Question

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

$\left(6,\frac{\mathrm{\pi }}{4},\frac{\mathrm{\pi }}{4}\right)$

Short answer

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

$\left(r,\theta ,z\right)=\left(3\sqrt{2},\frac{\mathrm{\pi }}{4},3\sqrt{2}\right)$

Step-by-step solution

Given Information

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

To find rectangular coordinates

As $\rho =6,\theta =\frac{\mathrm{\pi }}{4},\varphi =\frac{\mathrm{\pi }}{4}$

We know that

$x=\rho \sin \varphi \cos \theta ,y=\rho \sin \varphi \sin \theta ,z=\rho \cos \varphi$

Substituting values, we get

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

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

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

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

To find cylindrical coordinates

To find cylindrical coordinates from spherical coordinates, we use

$r=\rho \sin \varphi ,\theta =\theta ,z=\rho \cos \varphi$

Using given values of spherical coordinates, we get

$r=6\sin \frac{\mathrm{\pi }}{4},\theta =\frac{\mathrm{\pi }}{4},z=6\cos \frac{\mathrm{\pi }}{4}$

$r=6\frac{1}{\sqrt{2}},\theta =\frac{\mathrm{\pi }}{4},z=6\frac{1}{\sqrt{2}}$

$\Rightarrow r=3\sqrt{2},\theta =\frac{\mathrm{\pi }}{4},z=3\sqrt{2}$

Hence, cylindrical coordinates are$\left(r,\theta ,z\right)=\left(3\sqrt{2},\frac{\mathrm{\pi }}{4},3\sqrt{2}\right)$