Spherical Coordinates

in ${\mathbb{R}}^3$

$\boxed{{\displaystyle \begin{array}{rl}T& :(\rho ,\varphi ,\theta )\mapsto (x,y,z)\\ T^{-1}& :(x,y,z)\mapsto (\rho ,\varphi ,\theta )\\ T(\rho ,\varphi ,\theta )& =(\rho \sin \varphi \cos \theta ,\rho \sin \varphi \sin \theta ,\rho \cos \varphi )\\ & =(x(\rho ,\varphi ,\theta ),y(\rho ,\varphi ,\theta ),z(\rho ,\varphi ,\theta ))& & \text{transformation}\\ \\ x(\rho ,\varphi ,\theta )=r\cos \theta & =\rho \sin \varphi \cos \theta \\ y(\rho ,\varphi ,\theta )=r\sin \theta & =\rho \sin \varphi \sin \theta \\ z(\rho ,\varphi ,\theta )& =\rho \cos \varphi & & \text{transformation equations}\\ \\ \sqrt{x^2+y^2}=r& =\rho \sin \varphi \\ x^2+y^2+z^2& ={\rho }^2& & \text{Pythagorean identity}\\ dV& ={\rho }^2\sin \varphi d\rho d\varphi d\theta & & \text{Jacobian}\end{array}}}$

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Revised

02 Apr 2023

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Spherical Coordinate System

Spherical coordinates are good to use for regions with more than one axis of rotational symmetry.

In other words, more than one of the cross sections of the region is circular.

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Transformation Equations $(x,y,z)\leftrightarrow (\rho ,\varphi ,\theta )$

$\begin{array}{rl}\rho \ge 0& \text{distance from the origin}\\ 0\le \varphi \le \pi & \text{angle from the positive z-axis to the negative z-axis}\\ 0\le \theta \le 2\pi & \text{angle from the positive x-axis in the xy-plane}\end{array}$

If we are thinking about the Earth, then $\theta$ is longitude, $\varphi$ is latitude, and $\rho$ is the distance from the center of the Earth.

$\begin{array}{rl}x(\rho ,\varphi ,\theta )& =r\cos \theta =\rho \sin \varphi \cos \theta \\ y(\rho ,\varphi ,\theta )& =r\sin \theta =\rho \sin \varphi \sin \theta \\ z(\rho ,\varphi ,\theta )& =\rho \cos \varphi \\ \sqrt{x^2+y^2}& =r=\rho \sin \varphi \\ x^2+y^2+z^2& ={\rho }^2\end{array}$

Infinitesimals

$d\rho d\varphi d\theta$-voxels in the rectangular $\rho \varphi \theta$-grid are transformed into 3D wedges $dV$ in the spherical $xyz$-grid.

$\begin{array}{r}dV\ne d\rho d\varphi d\theta \end{array}$

How big is spherical $dV$ in relation to rectangular $d\rho d\varphi d\theta$-voxel?

The further the $d\rho d\varphi d\theta$-voxel is in the $\rho$-direction in the rectangular $d\rho d\varphi d\theta$-grid the larger the 3D wedge $dV$ is in the spherical $xyz$-grid.

$\begin{array}{rlr}dV={\rho }^2\sin \varphi d\rho d\varphi d\theta & & \text{from the Jacobian}\end{array}$

Jacobian

$\begin{array}{rl}\frac{\partial (x,y,z)}{\partial (\rho ,\varphi ,\theta )}& =\left|\begin{array}{ccc}{\displaystyle \frac{\partial x}{\partial \rho }}& {\displaystyle \frac{\partial x}{\partial \varphi }}& {\displaystyle \frac{\partial x}{\partial \theta }}\\ \\ {\displaystyle \frac{\partial y}{\partial \rho }}& {\displaystyle \frac{\partial y}{\partial \varphi }}& {\displaystyle \frac{\partial y}{\partial \theta }}\\ \\ {\displaystyle \frac{\partial z}{\partial \rho }}& {\displaystyle \frac{\partial z}{\partial \varphi }}& {\displaystyle \frac{\partial z}{\partial \theta }}\end{array}\right|=\left|\begin{array}{ccc}\sin \varphi \cos \theta & \rho \cos \varphi \cos \theta & -\rho \sin \varphi \sin \theta \\ \\ \sin \varphi \sin \theta & \rho \cos \varphi \sin \theta & \rho \sin \varphi \cos \theta \\ \\ \cos \varphi & -\rho \sin \varphi & 0\end{array}\right|=\cos \varphi \left|\begin{array}{cc}\rho \cos \varphi \cos \theta & -\rho \sin \varphi \sin \theta \\ \\ \rho \cos \varphi \sin \theta & \rho \sin \varphi \cos \theta \end{array}\right|+\rho \sin \varphi \left|\begin{array}{cc}\sin \varphi \cos \theta & -\rho \sin \varphi \sin \theta \\ \\ \sin \varphi \sin \theta & \rho \sin \varphi \cos \theta \end{array}\right|\\ & =\cos \varphi ({\rho }^2\cos \varphi \sin \varphi {\cos }^2\theta +{\rho }^2\cos \varphi \sin \varphi {\sin }^2\theta )+\rho \sin \varphi (\rho {\sin }^2\varphi {\cos }^2\theta +\rho {\sin }^2\varphi {\sin }^2\theta )\\ & ={\rho }^2{\cos }^2\varphi \sin \varphi ({\cos }^2\theta +{\sin }^2\theta )+{\rho }^2{\sin }^3\varphi ({\cos }^2\theta +{\sin }^2\theta )\\ & ={\rho }^2\sin \varphi ({\cos }^2\varphi +{\sin }^2\varphi )\\ & ={\rho }^2\sin \varphi \end{array}$

