Nature gives us the dimension of space, and we give the measurements of that space, and the different ways to measure space, gives us different coordinate systems. Spherical coordinate system is one of them.

Spherical Coordinate systems are used in describing three-dimensional coordinate systems.

These coordinates represent the points from a spherical global perspective and proves convenient for graphing surfaces in those spaces that have a point or center of symmetry.

Any point in spherical coordinate system is represented by the ordered triple ($\rho, \theta, \phi$).As the name suggests, spherical coordinate systems look like a sphere, as shown in the following diagram:

These coordinates represent the points from a spherical global perspective and proves convenient for graphing surfaces in those spaces that have a point or center of symmetry.

Any point in spherical coordinate system is represented by the ordered triple ($\rho, \theta, \phi$).As the name suggests, spherical coordinate systems look like a sphere, as shown in the following diagram:

1. $\rho$ is the distance from the origin to the point P, and $\rho$ $\geq $ 0.

2. $\theta$ is the angle generally used in the radians and measured from the x-axis till to point of projection on the x-y plane

3. $ \phi$ is the angle between the positive z-axis and the line from the origin to P.

4. Here, r $\geq$ 0, 0° = $\theta$ = 180° ($\pi$ radians) and 0° $\leq$ $ \phi$ < 360° (2$\pi$ radians).

Conversion from spherical coordinate system to Cartesian (rectangular) coordinate system:

x = $\rho$ sin$ \phi$ cos$\theta$

y = $\rho$ sin$ \phi$ sin$\theta$

z = $\rho$ cos$ \phi$

and $\rho$^{2} = x^{2} + y^{2} + z^{2}

Conversion from Cartesian coordinate system to Spherical coordinate system:

$r = \sqrt{x^{2}+y^{2}}$; $\rho$^{2} = x^{2} + y^{2} + z^{2}.

Cos$\theta$ = $\frac{x}{r}$, sin$\theta$ = $\frac{y}{r}$, tan$\theta$ =$\frac{y}{x}$ and cos$\rho$ = $\frac{z}{\rho}$

Example 1: Find the rectangular coordinates of the point with spherical coordinates $(2, \frac{\pi }{6},\frac{2 \pi}{3})$.

Solution: Rectangular coordinates are: ($2*\frac{\sqrt{3}}{2}*\frac{\sqrt{3}}{2}$; $2*\frac{\sqrt{3}}{2}*\frac{1}{2}$; $2*\frac{-1}{2}$)

= $(\frac{3}{2},\frac{\sqrt{3}}{2},-1)$ is the answer.

= $(\frac{3}{2},\frac{\sqrt{3}}{2},-1)$ is the answer.

Example 2: Identify the surface given by $\rho$ = 2.

Solution: Converting the above surface to Cartesian coordinates gives x^{2} + y^{2} + z^{2} = 4. Hence, it is a sphere centered at the origin, with radius 2.

This is important to remember. In general, the equation $\rho$ = c is the equation of the sphere of radius c centered at the origin in spherical coordinates.

This is important to remember. In general, the equation $\rho$ = c is the equation of the sphere of radius c centered at the origin in spherical coordinates.

Example 3: Convert (-1, 1 -$\sqrt{2}$) from Cartesian to Spherical coordinate system.

Solution: Here, $\rho$ = 2, from z = $\rho$ cos$\phi$, we get, cos$\phi$ = -$\frac{\sqrt{2}}{2}$ = 3 p /4.

Next, to find $\theta$, we will use the equation involving x or y, by using y. Now, sin$\theta$ = $\frac{y}{r}$ = 1 / 2 sin $\frac{3 \pi}{4}$ = $\frac{\sqrt{2}}{2}$.

Here, there could be various values for $\theta$, as $\frac{\pi}{4}$ or $\frac{3 \pi}{4}$.

However, the projection of P on the xy-plane is in the second quadrant, so we have $\theta$ = 3p/4.

Thus, the spherical coordinates are (2, $\frac{3 \pi}{4}$, $\frac{3 \pi}{4}$).