?
Paul's Online Notes
Home / Calculus II / 3-Dimensional Space / Spherical Coordinates
Show All Notes Hide All Notes

Section 12.13 : Spherical Coordinates

4. Convert the following equation written in Cartesian coordinates into an equation in Spherical coordinates.

\[{x^2} + {y^2} = 4x + z - 2\]

Show All Steps Hide All Steps

Start Solution

All we need to do here is plug in the following conversion formulas into the equation and do a little simplification.

\[x = \rho \sin \varphi \cos \theta \hspace{0.5in}y = \rho \sin \varphi \sin \theta \hspace{0.5in}z = \rho \cos \varphi \] Show Step 2

Plugging the conversion formula in gives,

\[{\left( {\rho \sin \varphi \cos \theta } \right)^2} + {\left( {\rho \sin \varphi \sin \theta } \right)^2} = 4\left( {\rho \sin \varphi \cos \theta } \right) + \rho \cos \varphi - 2\]

The first two terms can be simplified as follows.

\[\begin{align*}{\rho ^2}{\sin ^2}\varphi {\cos ^2}\theta + {\rho ^2}{\sin ^2}\varphi {\sin ^2}\theta & = 4\rho \sin \varphi \cos \theta + \rho \cos \varphi - 2\\ {\rho ^2}{\sin ^2}\varphi \left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) & = 4\rho \sin \varphi \cos \theta + \rho \cos \varphi - 2\\ {\rho ^2}{\sin ^2}\varphi & = 4\rho \sin \varphi \cos \theta + \rho \cos \varphi - 2\end{align*}\]