Section12.7Triple Integrals in Cylindrical and Spherical Coordinates
Motivating Questions
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 is the volume element in spherical coordinates? How does this inform us about evaluating a triple integral as an iterated integral in spherical coordinates?
In Section 12.5, we saw how advantageous it could be to use polar coordinates to write iterated integrals when trying to evaluate a double integral. In particular, we saw that double integrals with integrands and regions of integration that are simple to write in polar coordinates are the most convenient to convert, as well as how the area element would be equal to \(dA = r dr \, d\theta\) when using polar coordinates.
In Section 9.8, we introduced both cylindrical and spherical coordinate systems and explore converting points, equations, and regions between these and rectangular coordinate systems in three dimensions. Cylindrical coordinates are most commonly used in situations with rotational symmetry around the \(z\)-axis, whereas spherical coordinates are primarily used when there is rotational symmetry around the origin. We encourage you to review these elements, including the conversion equations between cylindrical and rectangular coordinates and conversion equations between spherical and rectangular coordinates. The following preview activity will include some descriptions of common surfaces given in either cylindrical or spherical coordinates.
In this Preview Activity, we will be reviewing the meaning of each of the cylindrical and spherical coordinates by looking at a description of a common surface in either cylindrical or spherical coordinates. For each task, you should draw a plot of the surface described by hand and write a few sentences describing how your plot relates to the cylindrical/spherical coordinates.
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 the surface 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 \(0 \leq z \leq 2\text{?}\)
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 is this a plot of?
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 \(0 \leq \theta \leq 2\pi\text{?}\)
In this section, we will look at converting triple integrals to use cylindrical and spherical coordinates and similar to our approach in Section 12.5, we will focus conceptually on converting
Subsection12.7.2Triple 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.
In particular, we need to look at the volume of a region created by a small step in each of the coordinates (\(\Delta r\text{,}\)\(\Delta \theta\text{,}\)\(\Delta z\)). You can see a plot of an example region in Figure 12.7.1 and you should recognize the shaded area in the \(xy\)-plane as the area element from our work on double integrals in polar coordinates. Recall that \(\Delta A\) was \(\frac{1}{2}(r_{i+1}+r_i) \Delta r \ \Delta \theta\) and since \(z\) is the perpendicular height of our region, we have
\begin{equation*}
\Delta V = \Delta A \ \Delta z = \frac{(r_{i+1}+r_i)}{2} \Delta r \, \Delta \theta \, \Delta z .
\end{equation*}
Remember that as \(\Delta r\) shrinks, \(\frac{(r_{i+1}+r_i)}{2}\) approaches the \(r\)-coordinate of this block. Therefore the volume element \(dV\) in cylindrical coordinates is given by \(dV = r \, dz \, dr \, d\theta\text{.}\)
We can also convert our integrand to polar coordinates by using \(x=r\cos(\theta)\) and \(y=r\sin(\theta)\text{.}\) Slicing our region into a radially simple or angularly simple description will provide the bounds of our iterated integrals used to evaluate the triple integral. Remember that cylindrical coordinates are really polar coordinates plus \(z\text{.}\)
While the volume element and the conversion of the integrand will work the same for every case, the task of writing an iterated integral in cylindrical coordinates will require us to slice the particular region in terms of simple bounds on \(\theta\text{,}\)\(r\text{,}\) and \(z\text{.}\) In the following activity, we explore how to do this in several situations where cylindrical coordinates are natural and advantageous.
In this activity, we will work with a few triple integrals and look at the advantages of utilizing cylindrical coordinates for the iterated integral calculation.
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 in Figure 12.7.2. Set up an iterated integral in cylindrical coordinates that gives the mass of the cone. You do not need to evaluate this integral.
Write 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 Figure 12.7.3. You do not need to evaluate this integral.
Subsection12.7.3Triple Integrals in Spherical Coordinates
Similar to our work with cylindrical and Cartesian coordinates, when it comes to thinking about particular surfaces in spherical 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.
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 \(dV\text{.}\) In Figure 12.7.4, we see a volume given by a change in each of the \(\rho, \theta, \phi\) coordinates. A conceptual description for where the volume element comes from is given in the proof to Key Idea 12.7.5.