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Examples

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[EXAMPLE]

An upper half ball of radius $\rho =3$ centered at the origin has a density $d$

$\begin{array}{r}d(x,y,z)=\frac{x^{2}z}{x^2+y^2+z^2}{\text{g cm}}^3\end{array}$

Calculate its mass.

$\boxed{{\displaystyle \text{mass}m=\frac{81\pi }{16}\text{g}}}$

$\begin{array}{r}\text{mass}m=\underset{R}{\iiint }d(x,y,z)dV=\underset{R}{\iiint }\frac{x^{2}z}{x^2+y^2+z^2}dxdydz\end{array}$

Draw the region in ${\mathbb{R}}^3$.

Find the new bounds.

$0\le \rho \le 3$

$\begin{array}{r}z\ge 0⟹\rho \cos \varphi \ge 0⟹\cos \varphi \ge 0⟹0\le \varphi \le \frac{\pi }{2}\text{where}0\le \varphi \le \pi \end{array}$

$0\le \theta \le 2\pi$

$\begin{array}{rl}\text{mass}m& =\underset{R}{\iiint }\frac{x^{2}z}{x^2+y^2+z^2}dxdydz\\ & =\underset{R}{\iiint }\frac{(\rho \sin \varphi \cos \theta {)}^2(\rho \cos \varphi )}{{\rho }^2}{\rho }^2\sin \varphi d\rho d\varphi d\theta =\underset{R}{\iiint }{\rho }^3{\sin }^3\varphi \cos \varphi {\cos }^2\theta d\rho d\varphi d\theta \\ & ={\int }_{\theta =0}^{\theta =2\pi }{\int }_{\varphi =0}^{\varphi =\frac{\pi }{2}}{\int }_{\rho =0}^{\rho =3}{\rho }^3{\sin }^3\varphi \cos \varphi {\cos }^2\theta d\rho d\varphi d\theta \\ & ={\int }_{\theta =0}^{\theta =2\pi }{\cos }^2\theta d\theta {\int }_{\varphi =0}^{\varphi =\frac{\pi }{2}}{\sin }^3\varphi \cos \varphi d\varphi {\int }_{\rho =0}^{\rho =3}{\rho }^{3}d\rho & & \text{by factoring}\\ & =\pi {\int }_{\varphi =0}^{\varphi =\frac{\pi }{2}}{\sin }^3\varphi \cos \varphi d\varphi {\int }_{\rho =0}^{\rho =3}{\rho }^{3}d\rho & & \pi ={\int }_{\theta =0}^{\theta =2\pi }{\cos }^2\theta d\theta \text{by trigonometric symmetry}\\ & =\frac{\pi }{4}{\int }_{\rho =0}^{\rho =3}{\rho }^{3}d\rho & & \frac{1}{4}={\int }_{\varphi =0}^{\varphi =\frac{\pi }{2}}{\sin }^3\varphi \cos \varphi d\varphi \text{by u-substitution}\\ & =\frac{81\pi }{16}& & \frac{3^4}{4}={\int }_{\rho =0}^{\rho =3}{\rho }^{3}d\rho \end{array}$

