Question

Give the rectangular and spherical coordinates for the point with cylindrical coordinates$\left(4,\frac{\mathrm{\pi }}{3},6\right)$

Short answer

$\left(x,y,z\right)=\left(2,2\sqrt{3},6\right)$

$\left(\rho ,\theta ,\varphi \right)=\left({\sqrt{52},\frac{\mathrm{\pi }}{3},\tan }^{-1}\frac{2}{63}\right)$

Step-by-step solution

Given Information

The given expression is$\left(r,\theta ,z\right)=\left(4,\frac{\mathrm{\pi }}{3},6\right)$

To find rectangular coordinates

As $r=4,\theta =\frac{\mathrm{\pi }}{3},z=6$

We know that for converting cylindrical to rectangular coordinates, we use

$x=r\cos \theta ,y=r\sin \theta ,z=z$

localid="1652288468915" $x=4\cos \frac{\mathrm{\pi }}{3},y=4\sin \frac{\mathrm{\pi }}{3},z=6$

$x=4\frac{1}{2},y=4\frac{\sqrt{3}}{2},z=6$

$x=2,y=2\sqrt{3},z=6$

Hence, $\left(x,y,z\right)=\left(2,2\sqrt{3},6\right)$

To find spherical coordinates

For converting cylindrical to spherical coordinates, we use

$\rho ={\sqrt{r^2+z^2},\theta =\theta ,\varphi =\tan }^{-1}\frac{r}{z}$

Using given values,

$\rho ={\sqrt{4^2+6^2},\theta =\frac{\mathrm{\pi }}{3},\varphi =\tan }^{-1}\frac{4}{6}$

$\rho ={\sqrt{52},\theta =\frac{\mathrm{\pi }}{3},\varphi =\tan }^{-1}\frac{2}{3}$

$\Rightarrow \rho ={2\sqrt{13},\theta =\frac{\mathrm{\pi }}{3},\varphi =\tan }^{-1}\left[\frac{2}{3}\right]$

Hence,$\left(\rho ,\theta ,\varphi \right)=\left(2\sqrt{13},{\frac{\mathrm{\pi }}{3},\tan }^{-1}\left[\frac{2}{3}\right]\right)$