Key Idea12.7.5.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
The latter expression is an iterated integral in spherical coordinates. Note that the volume element for spherical coordinates is \(dV= \rho^2 \sin(\phi) \, d\rho \, d\theta \, d\phi\)
The conversion of the function being integrated is straightforward because \(x= \rho\sin(\phi)\cos(\theta), y=\rho \sin(\phi) \sin(\theta), z=\rho \cos(\phi) \) does not require interpretation based on the coordinates used.
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 in Figure 12.7.4. 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. We have highlighted the three sides we will use in red, blue, and green.
Along the red side, both \(\theta\) and \(\phi\) are constant and \(\rho\) is changing. In fact, the length of the red side is simply the change in \(\rho\text{.}\)
Along the green side, both \(\theta\) and \(\rho\) are constant and \(\phi\) is changing. We can simplify our plot by looking along the plane \(\theta=\theta_0\) which will give us the following plot
Hopefully you recognize this as the same kind of picture as we used when looking at the area element in polar coordinates. The length of the green curve will be \(\rho_1 \Delta\phi\) by the same kind of argument.
\begin{equation*}
(\text{fraction of circle})(\text{circumference of circle})=\left(\frac{\Delta\phi}{2\pi}\right)\left(2\pi \rho_1 \right)= \rho_1 \Delta\phi
\end{equation*}
The blue curve occurs at a constant value of \(\rho\) and \(\phi\text{,}\) with \(\theta\) changing. In other words, the blue curve consists of \(x=\rho_1 \sin(\phi_0) \cos(\theta), y= \rho_1 \sin(\phi_0), z=\rho_1 \cos(\phi_0)\) which corresponds to a circle of radius \(\rho_1 \sin(\phi_0)\) at a height \(z=\rho_1 \cos(\phi_0)\text{.}\) We can use the same type of argument as above that the arc length of a curve corresponding to an angular change of \(\Delta \theta\) along a circle of radius \(\rho_1 \sin(\phi_0)\) will be \(\rho_1 \sin(\phi_0) \Delta \theta\text{.}\)
Remember that as we look at smaller and smaller changes in the coordinates, the difference between the values of the coordinates used will also shrink and \(\rho=0 \approx \rho_1 \approx \rho\text{,}\) for example. So we can get an idea of how large this volume will be by multiplying the lengths of the red, blue, and green segments to get
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 other coordinate systems and we will look a few simple cases in the next activities; The central idea in our process is to fix all but one coordinate and give the same bounding functions for the remaining coordinate, then project onto the remaining coordinates and repeat this process.
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 Figure 12.7.7.
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 of the form
\begin{equation*}
\iiint_S f(r\cos(\theta), r\sin(\theta), z) \, r \, dz \, dr \, d\theta.
\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 of the form
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.
Use cylindrical coordinates to evaluate the triple integral \(\displaystyle \int \!\! \int \!\!
\int_{\mathbf{E}} \, \sqrt{x^{2} + y^{2}} \, dV\text{,}\) where E is the solid bounded by the circular paraboloid \(z = 4 - 9 \left( x^{2} + y^{2}
\right)\) and the \(xy\) -plane.
Use spherical coordinates to evaluate the triple integral \(\displaystyle \int \!\! \int \!\!
\int_{\mathbf{E}} \, x^{2} + y^{2} + z^{2} \, dV\text{,}\) where E is the ball: \(x^{2} + y^{2} + z^{2} \leq 49\text{.}\)
Find the volume of the solid enclosed by the paraboloids \(z = 16 \left(
x^{2} + y^{2} \right)\) and \(z = 18 - 16 \left( x^{2} + y^{2}
\right)\text{.}\)
Suppose \(\displaystyle f(x,y,z) = \frac{1}{\sqrt{x^2+y^2+z^2}}\) and \(W\) is the bottom half of a sphere of radius \(2\text{.}\) Enter \(\rho\) as rho,\(\phi\) as phi, and \(\theta\) as theta.
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.
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.
