Convert cartesian inequalities to polar coordinates

Question

I have the following statement:

$\begin{cases} x ≥ 0, y ≥ 0, z ≥ 0 \\ x + y + z ≤ 1 \end{cases}$

I want to convert those inequalities to polar coordinates $\begin{cases} r,\theta, z \end{cases}$in order to calculate a triple integral.

I get $ z $ and $ \theta $ to:

$\begin{cases} 0 ≤ z ≤ 1\\ 0 ≤ \theta ≤ 2\pi \end{cases}$

But I don't understand how to convert $ r $

Any help would be greatly appreciated. Many thans:)

Answer 1

I think you mean cylindrical coordinates ? If yes : $x = r \cos \theta$ and $y = r \sin \theta$, so \begin{align} x \geqslant 0 &\Rightarrow cos \theta \geqslant 0 \rightarrow \theta \in [0, \pi/2] \cup [3\pi/2,2\pi] \\ y \geqslant 0 &\Rightarrow \sin \theta \geqslant 0 \Rightarrow \theta \in [0,\pi] \end{align} thus these two conditions state first that $\theta \in [0,\pi/2]$

then, you have \begin{align} x + y + z \geqslant 0 &\Rightarrow r\cos \theta + r\sin\theta + z \geqslant 0 \\ & \Rightarrow r(\cos \theta + \sin\theta) + z \geqslant 0 \\ &\Rightarrow r \times \sqrt{2}\sin(x+\pi/4) + z \geqslant 0 \end{align} the last term is because $\cos x = \sin(x +\pi/2)$ and because $\sin p + \sin q = \ldots$