$\begin{array}{rl}& {\int }_{\varphi =0}^{\varphi =\frac{\pi }{2}}{\sin }^3\varphi \cos \varphi d\varphi \\ u& =\sin \varphi \\ du& =\cos \varphi d\varphi \\ \varphi & =0⟹u=0\\ \varphi & =\frac{\pi }{2}⟹u=1\\ & {\int }_{\varphi =0}^{\varphi =\frac{\pi }{2}}{\sin }^3\varphi \cos \varphi d\varphi ={\int }_{u=0}^{u=1}{u}^{3}du={\left(\frac{1}{4}{u}^4\right)|}_{u=0}^{u=1}=\frac{1}{4}\end{array}$

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[EXAMPLE]

$\begin{array}{r}\frac{64\pi }{5}(\frac{1}{3}-\frac{1}{3\sqrt[3]{2}})=\underset{R}{\iiint }{z}^{2}dV\text{where}R=\{(x,y,z)\mid x^2+y^2+z^2\le 4,z\le \sqrt{x^2+y^2}\}\end{array}$

Draw the region in ${\mathbb{R}}^3$.

Cone with a spherical top

• Sphere of radius 2 at the origin $x^2+y^2+z^2=4$

• Cone $z=\sqrt{x^2+y^2}$

Find the new bounds.

$x^2+y^2+z^2\le 4⟹{\rho }^2\le 4⟹\rho \le 2\text{where}\rho \ge 0$

$\begin{array}{r}z\le \sqrt{x^2+y^2}⟹\rho \cos \varphi \le \rho \sin \varphi ⟹\cos \varphi \le \sin \varphi ⟹\cot \varphi \le 1\le \tan \varphi ⟹\varphi =\frac{\pi }{4}\text{where}0\le \varphi \le \pi \end{array}$

$0\le \theta \le 2\pi$

$\begin{array}{rl}\underset{R}{\iiint }{z}^{2}dV& =\underset{R}{\iiint }(\rho \cos \varphi {)}^2{\rho }^2\sin \varphi d\rho d\varphi d\theta =\underset{R}{\iiint }{\rho }^4{\cos }^2\varphi \sin \varphi d\rho d\varphi d\theta \\ & ={\int }_{\theta =0}^{\theta =2\pi }{\int }_{\varphi =0}^{\varphi =\frac{\pi }{4}}{\int }_{\rho =0}^{\rho =2}{\rho }^4{\cos }^2\varphi \sin \varphi d\rho d\varphi d\theta \\ & ={\int }_{\theta =0}^{\theta =2\pi }d\theta {\int }_{\varphi =0}^{\varphi =\frac{\pi }{4}}{\cos }^2\varphi \sin \varphi d\varphi {\int }_{\rho =0}^{\rho =2}{\rho }^{4}d\rho & & \text{by factoring}\\ & =\frac{64\pi }{5}{\int }_{\varphi =0}^{\varphi =\frac{\pi }{4}}{\cos }^2\varphi \sin \varphi d\varphi & & \frac{2^5}{5}={\int }_{\rho =0}^{\rho =2}{\rho }^{4}d\rho \\ & =\frac{64\pi }{5}(\frac{1}{3}-\frac{1}{3\sqrt[3]{2}})& & \frac{1}{3}-\frac{1}{3\sqrt[3]{2}}={\int }_{\varphi =0}^{\varphi =\frac{\pi }{4}}{\cos }^2\varphi \sin \varphi d\varphi \\ & \approx 8.6651\end{array}$

import numpy as np

64*np.pi/5*((1/3)-(1/(3*2**(3/2))))

8.66505352128086

$\begin{array}{rl}& {\int }_0^{\frac{\pi }{4}}{\cos }^2\theta \sin \theta d\theta \\ u& =\cos \theta \\ du& =-\sin \theta d\theta ⟺-du=\sin \theta d\theta \\ \theta & =0⟹u=1\\ \theta & =\frac{\pi }{4}⟹u=\frac{1}{\sqrt{2}}\\ & {\int }_0^{\frac{\pi }{4}}{\cos }^2\theta \sin \theta d\theta ={\int }_{\frac{1}{\sqrt{2}}}^1{u}^{2}du={\left(\frac{1}{3}{u}^3\right)|}_{\frac{1}{\sqrt{2}}}^1=\frac{1}{3}-\frac{1}{3}{\left(\frac{1}{\sqrt{2}}\right)}^3=\frac{1}{3}-\frac{1}{3\sqrt[3]{2}}\approx 0.2155\end{array}$

((1/3)-(1/(3*2**(3/2))))

0.21548220313557542

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[EXAMPLE]

$\begin{array}{r}\underset{R}{\iiint }\sqrt{x^2+y^2+z^2}dV\text{where}R=\{(x,y,z)\mid x^2+y^2+z^2\le 4,z\ge 1\}\end{array}$

Draw the region in ${\mathbb{R}}^3$.

