Precalculus

Analytic Geometry

Analytic Geometry Logo

Spherical Coordinates

To introduce spherical coordinates in space, consider three mutually perpendicular x-, y-, and z- axes with a common origin O. Let M be any point in space other than the point O and N be its projection onto the xy-plane. Further denote: ρ is the distance from the origin O to point M (the radial distance or radius), θ is the angle (also called the polar or zenith angle) that forms the directed line segment OM with the z-axis, and φ is the angle (azimuth) by which you need to turn the x-axis counterclockwise to coincide with the ray ON.

Spherical coordinate system
Figure 1.

Note that \(\rho\) is defined differently in polar and spherical coordinates. In polar (or cylindrical) coordinates, the distance \(\rho\) is defined in the \(xy-\)plane, while in the spherical coordinate system it is measured in space.

The three numbers \(\rho, \varphi, \theta\) are called the spherical coordinates of the point \(M.\) The name "spherical coordinates" is due to the fact that the coordinate surface \(\rho = const\) (that is, the surface with all points which are equidistant from the origin) is a sphere.

To provide a one-to-one correspondence between points in space and triplets of spherical coordinates \(\left({\rho, \varphi, \theta}\right),\) it is usually assumed that \(\rho\) and \(\varphi\) change within the following limits:

\[0 \le \rho \lt \infty,\;0 \le \varphi \lt 2\pi.\]

By definition, the \(\theta-\)coordinate lies in the range \(0 \le \theta \le \pi.\)

If point M coincides with point O, then \(\rho = 0.\) For point O, the \(\varphi-\) and \(\theta-\)coordinates are not defined.

Converting Between Spherical and Cartesian Coordinates

If you choose the axes of the Cartesian coordinate system as indicated above in the figure, then the Cartesian coordinates \(x, y, z\) of a point are related to its spherical coordinates \({\rho, \varphi, \theta}\) by the relations

\[x = \rho\sin\theta\cos\varphi,\;\;y = \rho\sin\theta\sin\varphi,\;\;z = \rho\cos\theta.\]

For converting a point from Cartesian coordinates to spherical coordinates, we use the formulas

\[\rho = \sqrt{x^2 + y^2 + z^2},\;\;\tan\varphi = \frac{y}{x},\;\;\cos\theta = \frac{z}{\rho} = \frac{z}{\sqrt{x^2 + y^2 + z^2}}.\]

Solved Problems

Click or tap a problem to see the solution.

Example 1

Find the spherical coordinates of each point given its Cartesian coordinates.

  1. \(A\left({1,0,0}\right)\)
  2. \(B\left({-4,-3,0}\right)\)
  3. \(C\left({2,2,1}\right)\)
  4. \(D\left({8,-4,1}\right)\)

Example 2

Convert from spherical to Cartesian coordinates.

  1. \(K\left({\sqrt{2},\frac{\pi}{4},\frac{\pi}{6}}\right)\)
  2. \(L\left({3\sqrt{2},\frac{3\pi}{4},\frac{\pi}{2}}\right)\)
  3. \(M\left({5,\frac{3\pi}{2},\frac{\pi}{2}}\right)\)

Example 1.

Find the spherical coordinates of each point given its Cartesian coordinates.

  1. \(A\left({1,0,0}\right)\)
  2. \(B\left({-4,-3,0}\right)\)
  3. \(C\left({2,2,1}\right)\)
  4. \(D\left({8,-4,1}\right)\)

Solution.

We use the following formulas:

