# 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.

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

### 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. $$M\left({5,\frac{3\pi}{2},\frac{\pi}{2}}\right)$$
$x = 5\cdot\sin\frac{\pi}{2}\cdot\cos\frac{3\pi}{2} = 5\cdot1\cdot0 = 0,$
$y = 5\cdot\sin\frac{\pi}{2}\cdot\sin\frac{3\pi}{2} = 5\cdot1\cdot\left({-1}\right) = -5,$
$z = 5\cdot\cos\frac{\pi}{2} = 5\cdot0 = 0;$
$M\left({5,\frac{3\pi}{2},\frac{\pi}{2}}\right) \mapsto M\left({0,-5,0}\right).$

### Example 3.

Express city's location in spherical coordinates given its latitude and longitude:

1. London $$51.5^{\circ}\,N, 0.1^{\circ} W$$
2. Rio de Janeiro $$22.9^{\circ}\,S, 43.2^{\circ} W$$

Solution.

1. London lies $$51.5^{\circ}$$ north of the equator. Assuming that the vertical axis is directed towards the North Pole, the $$\theta$$ angle is equal $$\theta = 90^\circ - 51.5^{\circ} = 38.5^{\circ}.$$ Movement to the west corresponds to a negative angle, so we have $$\varphi = -0.1^\circ$$. The radius of the earth is known and is $$\rho = 6370\,\text{km}.$$ Thus the spherical coordinates of London are
$\rho = 6370\,\text{km},\;\varphi = -0.1^\circ,\;\theta = 38.5^{\circ}.$
2. Rio de Janeiro lies $$22.9^{\circ}$$ south of the equator. Therefore we should add $$90^\circ$$ to its latitude. This gives $$\theta = 90^\circ + 22.9^{\circ} = 112.9^{\circ}.$$ The $$\varphi$$ angle is negative because it is measured in the west direction: $$\varphi = -43.2^\circ$$. Hence, the spherical coordinates of Rio de Janeiro are
$\rho = 6370\,\text{km},\;\varphi = -43.2^\circ,\;\theta = 112.9^{\circ}.$

### Example 4.

Find the spherical coordinates of point $$M$$ if the ray $$\mathbf{OM}$$ forms the angles $$\alpha = \frac{\pi}{4}$$ and $$\beta = \frac{\pi}{3}$$ with the $$x-$$ and $$y-$$axis, respectively, and the third coordinate of the point is $$z = 1.$$

Solution.

Let $$ON$$ be the projection of the ray $$\mathbf{OM}$$ onto the $$xy-$$plane. The angle $$\varphi$$ is the azimuth, so we can write:

$x_M = OL = ON\cos\varphi,\;\;y_M = OK = ON\sin\varphi.$

On the other hand, the right triangle $$OLM$$ implies that

$x_M = OL = OM\cos\alpha = \rho\cos\frac{\pi}{4} = \rho\frac{\sqrt{2}}{2}.$

Similarly, from the triangle $$OKM$$ we find

$y_M = OK = OM\cos\beta = \rho\cos\frac{\pi}{3} = \frac{\rho}{2}.$

Notice that

$ON = \sqrt{x_M^2 + y_M^2} = \sqrt{\rho^2 - z^2}.$

From this equation we determine the radius $$\rho:$$

$\sqrt{\left({\rho\frac{\sqrt{2}}{2}}\right)^2 + \left({\frac{\rho}{2}}\right)^2} = \sqrt{\rho^2 - 1^2}, \Rightarrow \sqrt{\frac{\rho^2}{2} + \frac{\rho^2}{4}} = \sqrt{\rho^2 - 1}, \Rightarrow \frac{3\rho^2}{4} = \rho^2 - 1, \Rightarrow \frac{\rho^2}{4} = 1, \Rightarrow \rho = 2.$

The Cartesian coordinates of the point $$M$$ are

$x_M = \rho\frac{\sqrt{2}}{2} = \sqrt{2},\;\;y_M = \frac{\rho}{2} = 1.$

Find the segment $$ON$$ and the angle $$\varphi:$$

$ON = \sqrt{\rho^2 - z^2} = \sqrt{2^2 - 1^2} = \sqrt{3},\;\;\cos\varphi = \frac{OL}{ON} = \frac{x_M}{ON} = \frac{\sqrt{2}}{\sqrt{3}} = \sqrt{\frac{2}{3}},$

that is $$\varphi = \arccos\sqrt{\frac{2}{3}}.$$

Calculate the $$\theta$$ angle:

$\cos\theta = \frac{z}{\rho} = \frac{1}{2}, \Rightarrow \theta = \arccos\frac{1}{2} = \frac{\pi}{3}.$

So the spherical coordinates of the point $$M$$ are

$\rho = 2,\;\varphi = \arccos\sqrt{\frac{2}{3}},\;\theta = \frac{\pi}{3}.$