Q 24. - Calculus

Question

Give the cylindrical and spherical coordinates for the point with rectangular coordinates $(-6,6,6)$

Step-by-Step Solution

Answer

$\left(\rho ,\theta ,\varphi \right)=\left(6\sqrt{3},\frac{\mathrm{\pi }}{4},{\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)\right)$

$(r,\theta ,\varphi )=(6\sqrt{2},-\frac{\mathrm{\pi }}{4},6)$

Step 1: Given Information

Given data is $x=-6, y=6, z=6$

Step 2:Finding the Cylindrical coordinates

It is given that $x=-6, y=6, z=6$

We know that 

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

Using given values, we get

$r=\sqrt{{\left(-6\right)}^2+6^2}, \theta ={\tan }^{-1}\frac{6}{-6}, z=6$

$\Rightarrow r=\sqrt{{\left(-6\right)}^2+6^2}, \theta ={\tan }^{-1}-1, z=6$

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

$\Rightarrow (r,\theta ,\varphi )=(6\sqrt{2},-\frac{\mathrm{\pi }}{4},6)$

Step 3: Finding the spherical coordinates

We know that

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

Using values

$\rho =\sqrt{{\left(-6\right)}^2+{\left(6\right)}^2+{\left(6\right)}^2}, \theta ={\tan }^{-1}\frac{6}{-6}, \varphi ={\cos }^{-1}\frac{6}{\sqrt{{\left(-6\right)}^2+{\left(6\right)}^2+{\left(6\right)}^2}}$

$\Rightarrow \rho =\sqrt{36+36+36}, \theta ={\tan }^{-1}-1, \varphi ={\cos }^{-1}\frac{6}{\sqrt{36+36+36}}$

$\Rightarrow \rho =6\sqrt{3}, \theta =\frac{\mathrm{\pi }}{4}, \varphi ={\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)$

$\left(\rho ,\theta ,\varphi \right)=\left(6\sqrt{3},\frac{\mathrm{\pi }}{4},{\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)\right)$