Question

. To convert from rectangular to spherical coordinates

$p=\_\_\_\_,\theta =\_\_\_\_,\varphi =\_\_\_\_$

Short answer

$$\begin{gathered} p=\sqrt{x^2+y^2+z^2} \\ \theta ={\tan }^{-1}\left(\frac{y}{x}\right) \\ \varphi ={\cos }^{-1}\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right) \end{gathered}$$

Step-by-step solution

Step 1:Given information

Given$\left(x,y,z\right)$are rectangular coodinates and$\left(p,\theta ,\varphi \right)$are spherical coordinates

Simplification

$\left(x,y,z\right)$are rectangular coordinates and $\left(p,\theta ,\varphi \right)$are spherical coordinates

The relation is

From the figure

$\Rightarrow \cos \varphi =\frac{z}{r},\sin \varphi =\frac{\rho }{r}$

Now

$x=\rho \cos \varphi$

$=r\sin \theta \cos \varphi$where $\left(\rho =r\sin \theta \right)$

similarly

$y=\rho \sin \varphi$

$=r\sin \theta \sin \varphi$

And $z=r\cos \varphi$

now

$$\begin{gathered} x^2+y^2+z^2=r^2\left[{\sin }^2\varphi \left({\sin }^2\theta +{\cos }^2\theta \right)+{\cos }^2\varphi \right] \\ \Rightarrow x^2+y^2+z^2=r^2\left[{\sin }^2\varphi +{\cos }^2\varphi \right] \\ \Rightarrow x^2+y^2+z^2=r^2 \\ \Rightarrow r=\sqrt{x^2+y^2+z^2} \end{gathered}$$

Similarly

$\Rightarrow \theta ={\tan }^{-1}\frac{y}{x}$

and we have $\cos \varphi =\frac{z}{r}$

$$\begin{gathered} \Rightarrow \varphi ={\cos }^{-1}\frac{z}{r} \\ \Rightarrow \varphi ={\cos }^{-1}\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right) \end{gathered}$$