}\) In the following activity, we explore how to do this in several situations where cylindrical coordinates are natural and advantageous.

Activity3.7.4

In this activity we work with triple integrals in cylindrical coordinates.

Let \(S\) be the solid bounded above by the graph of \(z = x^2+y^2\) 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 \(S\text{.}\) You do not need to evaluate this integral.
Suppose the density of the cone defined by \(r = 1 - z\text{,}\) with \(z \geq 0\text{,}\) is given by \(\delta(r, \theta, z) = z\text{.}\) A picture of the cone is shown at left in Figure3.7.3, and the projection of the cone onto the \(xy\)-plane in given at right in Figure3.7.3. Set up an iterated integral in cylindrical coordinates that gives the mass of the cone. You do not need to evaluate this integral.

Figure3.7.3The cylindrical cone \(r = 1-z\) and its projection onto the \(xy\)-plane.
Determine an iterated integral expression in cylindrical coordinates whose value is the volume of the solid bounded below by the cone \(z = \sqrt{x^2+y^2}\) and above by the cone \(z = 4 - \sqrt{x^2+y^2}\text{.}\) A picture is shown in Figure3.7.4. You do not need to evaluate this integral.
Figure3.7.4A solid bounded by the cones \(z = \sqrt{x^2+y^2}\) and \(z = 4 - \sqrt{x^2+y^2}\text{.}\)

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 \(0 \leq r\leq 1\) and \(0 \leq \theta \leq 2\pi\text{.}\)
2. We can think of the solid \(S\) as being bounded by three surfaces: above by the parabolic surface \(z = x^2+y^2\text{,}\) below by the unit disk in the \(xy\)-plane, and on the sides by a cylinder.
  - Parabolic surface: \(z=r^2\) with \(0\leq r\leq 1\) and \(0\leq \theta\leq 2\pi\text{.}\)
  - Unit disk in the \(xy\)-plane: \(z=0\) with \(0 \leq r\leq 1\) and \(0 \leq \theta \leq 2\pi\text{.}\)
  - Cylinder: \(r=1\) with \(0 \leq z\leq 1\) and \(0 \leq \theta \leq 2\pi\text{.}\)
3. To find the volume of a solid, we integrate 1 over the solid:
  
  \begin{equation*} \iiint_S 1 \,dV = \int_0^{2\pi} \int_0^1 \int_0^{r} 1 r \,dz \,dr \, d\theta. \end{equation*}
To find the mass of the cone, we integrate the density over the solid:

\begin{equation*} \iiint_S \delta(r, \theta, z) \,dV = \int_0^{2\pi} \int_0^1 \int_0^{1 - z} z r \,dr \,dz \, d\theta. \end{equation*}
To find the volume of a solid, we integrate 1 over the solid:

\begin{equation*} \iiint_S 1 \,dV = \int_0^{2\pi} \int_0^2 \int_{r}^{4-r} 1 r \,dz \,dr \, d\theta. \end{equation*}

Subsection3.7.3Spherical Coordinates

As we saw in Preview Activity3.7.1, the spherical coordinates of a point in 3-space have the form \((\rho, \theta, \phi)\text{,}\) where \(\rho\) is the distance from the point to the origin, \(\theta\) has the same meaning as in polar coordinates, and \(\phi\) 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 Figure3.7.1.

Figure3.7.5Converting from spherical to Cartesian coordinates.

The example in Preview Activity3.7.1 and Figure3.7.5 suggest how to convert between Cartesian and spherical coordinates.

Coverting between Cartesian and spherical coordinates

If we are given the Cartesian coordinates \((x,y,z)\) of a point \(P\text{,}\) then the spherical coordinates \((\rho,\theta,\phi)\) of \(P\) satisfy

\begin{equation*} \rho = \sqrt{x^2 + y^2 + z^2} \ \ \ \ \ \tan(\theta) = \frac{y}{x} \ \ \ \ \ \text{ and } \ \ \ \ \ \cos(\phi) = \frac{z}{\rho}, \end{equation*}

where in the latter two equations, we require \(x \ne 0\) and \(\rho \ne 0\text{.}\)
If we are given the spherical coordinates \((\rho,\theta,\phi)\) of a point \(P\text{,}\) then the Cartesian coordinates \((x,y,z)\) of \(P\) satisfy

\begin{equation*} x = \rho \sin(\phi) \cos(\theta) \ \ \ \ \ y = \rho \sin(\phi) \sin(\theta) \ \ \ \ \ \text{ and } \ \ \ \ \ z = \rho \cos(\phi). \end{equation*}

When it comes to thinking about particular surfaces in spherical coordinates, similar to our work with cylindrical and Cartesian coordinates, we usually write \(\rho\) as a function of \(\theta\) and \(\phi\text{;}\) 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 \(\theta\text{.}\) In spherical coordinates, we likewise often view \(\rho\) as a function of \(\theta\) and \(\phi\text{,}\) 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.

