How to Integrate in Spherical Coordinates

Last Updated: December 23, 2021

Integration in spherical coordinates is typically done when we are dealing with spheres or spherical objects. A massive advantage in this coordinate system is the almost complete lack of dependency amongst the variables, which allows for easy factoring in most cases.

This article will use the mathematician's convention of labeling coordinates $(\rho ,\theta ,\phi ),$ where $\rho$ is the radial distance, $\theta$ is the azimuthal angle, and $\phi$ is the polar angle. In physics, the angles are switched (but are still written out in that order).

Steps

1
Recall the coordinate conversions. Coordinate conversions exist from Cartesian to spherical and from cylindrical to spherical. Below is a list of conversions from Cartesian to spherical. Above is a diagram with point $P$ described in spherical coordinates.
- ${\begin{aligned}x&=\rho \sin \phi \cos \theta \\y&=\rho \sin \phi \sin \theta \\z&=\rho \cos \phi \\\rho ^{{2}}&=x^{{2}}+y^{{2}}+z^{{2}}\end{aligned}}$
- In the example where we calculate the moment of inertia of a ball, $x^{{2}}+y^{{2}}=\rho ^{{2}}\sin ^{{2}}\phi$ will be useful. Make sure you know why this is the case.
2
Set up the coordinate-independent integral. We are dealing with volume integrals in three dimensions, so we will use a volume differential ${\mathrm {d}}V$ and integrate over a volume $V.$
- $\int _{{V}}{\mathrm {d}}V$
- Most of the time, you will have an expression in the integrand. If so, make sure that it is in spherical coordinates.
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3
Set up the volume element.
- ${\mathrm {d}}V=\rho ^{{2}}\sin \phi {\mathrm {d}}\rho {\mathrm {d}}\phi {\mathrm {d}}\theta$
- Those familiar with polar coordinates will understand that the area element ${\mathrm {d}}A=r{\mathrm {d}}r{\mathrm {d}}\theta .$ This extra r stems from the fact that the side of the differential polar rectangle facing the angle has a side length of $r{\mathrm {d}}\theta$ to scale to units of distance. A similar thing is occurring here in spherical coordinates.
4
Set up the boundaries. Choose a coordinate system that allows for the easiest integration.
- Notice that $\phi$ has a range of $[0,\pi ],$ not $[0,2\pi ].$ This is because $\theta$ already has a range of $[0,2\pi ],$ so the range of $\phi$ ensures that we don't integrate over a volume twice.
5

Integrate. Once everything is set up in spherical coordinates, simply integrate using any means possible and evaluate.

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Method 1

Method 1 of 2:

Volume of a Sphere

1
Calculate the volume of a sphere of radius r.
- Choose a coordinate system such that the center of the sphere rests on the origin.
- ${\begin{aligned}\int _{{V}}\rho ^{{2}}\sin \phi {\mathrm {d}}\rho {\mathrm {d}}\phi {\mathrm {d}}\theta &=\int _{{0}}^{{r}}\rho ^{{2}}{\mathrm {d}}\rho \int _{{0}}^{{\pi }}\sin \phi {\mathrm {d}}\phi \int _{{0}}^{{2\pi }}{\mathrm {d}}\theta \\&=\left({\frac {1}{3}}r^{{3}}\right)(-\cos \pi +\cos 0)(2\pi )\\&={\frac {4}{3}}\pi r^{{3}}\end{aligned}}$

Method 2

Method 2 of 2:

Moment of Inertia of a Ball

1

Calculate the moment of inertia of a ball. Assume this ball has a mass $M,$ radius $R,$ and a constant density $\sigma .$ Most moment of inertia questions are written with answers in terms of $M$ and $R.$
2
Recall the moment of inertia formula.
- $I=\int _{{M}}r^{{2}}{\mathrm {d}}m,$ where $r={\sqrt {x^{{2}}+y^{{2}}}}$ is the perpendicular distance from the axis (we are choosing the z-axis) and we are integrating over the mass $M.$
3
Recall the relationship between mass, volume, and density when density is constant.
- $\sigma ={\frac {M}{V}}.$ Of course, we know the volume of the sphere, so $\sigma ={\frac {3M}{4\pi R^{{3}}}}.$
4
Rewrite the moment of inertia in terms of a volume integral, then solve. Note constants that factor out.
- ${\mathrm {d}}m=\sigma {\mathrm {d}}V,$ so therefore,
- ${\begin{aligned}I_{{z}}&=\int _{{V}}(x^{{2}}+y^{{2}}){\frac {3M}{4\pi R^{{3}}}}{\mathrm {d}}V\\&={\frac {3M}{4\pi R^{{3}}}}\int _{{0}}^{{R}}\rho ^{{2}}{\mathrm {d}}\rho \int _{{0}}^{{\pi }}\sin \phi {\mathrm {d}}\phi \int _{{0}}^{{2\pi }}{\mathrm {d}}\theta \rho ^{{2}}\sin ^{{2}}\phi \\&={\frac {3M}{4\pi R^{{3}}}}\int _{{0}}^{{R}}\rho ^{{4}}{\mathrm {d}}\rho \int _{{0}}^{{\pi }}\sin ^{{3}}\phi {\mathrm {d}}\phi \int _{{0}}^{{2\pi }}{\mathrm {d}}\theta \\&={\frac {3M}{4\pi R^{{3}}}}\left({\frac {1}{5}}R^{{5}}\right)(2\pi )\int _{{0}}^{{\pi }}(1-\cos ^{{2}}\phi )\sin \phi {\mathrm {d}}\phi ,u=\cos \phi \\&={\frac {3}{10}}MR^{{2}}\int _{{-1}}^{{1}}(1-u^{{2}}){\mathrm {d}}u\\&={\frac {3}{10}}MR^{{2}}(2)\left(1-{\frac {1}{3}}\right)\\&={\frac {2}{5}}MR^{{2}}\end{aligned}}$
- Note that in the step where the integral is written in terms of $u,$ the integrand is an even function. Therefore, we can factor out a 2 and set the lower boundary to 0 to simplify calculations.
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Tips

Do not use the variable of integration in the integral boundary, as this makes the process very unclear and will lead to confusion. Distinguish your variables with uppercase/lowercase letters, tilde “squiggly” signs above the variable of integration, or even different letters.

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One can make use of Jacobian change of variables to show that, with regards to the differential element, there exists an extra $\rho ^{{2}}\sin \phi$ in spherical coordinates.

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How to Integrate in Spherical Coordinates

Steps

Volume of a Sphere

Moment of Inertia of a Ball

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