Region bounded by

• Sphere of radius 2

• Plane parallel to the $xy$-plane at $z=1$

Find the new bounds.

$x^2+y^2+z^2\le 4⟺{\rho }^2\le 4⟺\rho \le 2\text{where}\rho \ge 0$

equation of the sphere $\rho =2$

$z\ge 1⟺\rho \cos \varphi \ge 1⟺\rho \ge \sec \varphi$

equation of the plane $\rho =\sec \varphi$

$\sec \varphi \le \rho \le 2$

$\begin{array}{r}2=\sec \varphi ⟺\cos \varphi =\frac{1}{2}⟺\varphi =\frac{\pi }{3}\end{array}$

$\begin{array}{r}0\le \varphi \le \frac{\pi }{3}\end{array}$

$0\le \theta \le 2\pi$

$\begin{array}{rl}\underset{R}{\iiint }\sqrt{x^2+y^2+z^2}dV& =\underset{R}{\iiint }{\rho }^3\sin \varphi d\rho d\varphi d\theta \\ & ={\int }_{\theta =0}^{\theta =2\pi }{\int }_{\varphi =0}^{\varphi =\frac{\pi }{3}}{\int }_{\rho =\sec \varphi }^{\rho =2}{\rho }^3\sin \varphi d\rho d\varphi d\theta \\ & ={\int }_{\theta =0}^{\theta =2\pi }d\theta {\int }_{\varphi =0}^{\varphi =\frac{\pi }{3}}{\int }_{\rho =\sec \varphi }^{\rho =2}{\rho }^3\sin \varphi d\rho d\varphi & & \text{by factoring}\\ \\ & =2\pi {\int }_{\varphi =0}^{\varphi =\frac{\pi }{3}}{\int }_{\rho =\sec \varphi }^{\rho =2}{\rho }^3\sin \varphi d\rho d\varphi \\ & =2\pi {\int }_{\varphi =0}^{\varphi =\frac{\pi }{3}}{(\frac{1}{4}{\rho }^4\sin \varphi )|}_{\rho =\sec \varphi }^{\rho =2}d\varphi \\ & =\frac{\pi }{2}{\int }_{\varphi =0}^{\varphi =\frac{\pi }{3}}(16\sin \varphi -{\sec }^4\varphi \sin \varphi )d\varphi \\ & =8\pi {\int }_{\varphi =0}^{\varphi =\frac{\pi }{3}}\sin \varphi d\varphi -\frac{\pi }{2}{\int }_{\varphi =0}^{\varphi =\frac{\pi }{3}}{\sec }^4\varphi \sin \varphi d\varphi \\ & =4\pi -\frac{\pi }{2}{\int }_{\varphi =0}^{\varphi =\frac{\pi }{3}}{\sec }^4\varphi \sin \varphi d\varphi & & \frac{1}{2}={\int }_{\varphi =0}^{\varphi =\frac{\pi }{3}}\sin \varphi d\varphi \\ & =4\pi -\frac{7\pi }{6}& & \frac{7}{3}={\int }_0^{\frac{\pi }{3}}{\sec }^4\theta \sin \theta d\theta \\ & =\frac{17\pi }{6}\end{array}$

$\begin{array}{rl}{\int }_0^{\frac{\pi }{3}}{\sec }^4\theta \sin \theta d\theta & ={\int }_0^{\frac{\pi }{3}}\frac{1}{{\cos }^4\theta }\sin \theta d\theta \\ u& =\cos \theta \\ du& =-\sin \theta d\theta ⟺-du=\sin \theta d\theta \\ \theta & =0⟹u=1\\ \theta & =\frac{\pi }{3}⟹u=\frac{1}{2}\\ {\int }_0^{\frac{\pi }{3}}{\sec }^4\theta \sin \theta d\theta & ={\int }_{\frac{1}{2}}^1{u}^{-4}du\\ & ={(-\frac{1}{3}{u}^{-3})|}_{\frac{1}{2}}^1=-\frac{1}{3}(1{)}^{-3}+\frac{1}{3}{\left(\frac{1}{2}\right)}^{-3}=\frac{7}{3}\end{array}$

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GeoGebra

R=3

surface(Rsin(p)cos(t),Rsin(p)sin(t),Rcos(p),p,0,P,t,0,T)

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Terms

• [W] Determinant

• [W] Fundamental Plane

• [W] Jacobian

• [W] Spherical Coordinate System

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