How to Integrate in Spherical Coordinates

Steps

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 {\displaystyle P} P described in spherical coordinates. x = ρ sin ⁡ ϕ cos ⁡ θ y = ρ sin ⁡ ϕ sin ⁡ θ z = ρ cos ⁡ ϕ ρ 2 = x 2 + y 2 + z 2 {\displaystyle {\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}}} {\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 = ρ 2 sin 2 ⁡ ϕ {\displaystyle x^{2}+y^{2}=\rho ^{2}\sin ^{2}\phi } x^{{2}}+y^{{2}}=\rho ^{{2}}\sin ^{{2}}\phi will be useful. Make sure you know why this is the case.

Set up the coordinate-independent integral. We are dealing with volume integrals in three dimensions, so we will use a volume differential d V {\displaystyle \mathrm {d} V} {\mathrm {d}}V and integrate over a volume V . {\displaystyle V.} V. ∫ V d V {\displaystyle \int _{V}\mathrm {d} 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.

Set up the volume element. d V = ρ 2 sin ⁡ ϕ d ρ d ϕ d θ {\displaystyle \mathrm {d} V=\rho ^{2}\sin \phi \mathrm {d} \rho \mathrm {d} \phi \mathrm {d} \theta } {\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 d A = r d r d θ . {\displaystyle \mathrm {d} A=r\mathrm {d} r\mathrm {d} \theta .} {\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 d θ {\displaystyle r\mathrm {d} \theta } r{\mathrm {d}}\theta to scale to units of distance. A similar thing is occurring here in spherical coordinates.

Set up the boundaries. Choose a coordinate system that allows for the easiest integration. Notice that ϕ {\displaystyle \phi } \phi has a range of [ 0 , π ] , {\displaystyle [0,\pi ],} [0,\pi ], not [ 0 , 2 π ] . {\displaystyle [0,2\pi ].} [0,2\pi ]. This is because θ {\displaystyle \theta } \theta already has a range of [ 0 , 2 π ] , {\displaystyle [0,2\pi ],} [0,2\pi ], so the range of ϕ {\displaystyle \phi } \phi ensures that we don't integrate over a volume twice.

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

Moment of Inertia of a Ball

Calculate the moment of inertia of a ball. Assume this ball has a mass M , {\displaystyle M,} M, radius R , {\displaystyle R,} R, and a constant density σ . {\displaystyle \sigma .} \sigma . Most moment of inertia questions are written with answers in terms of M {\displaystyle M} M and R . {\displaystyle R.} R.

Recall the moment of inertia formula. I = ∫ M r 2 d m , {\displaystyle I=\int _{M}r^{2}\mathrm {d} m,} I=\int _{{M}}r^{{2}}{\mathrm {d}}m, where r = x 2 + y 2 {\displaystyle r={\sqrt {x^{2}+y^{2}}}} 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 . {\displaystyle M.} M.

Recall the relationship between mass, volume, and density when density is constant. σ = M V . {\displaystyle \sigma ={\frac {M}{V}}.} \sigma ={\frac {M}{V}}. Of course, we know the volume of the sphere, so σ = 3 M 4 π R 3 . {\displaystyle \sigma ={\frac {3M}{4\pi R^{3}}}.} \sigma ={\frac {3M}{4\pi R^{{3}}}}.

Rewrite the moment of inertia in terms of a volume integral, then solve. Note constants that factor out. d m = σ d V , {\displaystyle \mathrm {d} m=\sigma \mathrm {d} V,} {\mathrm {d}}m=\sigma {\mathrm {d}}V, so therefore, I z = ∫ V ( x 2 + y 2 ) 3 M 4 π R 3 d V = 3 M 4 π R 3 ∫ 0 R ρ 2 d ρ ∫ 0 π sin ⁡ ϕ d ϕ ∫ 0 2 π d θ ρ 2 sin 2 ⁡ ϕ = 3 M 4 π R 3 ∫ 0 R ρ 4 d ρ ∫ 0 π sin 3 ⁡ ϕ d ϕ ∫ 0 2 π d θ = 3 M 4 π R 3 ( 1 5 R 5 ) ( 2 π ) ∫ 0 π ( 1 − cos 2 ⁡ ϕ ) sin ⁡ ϕ d ϕ , u = cos ⁡ ϕ = 3 10 M R 2 ∫ − 1 1 ( 1 − u 2 ) d u = 3 10 M R 2 ( 2 ) ( 1 − 1 3 ) = 2 5 M R 2 {\displaystyle {\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}}} {\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 , {\displaystyle u,} u, the integrand is an even function. Therefore, we can factor out a 2 and set the lower boundary to 0 to simplify calculations.