Activity3.7.5

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.

What familiar surface is described by the points in spherical coordinates with \(\rho = 1\text{,}\) \(0 \leq \theta \leq 2\pi\text{,}\) and \(0 \leq \phi \leq \pi\text{?}\) How does this particular example demonstrate the reason for the name of this coordinate system? What if we restrict \(\phi\) to \(0 \leq \phi \leq \frac{\pi}{2}\text{?}\)
What familiar surface is described by the points in spherical coordinates with \(\phi = \frac{\pi}{3}\text{,}\) \(0 \leq \rho \leq 1\text{,}\) and \(0 \leq \theta \leq 2\pi\text{?}\)
What familiar shape is described by the points in spherical coordinates with \(\theta = \frac{\pi}{6}\text{,}\) \(0 \leq \rho \leq 1\text{,}\) and \(0 \leq \phi \leq \pi\text{?}\)
Plot the graph of \(\rho = \theta\text{,}\) for \(0 \leq \phi \leq \pi\) and \(0 \leq \theta \leq 2 \pi\text{.}\) How does the resulting surface appear?

Answer

The points in spherical coordinates with \(\rho = 1\text{,}\) \(0 \leq \theta \leq 2\pi\text{,}\) and \(0 \leq \phi \leq \pi\) 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 \(\phi\) to \(0 \leq \phi \leq \frac{\pi}{2}\) we get the upper hemisphere.
The points in spherical coordinates with \(\phi = \frac{\pi}{3}\text{,}\) \(0 \leq \rho \leq 1\text{,}\) and \(0 \leq \theta \leq 2\pi\) give a cone (the curved face, not the disk).
The points in spherical coordinates with \(\theta = \frac{\pi}{6}\text{,}\) \(0 \leq \rho \leq 1\text{,}\) and \(0 \leq \phi \leq \pi\) give a semicircle.
The graph of \(\rho = \theta\text{,}\) for \(0 \leq \phi \leq \pi\) and \(0 \leq \theta \leq 2 \pi\) looks like a seashell.

As the name and Activity3.7.5 indicate, spherical coordinates are particularly useful for describing surfaces that are spherical in nature; they are also convenient for working with certain conical surfaces.

Subsection3.7.4Triple Integrals in Spherical Coordinates

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

Activity3.7.6

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 \(\rho_1 \leq \rho \leq \rho_2\) (with \(\Delta \rho = \rho_2-\rho_1)\text{,}\) \(\phi_1 \leq \phi \leq \phi_2\) (with \(\Delta \phi = \phi_2-\phi_1\)), and \(\theta_1 \leq \theta \leq \theta_2\) (with \(\Delta \theta = \theta_2-\theta_1\)). An illustration of such a box is given at left in Figure3.7.6. 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 \(\Delta V\) than to compute it directly. Here we can approximate the volume \(\Delta V\) 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,

\begin{equation*} \Delta V \approx |PS| \ |\overset{\frown}{PR}| \ |\overset{\frown}{PQ}|, \end{equation*}

where \(|\overset{\frown}{PR}|\) denotes the length of the circular arc from \(P\) to \(R\text{.}\)

Figure3.7.6Left: A spherical box. Right: A spherical volume element.

What is the length \(|PS|\) in terms of \(\rho\text{?}\)
What is the length of the arc \(\overset{\frown}{PR}\text{?}\) (Hint: The arc \(\overset{\frown}{PR}\) is an arc of a circle of radius \(\rho_2\text{,}\) and arc length along a circle is the product of the angle measure (in radians) and the circle's radius.)
What is the length of the arc \(\overset{\frown}{PQ}\text{?}\) (Hint: The arc \(\overset{\frown}{PQ}\) lies on a horizontal circle as illustrated at right in Figure3.7.6. What is the radius of this circle?)
Use your work in (a), (b), and (c) to determine an approximation for \(\Delta V\) in spherical coordinates.

