I want to compute the volume of the 3-dimensional unit ball.
$A := \{(x,y,z) \in \mathbb{R}^3: x^2 + y^2 + z^2 \leq 1$
From this I want to get the normal domain as:
$A = \{(x,y,z): x \in [-1,1], y \in [-\sqrt{1-x^2}, \sqrt{1-x^2}], z \in [-\sqrt{1-x^2-y^2},\sqrt{1-x^2-y^2}]\}$
The question is, how do I get the x, y, z intervals from the inequality?
The more concrete (but less 'exact') question is: Is it ok to 'leave out' z in the definition of the y interval, i. e. disregard z?
I found the answer (any corrections or improvements appreciated):
$$ x^2 + y^2 + z^2 \leq 1 $$
Because $x^2 \geq 0$, $y^2 \geq 0$, $z^2 \geq 0$:
$$ 0 \leq x^2 + y^2 + z^2 \leq 1 $$
From that we can say, because $z^2 \geq 0$ and $0 \leq x^2 + y^2 + z^2 \leq 1$ that leaving out $z$ will not change the in equality and get
$$ 0 \leq x^2 + y^2 \leq 1 $$
From this we can split off
$$ x^2 + y^2 \leq 1 $$
which can be changed to express $y$ with $x$.
The same holds for expressing $z$ with $x$ and $y$.