# Identify the surface whose equation is given as

$$\rho=\sin\theta\sin\phi$$.

The objective of this question is to find a type of surface represented by the given equation.

A surface can be regarded as a geometrical shape which is like a deformed plane. The boundaries of solid objects in a usual 3-D Euclidean space, such as spheres, are common examples of surfaces.

In other words, it is a 2-D collection of points, that is, a flat surface, a 3-D collection of points having a curve as its cross-section, that is, a curved surface, or a boundary of 3-D solid. More generally, a surface can be defined as a continuous boundary that divides a 3-D space into two regions.

We know that the Cartesian coordinates can be represented into spherical coordinates in the following way:

$x=\rho\sin\phi\cos\theta$                  (1)

$y=\rho \sin\phi \sin\theta$                 (2)

$z=\rho\cos\theta$                                  (3)

Now, multiply both sides of the given equation by $\rho$ to get:

$\rho^2=\rho\sin\theta\sin\phi$

Since $\rho^2=x^2+y^2+z^2$, and from (2) $y=\rho\sin\theta\sin\phi$:

This implies that $y=\rho^2$.

And hence:

$x^2+y^2+z^2=y$

$\implies x^2+y^2-y+z^2=0$

Completing the square for the term involving $y$:

$x^2+\left(y-\dfrac{1}{2}\right)^2+z^2=\dfrac{1}{4}$

or $(x-0)^2+\left(y-\dfrac{1}{2}\right)^2+(z-0)^2=\left(\dfrac{1}{2}\right)^2$

So the above equation represents a sphere of  radius $\dfrac{1}{2}$ with center at $\left(0,\dfrac{1}{2},0\right)$.

## Example 1

Given an equation in spherical coordinates as $\rho=2\sin\phi\cos\theta$, determine the surface represented by the equation.

### Solution

Now multiply both sides of the given equation by $\rho$ to get:

$\rho^2=2\rho\sin\phi\cos\theta$

Since $\rho^2=x^2+y^2+z^2$, and from (1) $x=\rho\sin\phi\cos\theta$:

This implies that $\dfrac{x}{2}=\rho^2$.

And hence:

$x^2+y^2+z^2=\dfrac{x}{2}$

$\implies x^2-\dfrac{x}{2}+y^2+z^2=0$

Completing the square for the term involving $x$:

$\left(x-\dfrac{1}{4}\right)^2+y^2+z^2=\dfrac{1}{16}$

or $\left(x-\dfrac{1}{4}\right)^2+\left(y-0\right)^2+(z-0)^2=\left(\dfrac{1}{4}\right)^2$

So the above equation represents a sphere of radius $\dfrac{1}{4}$ with center at $\left(\dfrac{1}{4},0,0\right)$.

## Example 2

Given an equation in spherical coordinates as $\rho=\cos\phi$, determine the surface represented by the equation.

### Solution

Now multiply both sides of the given equation by $\rho$ to get:

$\rho^2=\rho\cos\phi$

Since $\rho^2=x^2+y^2+z^2$, and from (3) $z=\rho\cos\phi$:

This implies that $z=\rho^2$.

And hence:

$x^2+y^2+z^2=z$

$\implies x^2+y^2+z^2-z=0$

Completing the square for the term involving $z$:

$x^2+y^2+\left(z-\dfrac{1}{2}\right)^2=\dfrac{1}{4}$

or $x^2+y^2+\left(z-\dfrac{1}{2}\right)^2=\left(\dfrac{1}{2}\right)^2$

So the above equation represents a sphere of radius $\dfrac{1}{2}$ with center at $\left(0,0,\dfrac{1}{2}\right)$.