Answer

The length of \(PS\) is the change in \(\rho\text{,}\) so \(|PS| = \Delta\rho\text{.}\)
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 \(\overset{\frown}{PR}\) is \(|\overset{\frown}{PR}|=\rho_2\,\Delta\phi\text{.}\)
What is the radius of the circle in Figure3.7.6 is \(\rho_2\sin(\phi_2)\text{,}\) so we can do another arc length calculation to get \(|\overset{\frown}{PQ}|=\rho_2\sin(\phi_2)\,\Delta\theta\text{.}\)
We can then approximate

\begin{align*} \Delta V \amp \approx |PS| \ |\overset{\frown}{PR}| \ |\overset{\frown}{PQ}|\\ \amp = (\Delta\rho) (\rho_2\,\Delta\phi) (\rho_2\sin(\phi_2)\,\Delta\theta)\\ \amp = \rho_2^2 \sin(\phi_2)\,\Delta\rho\,\Delta\theta\,\Delta\phi. \end{align*}

Letting \(\Delta \rho\text{,}\) \(\Delta \theta\text{,}\) and \(\Delta \phi\) go to 0, it follows from the final result in Activity3.7.6 that \(dV = \rho^2 \, \sin(\phi) \, d\rho \, d\theta \, d\phi \) 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 \(\R^3\text{,}\) the triple integral \(\iiint_S f(x,y,z) \, dV\) is converted to the integral

\begin{equation*} \iiint_S f(\rho\sin(\phi)\cos(\theta), \rho \sin(\phi) \sin(\theta), \rho \cos(\phi)) \, \rho^2 \sin(\phi) \, d\rho \, d\theta \, d\phi \end{equation*}

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 \(\phi\text{,}\) \(\theta\text{,}\) and \(\rho\text{.}\) The process is similar to our earlier work in the other two coordinate systems.

Activity3.7.7

We can use spherical coordinates to help us more easily understand some natural geometric objects.

Recall that the sphere of radius \(a\) has spherical equation \(\rho = a\text{.}\) Set up and evaluate an iterated integral in spherical coordinates to determine the volume of a sphere of radius \(a\text{.}\)
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 \(\phi=\frac{\pi}{4}\) from the sphere \(\rho = 2\) if the density \(\delta\) at the point \((x,y,z)\) is \(\delta(x,y,z) = \sqrt{x^2+y^2+z^2}\text{.}\) An illustration of the solid is shown in Figure3.7.7.
Figure3.7.7The solid cut from the sphere \(\rho = 2\) by the cone \(\phi=\frac{\pi}{4}\text{.}\)

Answer

Recall that the sphere of radius \(a\) has spherical equation \(\rho = a\text{.}\) An iterated integral in spherical coordinates to determine the volume of a sphere of radius \(a\) is:

\begin{equation*} \iiint_S 1 \Delta V =\int_0^\pi \int_0^{2\pi} \int_0^a 1 \rho^2\sin(\phi) \, d\rho \, d\theta \, d\phi. \end{equation*}
Since we want to work in spherical coordinates, we should change the density function to be in spherical coordinates:

\begin{align*} \delta(\rho\sin(\phi) \amp \, \cos(\theta), \rho \sin(\phi) \sin(\theta), \rho \cos(\phi))\\ \amp = \sqrt{\rho^2\sin^2(\phi)\cos^2(\theta)+ \rho^2 \sin^2(\phi) \sin^2(\theta) + \rho^2 \cos^2(\phi)}\\ \amp = \sqrt{\rho^2(\sin^2(\phi)(\cos^2(\theta)+ \sin^2(\theta)) + \cos^2(\phi))}\\ \amp = \rho\sqrt{\sin^2(\phi)(1) + \cos^2(\phi)}\\ \amp = \rho\sqrt{1}\\ \amp = \rho. \end{align*}

We can now find the mass by integrating the density over the solid:

\begin{align*} \iiint_S \delta(\rho\sin(\phi) \amp \, \cos(\theta), \rho \sin(\phi) \sin(\theta), \rho \cos(\phi)) \Delta V\\ \amp =\int_{\pi/4}^\pi \int_0^{2\pi} \int_0^2 \rho \, \rho^2 \sin(\phi) \, d\rho \, d\theta \, d\phi\\ \amp =\int_{\pi/4}^\pi \int_0^{2\pi} \int_0^2 \rho^3 \sin(\phi) \, d\rho \, d\theta \, d\phi. \end{align*}

Subsection3.7.5Summary

The cylindrical coordinates of a point \(P\) are \((r,\theta,z)\) where \(r\) is the distance from the origin to the projection of \(P\) onto the \(xy\)-plane, \(\theta\) 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 \((x,y,z)\text{,}\) it follows that its cylindrical coordinates are given by

\begin{equation*} r^2 = x^2 + y^2, \ \ \ \ \ \tan(\theta) = \frac{y}{x}, \ \ \ \ \ z = z. \end{equation*}

