Triple Integrals in Cylindrical and Spherical Coordinates

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Objectives:

1.: Be comfortable setting up and computing triple integrals in cylindrical and spherical coordinates.
2.: Understand the scaling factors for triple integrals in cylindrical and spherical coordinates, as well as where they come from.
3.: Be comfortable picking between cylindrical and spherical coordinates.

Cylindrical Coordinates

Recap Video

Here is a video highlights the main points of the section.

To summarize:

The transformation $\textbf {r}(r,\theta ,z) = (r\cos (\theta ),r\sin (\theta ),z)$ is called cylindrical coordinates.

It is unimportant which variable stays the same. It is one variable the same with polar in the other two.

Let $R$ be a region in Cartesian space which corresponds to a region $D$ in cylindrical coordinates. Then $\iiint _R f(x,y,z)dV = \iiint _R f(\textbf {r}(r,\theta ,z))rdz\,dr\,d\theta .$

The extra $r$ in the formula appears for the same reason it appeared in the polar coordinate integrals.

Example Video

Here is an example of setting up the bounds for a triple integral in cylindrical coordinates.

If $R$ is the solid bounded by $z = x^2+y^2$ , $z=1$ and $z=4$ . Set up $\displaystyle \iiint _R (2y+z)dV$ in cylindrical coordinates.

Problems

Evaluate $\displaystyle \iiint _R \sqrt {x^2+y^2}dV$ , where $R$ is the region $x^2+y^2 \leq 9$ , $0 \leq z \leq 5$ .

The region $R$ is a solid cylinder, hence cylindrical coordinates. In this case, that means polar in $(x,y)$ and $z$ stays the same. Therefore, in cylindrical coordinates, the region can be described as $\answer {0} \leq z \leq \answer {5},$ $\answer {0} \leq r \leq \answer {3},$ $0 \leq \theta \leq \answer {2\pi }.$

The integrand $\sqrt {x^2+y^2}$ in cylindrical coordinates is $x^2+y^2 = \answer {r}.$ The integral therefore becomes

$\begin{eqnarray*} \iiint_R \sqrt{x^2+y^2}dV &=& \iiint_D f(\textbf{r}(r,\theta,z))rdz\,dr\,d\theta \\ &=& \int_{\answer{0}}^{\answer{2\pi}}\int_{\answer{0}}^{\answer{3}} \int_{\answer{0}}^{\answer{5}} \answer{r^2}dz\,dr\,d\theta \\ &=& \answer{90\pi}. \end{eqnarray*}$

Let $R$ be the region given by $0 \leq z \leq x^2+y^2$ , $x^2+y^2 \leq 3$ . Evaluate $\displaystyle \iiint _R (x+2)dV$ .

This is the region under a paraboloid and inside a cylinder. The reason cylindrical coordinates would be a good coordinate system to pick is that the condition $x^2+y^2 \leq 3$ means we will probably go to polar later anyway, so we can just go there now with cylindrical coordinates.

The paraboloid’s equation in cylindrical coordinates (i.e. in terms of $r$ , $\theta$ , and $z$ ) is $z = x^2+y^2 = \answer {r^2}.$ Thus, our bounds for $z$ will be $\answer {0} \leq z \leq \answer {r^2}.$ Now that we have $z$ , we can look at the $xy$ -plane for our polar bounds. The disc $x^2+y^2 \leq 3$ in polar is $\answer {0} \leq r \leq \answer {\sqrt {3}}$ $0\leq \theta \leq \answer {2\pi }.$ Therefore, the integral becomes

$\begin{eqnarray*} \iiint_R (x+2)dV &=& \iiint_D f(\textbf{r}(r,\theta,z))rdz\,dr\,d\theta \\ &=& \int_{\answer{0}}^{\answer{2\pi}} \int_{\answer{0}}^{\answer{\sqrt{3}}} \int_{\answer{0}}^{\answer{r^2}} \answer{r^2\cos(\theta)+2r} dz\,dr\,d\theta \\ &=& \answer{9\pi}. \end{eqnarray*}$

Spherical Coordinates

Throughout this section, you will need to use “ $p$ ” to denote “ $\phi$ .”

Recap Video

Here is a video highlights the main points of the section.

