# Describe in words the surface whose equation is given. φ = π/6

The aim of the question is to learn how to visualize a given equation by comparing with the standard shape equations.

The equation of the cone (for example) is given by the following formula:

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

Similarly, the equation of the circle (in xy-plane) is given by the following formula:

$x^2 \ + \ y^2 \ = \ R^2$

Where x, y, z are the cartesian coordinates and R is the radius of the circle.

Given:

$\phi \ = \ \dfrac{ \pi }{ 6 }$

The cartesian coordinates can be calculated using the following formulas:

$x \ = \ R \ cos( \theta ) \ sin( \phi ) \ = \ \dfrac{ 1 }{ \sqrt{ 2 } } R \ cos( \theta )$

$y \ = \ R \ sin( \theta ) \ sin( \phi ) \ = \ \dfrac{ 1 }{ \sqrt{ 2 } } R \ sin( \theta )$

$z \ = \ R \ cos( \phi ) \ = \ \dfrac{ 1 }{ \sqrt{ 2 } } R$

Lets find $x^2 \ + \ y^2$:

$x^2 \ + \ y^2 \ = \ \bigg ( \dfrac{ 1 }{ \sqrt{ 2 } } R \ cos( \theta ) \bigg )^2 \ + \ \bigg ( \dfrac{ 1 }{ \sqrt{ 2 } } R \ sin( \theta ) \bigg )^2$

$x^2 \ + \ y^2 \ = \ \dfrac{ 1 }{ 2 } R^2 \ \bigg ( cos^2( \theta ) \ + \ sin^2( \theta ) \bigg )$

Since $cos^2( \theta ) \ + \ sin^2( \theta ) \ = \ 1$:

$x^2 \ + \ y^2 \ = \ \dfrac{ 1 }{ 2 } R^2$

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

The above equation represents a cone centered at the origin along the z-axis.

To find the direction of this cone, we solve the above equation for z:

$z \ = \ \pm \sqrt{ x^2 + y^2 }$

Since R is always positive, z must also be always positive:

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

Hence, the cone is located along the positive z-axis.

## Numerical Result

The given equation represents a cone with vertex at the origin directed along the positive z-axis.

## Example

Describe the following equation in words:

$\phi \ = \ \dfrac{ \pi }{ 2 }$

The cartesian coordinates of this equation are:

$x \ = \ R \ cos( \theta ) \ sin( \phi ) \ = \ R \ cos( \theta )$

$y \ = \ R \ sin( \theta ) \ sin( \phi ) \ = \ R \ sin( \theta )$

$z \ = \ R \ cos( \phi ) \ = \ 0$

Lets find $x^2 \ + \ y^2$:

$x^2 \ + \ y^2 \ = \ \bigg ( R \ cos( \theta ) \bigg )^2 \ + \ \bigg ( \dfrac{ 1 }{ \sqrt{ 2 } } R \ sin( \theta ) \bigg )^2$

$x^2 \ + \ y^2 \ = \ R^2 \ \bigg ( cos^2( \theta ) \ + \ sin^2( \theta ) \bigg )$

$x^2 \ + \ y^2 \ = \ R^2$

The above equation represents a circle centered at the origin in the xy-plane with radius R.