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

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. \(\displaystyle\int_0^{2\pi} \displaystyle\int_0^{\pi/6} \displaystyle\int_0^{2\sqrt{3}/\cos\phi} \rho^2\sin\phi\ d\rho\ d\phi\ d\theta = \dfrac{8\pi}{3}(9\sqrt{3}-8)\)