To summarize:

The transformation $\textbf {r}(\rho ,\theta ,\phi ) = (\rho \cos (\theta )\sin (\phi ),\rho \sin (\theta )\sin (\phi ),\rho \cos (\phi ))$ is called spherical coordinates.

Let $R$ be a region in Cartesian space which corresponds to a region $D$ in spherical coordinates. Then $\iiint _R f(x,y,z)dV = \iiint _R f(\textbf {r}(\rho ,\theta ,\phi ))\rho ^2\sin (\phi )d\rho \,d\phi \,d\theta .$

Example Video

Here is an example of setting up the bounds for a triple integral in spherical coordinates.

If $R$ is the solid given by $x^2+y^2+z^2 \leq 4$ , $x^2+y^2+z^2 \geq 2z$ , and $z \geq 0$ , set up $\displaystyle \iiint _R zdV$ in spherical coordinates.

Problems

NOTE: Use “p” instead of $\phi$ in your answer inputs. For example input $\sin (p)$ for $\sin (\phi )$ .

Let $R$ be the portion of the sphere $x^2+y^2+z^2 \leq 4$ with $z \geq 0$ . Evaluate $\displaystyle \iiint _R ydV$ .

The region $R$ is only a part of a sphere, so we can use spherical coordinates. In spherical, the sphere becomes $\rho ^2 \leq 4$ , i.e. $\rho \leq 2$ . So the bounds for $\rho$ are $\answer {0} \leq \rho \leq \answer {2}.$ The bounds for $\theta$ and $\phi$ are $0 \leq \theta \leq \answer {2\pi },$ $0\leq \phi \leq \answer {\pi /2}.$ Therefore, the integral becomes

$\begin{eqnarray*} \iiint_R ydV &=& \iiint_D f(\textbf{r}(\rho,\theta,\phi))\rho^2\sin(\phi)d\rho\,d\phi\,d\theta \\ &=& \int_{\answer{0}}^{\answer{2\pi}} \int_{\answer{0}}^{\answer{\pi/2}} \int_{\answer{0}}^{\answer{2}} \answer{\rho^3\sin(\theta)\sin^2(p)} d\rho\,d\phi\,d\theta \\ &=& \answer{0}. \end{eqnarray*}$

Let $R$ be the region $z \geq \sqrt {x^2+y^2}$ and $x^2+y^2+z^2\leq 4$ . Find the volume of $R$ .

Let’s use spherical again. In this case, to find the volume of $R$ , we need to integrate $f(x,y,z) = \answer {1}$ . The region is above a cone and inside a sphere. In spherical the equation of the sphere is $\rho = \answer {2}$ . If we examine the cross sections, we get that the bounds for $\rho$ will be $\answer {0} \leq \rho \leq \answer {2}.$ To get $\phi$ , notice that the cone $z = \sqrt {x^2+y^2}$ in spherical has equation $\phi = \answer {\pi /4}$ . Since $z$ is above the cone, $\phi$ will go from $0$ to the cone, i.e. $\answer {0} \leq \phi \leq \answer {\pi /4}.$ Finally, $\answer {0} \leq \theta \leq 2\pi .$ Therefore, the integral becomes

$\begin{eqnarray*} \iiint_R 1dV &=& \iiint_D f(\textbf{r}(\rho,\theta,\phi))\rho^2\sin(\phi)d\rho\,d\phi\,d\theta \\ &=& \int_{\answer{0}}^{\answer{2\pi}} \int_{\answer{0}}^{\answer{\pi/4}} \int_{\answer{0}}^{\answer{2}} \answer{\rho^2\sin(p)} d\rho\,d\phi\,d\theta \\ &=& \answer{\frac{8}{3}\pi(2-\sqrt{2})}. \end{eqnarray*}$

Picking a Coordinate System

For the following problems, you should decide whether to use Cartesian, cylindrical, or spherical coordinates to evaluate the triple integral. It would be good practice for you to try all three and see which ones are realistic options for each problem.

You once did the next problem as a double integral. Now do it as a triple integral and convince yourself it is the same thing.

