Paul's Online Notes
Home / Calculus III / 3-Dimensional Space / Spherical Coordinates
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

### Section 12.13 : Spherical Coordinates

5. Convert the equation written in Spherical coordinates into an equation in Cartesian coordinates.

${\rho ^2} = 3 - \cos \varphi$

Show All Steps Hide All Steps

Start Solution

There really isn’t a whole lot to do here. All we need to do is to use the following conversion formulas in the equation where (and if) possible

$\begin{array}{c}x = \rho \sin \varphi \cos \theta \hspace{0.5in}y = \rho \sin \varphi \sin \theta \hspace{0.5in}z = \rho \cos \varphi \\ {\rho ^2} = {x^2} + {y^2} + {z^2}\end{array}$ Show Step 2

To make this problem a little easier let’s first multiply the equation by $$\rho$$. Doing this gives,

${\rho ^3} = 3\rho - \rho \cos \varphi$

Doing this makes recognizing the right most term a little easier.

Show Step 3

Using the appropriate conversion formulas from Step 1 gives,

$\require{bbox} \bbox[2pt,border:1px solid black]{{{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{\frac{3}{2}}} = 3\sqrt {{x^2} + {y^2} + {z^2}} - z}}$