How do I know if these two integrals are the same? I need to find out if it is true or false that these integrals are the same
$$\int_{0}^{1}\int_{0}^{1}\int_{0}^{1-x^2} f(x,y,z)dz dy dx $$
$$\int_{0}^{1}\int_{0}^{1}\int_{-\sqrt{1-z}}^{\sqrt{1-z}}f(x,y,z) dx dy dz$$
I know that $0\leq z\leq 1-x^2$
From just the $z\leq 1-x^2$ part, I know that $-\sqrt{1-z}\leq x \leq \sqrt{1-z}$
But I'm not sure where the bounds for z would come from i.e. how does the second integral shown have 0 and 1 as the limits of integration for z?
Also, isn't it technically that
$0\leq z \leq 1-x^2 \implies$
$x^2 \leq z+x^2 \leq 1 $
$x^2-z\leq x^2 \leq 1-z$
$\pm\sqrt{x^2-z}\leq x \leq \pm \sqrt{1-z}$
Do I just ignore the left side of the equation?
 A: If they are equal, then they should be equal when $f=1$.
But when $f=1$ the first one is equal to
$$
\int_0^1(1-x^2)\;dx=\frac23
$$
while the second one is
$$
\int_0^12\sqrt{1-z}\;dz = \frac43.
$$

To figure out the exact bounds when changing the order of the first iterated integral, it is helpful to draw a picture of the region.
Alternatively, observe that the region of the first integral is given by
$$
E=\{(x,y,z)\mid 0\le x\le 1,\quad  0\le y\le 1,\quad  0\le z\le 1-x^2\}
$$
which is a normal domain in $\mathbf{R}^3$:
$$
\displaystyle E=\{(x,y,z)\mid 0\le y\le 1, (x,z)\in D\} 
$$
where $D:=\{(x,z)\mid 0\le x\le 1, 0\le z\le 1-x^2\}$ is a 2-dimensional region, which itself is a normal domain in $\mathbf{R}^2$.
Now we have already reduced the problem to a 2-dimensional one. Drawing a picture (see below), you can see that $D$ is nothing but a region bounded by the graph of the function $z=1-x^2$ ($0\le x\le 1$) and three other lines. It is equivalent to
$$
D=\{(x,z)\mid 0\le z\le 1, 0\le x\le \sqrt{1-z}\}\tag{0}
$$
If you want to find (0) algebraically, note that $0\le z\le 1-x^2$ implies
$$
-\sqrt{1-z}\le x\le\sqrt{1-z},\quad z\ge 0\tag{1}
$$
which together with the condition $0\le x\le 1$ gives
$$
0\le x\le \sqrt{1-z}\tag{2}
$$
On the other hand, $0\le x\le 1$ and $0\le z\le 1-x^2$ together imply that $0\le z\le 1$.

