Q 26. - Calculus

Question

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

Step-by-Step Solution

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 1: Given Information

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

Step 2: 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

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

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

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

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

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

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