Question

Give the rectangular and spherical coordinates for the point with cylindrical coordinates:

$\left(\sqrt{48},\frac{\mathrm{\pi }}{3},4\right)$

Short answer

Rectangular coordinates are$\left(2\sqrt{3},6,4\right)$and spherical coordinates are$\left(8,\frac{\mathrm{\pi }}{3},\frac{\mathrm{\pi }}{3}\right)$

Step-by-step solution

Given Information

The given expression is$\left(r,\theta ,\varphi \right)=\left(\sqrt{48},\frac{\mathrm{\pi }}{3},4\right)$

Finding the rectangular coordinates

As $r=\sqrt{48},\theta =\frac{\mathrm{\pi }}{3},\varphi =4$

To find out rectangular coordinates, we will use

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

Using the values

$x=\sqrt{48}\cos \frac{\mathrm{\pi }}{3},y=\sqrt{48}\sin \frac{\mathrm{\pi }}{3},z=4$

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

$\left(x,y,z\right)=\left(2\sqrt{3},6,4\right)$

Finding the spherical coordinates

To convert rectangular to spherical coordinates

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

Putting the values

$\rho ={\sqrt{{\left(\sqrt{48}\right)}^2+4^2},\theta =\frac{\mathrm{\pi }}{3},\varphi =\tan }^{-1}\frac{\sqrt{48}}{4}$

$\rho ={\sqrt{48+16},\theta =\frac{\mathrm{\pi }}{3},\varphi =\tan }^{-1}\sqrt{3}$

$\Rightarrow \rho =\sqrt{64},\theta =\frac{\mathrm{\pi }}{3},\varphi =\frac{\mathrm{\pi }}{3}$

$\Rightarrow \rho =8,\theta =\frac{\mathrm{\pi }}{3},\varphi =\frac{\mathrm{\pi }}{3}$

$\Rightarrow \left(\rho ,\theta ,\varphi \right)=\left(8,\frac{\mathrm{\pi }}{3},\frac{\mathrm{\pi }}{3}\right)$