Question

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

Short 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-by-step solution

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

localid="1652247726196" $r=\sqrt{{\left(-6\right)}^2+6^2},\theta ={\tan }^{-1}\frac{6}{-6},z=6$

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

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

localid="1652247754695" $\Rightarrow (r,\theta ,\varphi )=(6\sqrt{2},-\frac{\mathrm{\pi }}{4},6)$

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

role="math" localid="1652247891309" $\rho =\sqrt{{\left(-6\right)}^2+{\left(6\right)}^2+{\left(6\right)}^2},\theta ={\tan }^{-1}\frac{6}{-6},\varphi ={\cos }^{-1}\frac{6}{\sqrt{{\left(-6\right)}^2+{\left(6\right)}^2+{\left(6\right)}^2}}$

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

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

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