When \(P\) has given cylindrical coordinates \((r,\theta,z)\text{,}\) its rectangular coordinates are

\begin{equation*} x = r \cos(\theta), \ \ \ \ \ y = r \sin(\theta), \ \ \ \ \ z = z. \end{equation*}
The volume element \(dV\) in cylindrical coordinates is \(dV = r \, dz \, dr \, d\theta\text{.}\) Hence, a triple integral \(\iiint_S f(x,y,z) \, dA\) can be evaluated as the iterated integral

\begin{equation*} \iiint_S f(r\cos(\theta), r\sin(\theta), z) \, r \, dz \, dr \, d\theta. \end{equation*}
The spherical coordinates of a point \(P\) in 3-space are \(\rho\) (rho), \(\theta\text{,}\) and \(\phi\) (phi), where \(\rho\) is the distance from \(P\) to the origin, \(\theta\) is the angle that the projection of \(P\) onto the \(xy\)-plane makes with the positive \(x\)-axis, and \(\phi\) is the angle between the positive \(z\) axis and the vector from the origin to \(P\text{.}\) When \(P\) has Cartesian coordinates \((x,y,z)\text{,}\) the spherical coordinates are given by

\begin{equation*} \rho^2 = x^2 + y^2 + z^2, \ \ \ \ \ \tan(\theta) = \frac{y}{x}, \ \ \ \ \ \cos(\phi) = \frac{z}{\rho}. \end{equation*}

Given the point \(P\) in spherical coordinates \((\rho, \phi, \theta)\text{,}\) its rectangular coordinates are

\begin{equation*} x = \rho \sin(\phi) \cos(\theta), \ \ \ \ \ y = \rho \sin(\phi) \sin(\theta), \ \ \ \ \ z = \rho \cos(\phi). \end{equation*}
The volume element \(dV\) in spherical coordinates is \(dV = \rho^2 \sin(\phi) \, d\rho \, d\theta \, d\phi\text{.}\) Thus, a triple integral \(\iiint_S f(x,y,z) \, dA\) can be evaluated as the iterated integral

\begin{equation*} \iiint_S f(\rho\sin(\phi)\cos(\theta), \rho \sin(\phi) \sin(\theta), \rho \cos(\phi)) \, \rho^2 \sin(\phi) \, d\rho \, d\theta \, d\phi. \end{equation*}

Subsection3.7.6Exercises

1

2

3

4

5

6

7

8

9

10

11

12

13

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.

Find the volume of the cap cut from the solid sphere \(x^2 + y^2 + z^2 = 4\) by the plane \(z=1\text{,}\) as well as the \(z\)-coordinate of its centroid.
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 \(x\text{,}\) \(y\text{,}\) and \(z\) are all nonnegative). Assume that the density of the solid given by \(\delta(x,y,z) = \frac{1}{1+x^2+y^2+z^2}\text{.}\)
Find the volume of the solid bounded below by the \(xy\)-plane, on the sides by the sphere \(\rho=2\text{,}\) and above by the cone \(\phi = \pi/3\text{.}\)
Find the \(z\) coordinate of the center of mass of the region that is bounded above by the surface \(z = \sqrt{\sqrt{x^2 + y^2}}\text{,}\) on the sides by the cylinder \(x^2 + y^2 = 4\text{,}\) and below by the \(xy\)-plane. Assume that the density of the solid is uniform and constant.
Find the volume of the solid that lies outside the sphere \(x^2 + y^2 + z^2 = 1\) and inside the sphere \(x^2 + y^2 + z^2 = 2z\text{.}\)

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.

\(\int_{0}^{1} \int_{0}^{\sqrt{1-x^2}} \int_{\sqrt{x^2 + y^2}}^{\sqrt{2-x^2-y^2}} xy\, dz \, dy \, dx\)
\(\int_{0}^{\pi/2} \int_{0}^{\pi} \int_{0}^{1} \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta\)
\(\int_{0}^{2\pi} \int_{0}^{1} \int_{r}^{1} r^2 \cos(\theta) \, dz \, dr \, d\theta\)

16

Consider the solid region \(S\) bounded above by the paraboloid \(z = 16 - x^2 - y^2\) and below by the paraboloid \(z = 3x^2 + 3y^2\text{.}\)

Describe parametrically the curve in \(\R^3\) in which these two surfaces intersect.
In terms of \(x\) and \(y\text{,}\) write an equation to describe the projection of the curve onto the \(xy\)-plane.
What coordinate system do you think is most natural for an iterated integral that gives the volume of the solid?
Set up, but do not evaluate, an iterated integral expression whose value is average \(z\)-value of points in the solid region \(S\text{.}\)
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.