Convert $(-\sqrt{2},1,0)$ in to cyclidrical and spherical coordinates

Question

So for my answers I'm following the formulas given but I get stuck at finding $\theta$ because it equals $\pi + tan^{-1}(-\frac{1}{\sqrt{2}})$. Does that sound right?

Answer 1

Cylindrical coordinates map $(x,y)$ to $(r,\theta)$ and leave $z$ as-is.

Calculate $r$ as the length of the $x,y$ projection: $r=\sqrt{(-\sqrt{2})^2 + 1^2} = \sqrt{3}.$ Calculate $\theta$ as $\theta = \tan^{-1}\frac{y}{x} = \tan^{-1}\frac{1}{-\sqrt{2}}.$ Since your $x$ is negative and your $y$ is positive, it will be in the second quadrant: $\theta \approx 144.7^{\circ}.$

Spherical coordinates are straightforward for this one. Since $z=0$, $\theta=0$. Your value of $r$ is the same, and your value of $\phi$ is the value of $\theta$ you found for the cylindrical coordinates.