Q 28. - Calculus

Question

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

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

Step-by-Step Solution

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 1: Given Information

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

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

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