$$\def\ans#1{\bbox[border:1px solid green,6pt]{#1}}$$

# Triple Integrals in Spherical Coordinates

## Introduction

Recall what we know about spherical coordinates:

\begin{align*} x &= \rho \cos \theta \sin \phi\\ y &= \rho \sin \theta \sin \phi\\ z &= \rho \cos \phi\\ \\ \rho^2 &= x^2+y^2+y^2 \end{align*}

Therefore, a triple integral in rectangular coordinates can be rewritten in terms of spherical coordinates:

$\iiint_D f(x,y,z)\ dV = \iiint_D f(\rho, \phi, \theta)\ \rho^2 \sin \phi\ d\rho\ d\phi\ d\theta$

We'll tend to use spherical coordinates when we encounter a triple integral with $$x^2+y^2+z^2$$ somewhere.

## Examples

Convert the following integral to spherical coordinates and evaluate. $\iiint_D (x^2+y^2+z^2)^{-3/2}\ dV$ where $$D$$ is the region in the first octant between two spheres of radius 1 and 2 centered at the origin.

#### Solution

\begin{align} \iiint_D &(x^2+y^2+z^2)^{-3/2}\ dV\\ &= \int_0^{\pi/2} \int_0^{\pi/2} \int_1^2 \rho^{-3} \cdot \rho^2 \sin\phi\ d\rho\ d\phi\ d\theta\\ &= \int_0^{\pi/2} \int_0^{\pi/2} \sin \phi(\ln \rho) \bigg|_1^2 \ d\phi\ d\theta\\ &= \int_0^{\pi/2} \int_0^{\pi/2} \sin \phi \cdot \ln 2 \ d\phi\ d\theta\\ &= \int_0^{\pi/2} -\ln 2 \cdot \cos \phi\bigg|_0^{\pi/2} \ d\theta\\ &= \int_0^{\pi/2} \ln 2 \ d\theta = \ans{\dfrac{\pi}{2}\cdot \ln 2} \end{align}

Find the volume of the solid $$D$$ that lies inside the cone $$\phi = \dfrac{\pi}{6}$$ and inside the sphere $$\rho = 4$$.

#### Solution

\begin{align} V &= \int_0^{2\pi} \int_0^{\pi/6} \int_0^4 \rho^2 \sin\phi\ d\rho\ d\phi\ d\theta\\ &= \int_0^{2\pi} \int_0^{\pi/6} \dfrac{64}{3} \sin\phi\ d\phi\ d\theta\\ &= \int_0^{2\pi} -\dfrac{64}{3}\left(\dfrac{\sqrt{3}}{2}-1\right)\ d\theta\\ &= \ans{\dfrac{128\pi}{3}-\dfrac{64\pi\sqrt{3}}{3}} \end{align}

#### Try it yourself:

(click on a problem to show/hide its answer)

1. Use spherical coordinates to evaluate $$\displaystyle\iiint_D (x^2+y^2+z^2)^{5/2}\ dV$$, where $$D$$ is the unit ball.
2. $$\displaystyle\int_0^{2\pi} \displaystyle\int_0^{\pi} \displaystyle\int_0^{1} \rho^5\cdot\rho^2\sin\phi\ d\rho\ d\phi\ d\theta = \dfrac{\pi}{2}$$

3. Use spherical coordinates to evaluate $$\displaystyle\iiint_D \dfrac{1}{(x^2+y^2+z^2)^{3/2}}\ dV$$, where $$D$$ lies between spheres of radius 1 and 2 centered at the origin.
4. $$\displaystyle\int_0^{2\pi} \displaystyle\int_0^{\pi} \displaystyle\int_1^{2} \rho^{-3}\cdot\rho^2\sin\phi\ d\rho\ d\phi\ d\theta = 4\pi \ln 2$$

5. Use spherical coordinates to evaluate $$\displaystyle\int_{0}^{\pi} \displaystyle\int_{0}^{\pi/6} \displaystyle\int_{2\sec \phi}^{4} \rho^2\sin\phi\ d\rho\ d\phi\ d\theta$$.
6. $$\pi\left(\dfrac{188}{9}-\dfrac{32\sqrt{3}}{3}\right)$$

7. Use spherical coordinates to find the volume of the solid bounded by the sphere $$\rho=2\cos\phi$$ and the hemisphere $$\rho=1$$, $$z \geq 0$$.
8. $$\displaystyle\int_0^{2\pi} \displaystyle\int_0^{\pi/3} \displaystyle\int_0^{1} \rho^2\sin\phi\ d\rho\ d\phi\ d\theta = \dfrac{5\pi}{12}$$

9. Use spherical coordinates to find the volume of the solid outside the cone $$\phi=\dfrac{\pi}{4}$$ and inside the sphere $$\rho=4 \cos \phi$$.
10. $$\displaystyle\int_0^{2\pi} \displaystyle\int_{\pi/4}^{\pi/2} \displaystyle\int_0^{4\cos\phi} \rho^2\sin\phi\ d\rho\ d\phi\ d\theta = \dfrac{8\pi}{3}$$

11. Use spherical coordinates to find the volume of the part of the ball $$\rho \leq 4$$ that lies between the planes $$z=2$$ and $$z=2\sqrt{3}$$.
12. $$\displaystyle\int_0^{2\pi} \displaystyle\int_0^{\pi/6} \displaystyle\int_0^{2\sqrt{3}/\cos\phi} \rho^2\sin\phi\ d\rho\ d\phi\ d\theta = \dfrac{8\pi}{3}(9\sqrt{3}-8)$$