Let $R$ be the region bounded above by $z = 8-x^2-y^2$ and below by $z = x^2+y^2$ . Then the volume of $R$ is: $\answer {16\pi }$

This is easiest in cylindrical coordinates. The one thing you need is the intersection of the two paraboloids, which works out to be $8-x^2-y^2 = x^2+y^2$ , or $x^2+y^2=4$ . This tells us what the projection onto the $xy$ -plane will look like so we can set up the polar integral. The integral then becomes $\int _0^{2\pi } \int _0^2 \int _{r^2}^{8-r^2} r dz\,dr\,d\theta = 16\pi .$

Let $R$ be the portion of $x^2+y^2+z^2 = 4$ above the plane $z=1$ . Consider the integral $\displaystyle \iiint _R zdV$ .

In cylindrical coordinates, the integral would be $\int _{\answer {0}}^{\answer {2\pi }} \int _{\answer {0}}^{\answer {\sqrt {3}}} \int _{\answer {1}}^{\answer {\sqrt {4-r^2}}} \answer {zr} dz\,dr\,d\theta .$
In spherical, the integral would be (input “p” for phi): $\int _{\answer {0}}^{\answer {2\pi }}\int _{\answer {0}}^{\answer {\pi /3}} \int _{\answer {\sec (p)}}^{\answer {2}} \rho ^3\cos (p)\sin (p) d\rho \,d\phi \,d\theta .$
Evaluating either way gives an answer of: $\answer {\frac {9\pi }{4}}$ .

You could set this up in cylindrical or spherical. Notice the intersection is the circle $x^2+y^2=3$ . In cylindrical, the setup would be $\int _0^{2\pi } \int _0^{\sqrt {3}}\int _1^{\sqrt {4-r^2}}zrdz\,dr\,d\theta .$ This is actually solvable because the square root will disappear after integrating with respect to $z$ .

In spherical, the plane has equation $\rho \cos (\phi ) = 1$ , or $\rho = \sec (\phi )$ . The sphere has equation $\rho =2$ . These give the $\rho$ bounds. To get $\phi$ , notice that the region starts at $\phi = 0$ , and it will go up to the intersection of the plane with the sphere, which is $\sec (\phi ) = 2$ , or $\phi = \pi /3$ . This gives the $\phi$ bounds. The setup is therefore $\int _0^{2\pi } \int _0^{\pi /3} \int _{\sec (\phi )}^2 \rho ^3\cos (\phi )\sin (\phi )d\rho \,d\phi \,d\theta .$

Let $R$ be the region bounded by $z = x+y$ , $z=0$ , $y = 0$ , $y=x$ , and $x=2$ . Then $\displaystyle \iiint _R (y+2z)dV = \answer {\frac {38}{3}}$ .

Easiest in Cartesian. One possible setup is $\int _0^2 \int _0^x \int _0^{x+y} (y+2z)dz\,dy\,dx.$

Let $R$ be the region in the first octant bounded by $z = x^2+y^2$ and $z = \sqrt {x^2+y^2}$ . Then $\displaystyle \iiint _R ydV = \answer {\frac {1}{20}}$

Easiest in cylindrical. It is tricky to find the intersection, particularly if you do $x^2+y^2 = \sqrt {x^2+y^2}$ . Another way is to substitute $z = \sqrt {x^2+y^2}$ into the other equation to get $z = z^2$ , so $z = 0$ or $z=1$ . In this case this means $0 \leq x^2+y^2 \leq 1$ . Thus, in cylindrical, we get $\int _0^{\pi /2} \int _0^1 \int _{r^2}^r r^2\sin (\theta )dz\,dr\,d\theta .$

Let $R$ be the solid given by $z \leq \sqrt {\frac {1}{3}(x^2+y^2)}$ and $x^2+y^2+z^2 \leq 2$ . The volume of $R$ is: $\answer {2\pi \sqrt {2}}$

The setup is $\int _0^{2\pi } \int _{\pi /3}^{\pi } \int _0^{\sqrt {2}} \rho ^2\sin (\phi ) d\rho \,d\phi \,d\theta .$

Let $R$ be the region given by $x^2+y^2+z^2 \leq 4z$ and $z \leq \sqrt {x^2+y^2}$ . Then $\displaystyle \iiint _R zdV =\answer {\frac {8\pi }{3}}$ .

The sphere has equation $\rho ^2 = 4\rho \cos (\phi )$ , or $\rho =4\cos (\phi )$ . Thus, the setup is $\int _0^{2\pi } \int _{\pi /4}^{\pi /2} \int _0^{4\cos (\phi )} \rho ^3\cos (\phi )\sin (\phi )d\rho \,d\phi \,d\theta .$