\[\rho = \sqrt{x^2 + y^2 + z^2},\;\;\tan\varphi = \frac{y}{x},\;\;\cos\theta = \frac{z}{\rho}.\]
  1. \(A\left({1,0,0}\right)\)
    \[\rho = \sqrt{1^2 + 0^2 + 0^2} = 1,\;\tan\varphi = \frac{0}{1} = 0, \Rightarrow \varphi = 0,\;\cos\theta = \frac{0}{1} = 0, \Rightarrow \theta = \frac{\pi}{2};\]
    \[A\left({1,0,0}\right) \mapsto A\left({1,0,\frac{\pi}{2}}\right);\]
  2. \(B\left({-4,-3,0}\right)\)
    \[\rho = \sqrt{\left({-4}\right)^2 + \left({-3}\right)^2 + 0^2} = 5,\;\tan\varphi = \frac{-3}{-4} = \frac{3}{4}, \Rightarrow \varphi = \arctan\frac{3}{4},\;\cos\theta = \frac{0}{5} = 0, \Rightarrow \theta = \frac{\pi}{2};\]
    \[B\left({-4,-3,0}\right) \mapsto B\left({5,\arctan\frac{3}{4},\frac{\pi}{2}}\right);\]
  3. \(C\left({2,2,1}\right)\)
    \[\rho = \sqrt{2^2 + 2^2 + 1^2} = 3,\;\tan\varphi = \frac{2}{2} = 1, \Rightarrow \varphi = \frac{\pi}{4},\;\cos\theta = \frac{1}{3}, \Rightarrow \theta = \arccos\frac{1}{3};\]
    \[C\left({2,2,1}\right) \mapsto C\left({3,\frac{\pi}{4},\arccos\frac{1}{3}}\right);\]
  4. \(D\left({8,-4,1}\right)\)
    \[\rho = \sqrt{8^2 + \left({-4}\right)^2 + 1^2} = 9,\;\tan\varphi = \frac{-4}{8} = -\frac{1}{2}, \Rightarrow \varphi = \arctan\left({-\frac{1}{2}}\right) = -\arctan\frac{1}{2},\;\cos\theta = \frac{1}{9}, \Rightarrow \theta = \arccos\frac{1}{9};\]
    \[D\left({8,-4,1}\right) \mapsto D\left({9,-\arctan\frac{1}{2},\arccos\frac{1}{9}}\right).\]

Example 2.

Convert from spherical to Cartesian coordinates.

  1. \(K\left({\sqrt{2},\frac{\pi}{4},\frac{\pi}{6}}\right)\)
  2. \(L\left({3\sqrt{2},\frac{3\pi}{4},\frac{\pi}{2}}\right)\)
  3. \(M\left({5,\frac{3\pi}{2},\frac{\pi}{2}}\right)\)

Solution.

To translate between spherical and cartesian (rectangular) coordinates, we use the formulas

\[x = \rho\sin\theta\cos\varphi,\;\;y = \rho\sin\theta\sin\varphi,\;\;z = \rho\cos\theta.\]
  1. \(K\left({\sqrt{2},\frac{\pi}{4},\frac{\pi}{6}}\right)\)
    \[x = \sqrt{2}\cdot\sin\frac{\pi}{6}\cdot\cos\frac{\pi}{4} = \sqrt{2}\cdot\frac{1}{2}\cdot\frac{\sqrt{2}}{2} = \frac{1}{2},\]
    \[y = \sqrt{2}\cdot\sin\frac{\pi}{6}\cdot\sin\frac{\pi}{4} = \sqrt{2}\cdot\frac{1}{2}\cdot\frac{\sqrt{2}}{2} = \frac{1}{2},\]
    \[z = \sqrt{2}\cdot\cos\frac{\pi}{6} = \sqrt{2}\cdot\frac{\sqrt{3}}{2} = \frac{\sqrt{6}}{2};\]
    \[K\left({\sqrt{2},\frac{\pi}{4},\frac{\pi}{6}}\right) \mapsto K\left({\frac{1}{2},\frac{1}{2},\frac{\sqrt{6}}{2}}\right);\]
  2. \(L\left({3\sqrt{2},\frac{3\pi}{4},\frac{\pi}{2}}\right)\)
    \[x = 3\sqrt{2}\cdot\sin\frac{\pi}{2}\cdot\cos\frac{3\pi}{4} = 3\sqrt{2}\cdot1\cdot\left({-\frac{\sqrt{2}}{2}}\right) = -3,\]
    \[y = 3\sqrt{2}\cdot\sin\frac{\pi}{2}\cdot\sin\frac{3\pi}{4} = 3\sqrt{2}\cdot1\cdot\frac{\sqrt{2}}{2} = 3,\]
    \[z = 3\sqrt{2}\cdot\cos\frac{\pi}{2} = 3\sqrt{2}\cdot0 = 0;\]
    \[L\left({3\sqrt{2},\frac{3\pi}{4},\frac{\pi}{2}}\right) \mapsto L\left({-3,3,0}\right);\]
  3. \(M\left({5,\frac{3\pi}{2},\frac{\pi}{2}}\right)\)
    \[x = 5\cdot\sin\frac{\pi}{2}\cdot\cos\frac{3\pi}{2} = 5\cdot1\cdot0 = 0,\]
    \[y = 5\cdot\sin\frac{\pi}{2}\cdot\sin\frac{3\pi}{2} = 5\cdot1\cdot\left({-1}\right) = -5,\]
    \[z = 5\cdot\cos\frac{\pi}{2} = 5\cdot0 = 0;\]
    \[M\left({5,\frac{3\pi}{2},\frac{\pi}{2}}\right) \mapsto M\left({0,-5,0}\right).\]