# Describe in words the surface whose equation is given as:

– $\phi \space = \space \frac {\pi}{3}$

The main objective of this question is to visualize the given equation.

This question uses the concept of visualizing the given equation by comparing it to the equations of the standard shapes along with the concept of the Cartesian coordinate system and spherical coordinate system.

We are given that Spherical Coordinates are $\phi = \dfrac{\pi}{3}$:

$cos\phi \space = \space cos \left( \dfrac{\pi}{3}\right) \space = \space \dfrac{1}{2} \hspace{3ex}$

$x \space = \space \rho sin\phi cos\theta \hspace{3ex}$

$cos^2 \phi \space = \space \dfrac{1}{4} \hspace{3ex}$

$y \space = \space \rho sin\phi sin\theta \hspace{3ex}$

$\rho^2cos^2\theta \space = \space \dfrac{1}{4} \rho^2 \hspace{3ex}$

$z^2 \space = \space \dfrac{1}{4}(x^2 + y^2 + z^2) \hspace{3ex}$

$x^2 + y^2 + z^2 \space = \space \rho^2 \hspace{3ex}$

$4z^2 \space = \space x^2 + y^2 + z^2 \hspace{3ex}$

$3z^2 \space = \space x^2 + y^2 \hspace{3ex}$

So:

$3z^2 = x^2 + y^2$ is a double cone.

The given equation represents a double cone.

## Example

Describe the surface area for the three given equations.

$\phi = \dfrac{ \pi }{ 5 }, \space \phi = \dfrac{ \pi }{ 7 } \space and \space \phi = \dfrac{ \pi }{ 9 }$

In this question, we have to visualize the given expression.

We are given that Spherical Coordinates are  $\phi = \dfrac{\pi}{5}$.

We know that:

$cos\phi \space = \space cos \left( \dfrac{\pi}{5}\right) \space = \space 0.8090 \hspace{3ex}$

$x \space = \space \rho sin\phi cos\theta \hspace{3ex}$

Squaring $cos$ value will result in:

$cos^2 \phi \space = \space 0.654481 \hspace{3ex}$

$y \space = \space \rho sin\phi sin\theta \hspace{3ex}$

$\rho^2cos^2\theta \space = \space 0.654481 \rho^2 \hspace{3ex}$

$z^2 \space = \space 0.654481(x^2 + y^2 + z^2) \hspace{3ex}$

$x^2 + y^2 + z^2 \space = \space \rho^2 \hspace{3ex}$

$0.654481z^2 \space = \space x^2 + y^2 + z^2 \hspace{3ex}$

Now solving for $\phi = \dfrac{ \pi }{ 7 }$.

We are given that Spherical Coordinates are  $\phi = \dfrac{\pi}{7}$.

We know that:

$cos\phi \space = \space cos \left( \dfrac{\pi}{7}\right) \space = \space 0.900 \hspace{3ex}$

$x \space = \space \rho sin\phi cos\theta \hspace{3ex}$

Squaring $cos$ value will result in:

$cos^2 \phi \space = \space 0.81 \hspace{3ex}$

$y \space = \space \rho sin\phi sin\theta \hspace{3ex}$

$\rho^2cos^2\theta \space = \space 0.81 \rho^2 \hspace{3ex}$

$z^2 \space = \space 0.81(x^2 + y^2 + z^2) \hspace{3ex}$

$x^2 + y^2 + z^2 \space = \space \rho^2 \hspace{3ex}$

$0.81z^2 \space = \space x^2 + y^2 + z^2 \hspace{3ex}$

asa

Now solving for $\phi = \dfrac{ \pi }{ 9 }$.

We are given that Spherical Coordinates are  $\phi = \dfrac{\pi}{9}$.

We know that:

$cos\phi \space = \space cos \left( \dfrac{\pi}{9}\right) \space = \space 0.939 \hspace{3ex}$

$x \space = \space \rho sin\phi cos\theta \hspace{3ex}$

Squaring $cos$ value will result in:

$cos^2 \phi \space = \space 0.81 \hspace{3ex}$

$y \space = \space \rho sin\phi sin\theta \hspace{3ex}$

$\rho^2cos^2\theta \space = \space 0.881 \rho^2 \hspace{3ex}$

$z^2 \space = \space 0.881(x^2 + y^2 + z^2) \hspace{3ex}$

$x^2 + y^2 + z^2 \space = \space \rho^2 \hspace{3ex}$

$0.881z^2 \space = \space x^2 + y^2 + z^2 \hspace{3ex}$