Question:Find formulas for $\mathit{r}\mathbf{,}\mathit{\theta }\mathbf{,}\mathit{\varphi }$ in terms of x, y, z (the inverse, in other words, of Eq. 1.62)

The spherical coordinates are defined in terms of $\mathit{r}\mathbf{,}\mathit{\theta }\mathbf{,}\mathit{\varphi }$, where r is the distance from origin, $\mathit{\theta }$is the polar angle and $\mathit{\varphi }$is the azimuthal angle.

The spherical coordinates is drawn as,

In the triangle OMP, the angle M is $90°$and the length of OP is r . From the figure, it can be written as $OM=r\sin \theta$. Also from the figure, we can write as,

$$\begin{gathered} x=\left(r\sin \theta \right)\cos \varphi \\ y=\left(r\sin \theta \right)\sin \varphi \\ r=r\cos \theta \end{gathered}$$

Substitute localid="1657524559006" $\left(r\sin \theta \right)\cos \varphi$for $x$, $\left(r\sin \theta \right)\sin \varphi$for $y$and $r\cos \theta$for into $x^2+y^2+z^2$.

$$\begin{gathered} x^2+y^2+z^2={\left[(r\sin \theta )\cos \theta \right]}^2+{\left[(r\sin \theta )\sin \theta \right]}^2+{\left[(r\cos \theta )\right]}^2 \\ =r^3\left[{\sin }^2\theta ({\cos }^2\varphi +{\sin }^2\theta )+\left({\cos }^2\varphi \right)\right] \\ =r^2\left[{\sin }^2\theta \left(1\right)+{\cos }^2\varphi \right] \\ =r^2 \end{gathered}$$

Thus, the formula of $r$is obtained to be equal to$r=\sqrt{x^2+y^2+z^2}$.

The formula for z is$z=r\cos \theta$, which can be rearranged as $\theta ={\cos }^{-1}\left(\frac{z}{r}\right)$.

Substitute $r=\sqrt{x^2+y^2+z^2}$ for $r$ into$\theta ={\cos }^{-1}\left(\frac{z}{r}\right)$.

localid="1657529438426" $\theta ={\cos }^{-1}\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right)$

Thus formula for $\theta$is obtained as $\theta ={\cos }^{-1}\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right)$.

Now, Divide the formula of$y=(r\sin \theta )\sin \varphi \mathrm{by}x=(r\sin \theta )\cos \varphi \mathrm{as},$

$$\begin{gathered} \frac{y}{x}=\frac{(r\sin \theta )\sin \varphi }{\mathbf{(}rsin\theta \mathbf{)}cos\varphi } \\ =\frac{\sin \varphi }{\cos \varphi } \\ =\tan \varphi \end{gathered}$$

Therefore, $\frac{y}{x}=\tan \varphi$ , then value of $\varphi$can be obtained as,

$$\begin{gathered} \frac{y}{x}=\tan \varphi \\ \varphi ={\tan }^{-1}\left(\frac{y}{x}\right) \end{gathered}$$

Therefore, the value of $\varphi$is obtained as $\varphi ={\tan }^{-1}\left(\frac{y}{x}\right)$.