# Convert the following to spherical coordinates. (a) 3x^2 - 4x + 3y^2 + 3z^2 = 0 (b) x + 5y +...

## Question:

Convert the following to spherical coordinates.

(a) {eq}3x^2 - 4x + 3y^2 + 3z^2 = 0 {/eq}

(b) {eq}x + 5y + z = 1 {/eq}

## Spherical coordinates:

Most of the surfaces that we know are expressed in a comfortable way in rectangular coordinates but this is not always true.

There are certain surfaces that can be expressed in a more convenient way in other coordinate systems such as spherical coordinates (example: spheres, cones with vertex at the origin, etcetera).

Using the change of rectangular coordinates to spherical coordinates:

{eq}\left\{ {\begin{array}{*{20}{c}} {x = \rho \cos \alpha \,sin\,\beta }\\ {y = \rho sin\,\alpha sin\,\beta }\\ {z = \rho \cos \beta } \end{array}} \right.\;\rho \ge 0,\;0 \le \alpha \le 2\pi ,\;0 \le \beta \le \pi {/eq}

(a) Taking into account: {eq}{x^2} + {y^2} + {z^2} = {\rho ^2} {/eq}

We have:

{eq}3{x^2} - 4x + 3{y^2} + 3{z^2} = 0\\ 3{x^2} + 3{y^2} + 3{z^2} - 4x = 0\\ 3{\rho ^2} - 4\rho \cos \alpha \,sin\,\beta = 0 {/eq}

(b) Substituting in the plane equation:

{eq}x + 5y + z = 1 {/eq}

We have:

{eq}x + 5y + z = 1\\ \rho \cos \alpha \,sin\,\beta + 5\rho sin\,\alpha sin\,\beta + \rho \cos \beta = 1 {/eq}