On setting the limits of integration of a triple integral I have been struggling setting the limits of integration and I have this integral
$$\iiint_D z^2 dV$$
where $D$ is bounded by (1)  $z=0\quad$ (2) $x^2+z=1\quad$ (3)$y^2+z=1$
I am not sure on how to identify the limits, I know that $z$ varies from $0$ to $1$ but I am not sure on how to delimitate on $y$ and $x$
so far $\iint\int_0^1 z^2 \,dz\, dy\,dx$ how can I proceed?
 A: We can rewrite bound (2) as $x^2 = 1-z$. If $0 \le z \le 1$, then $0 \le 1-z \le 1$, so $0 \le x^2 \le 1$, which says that $0 \le x \le 1$. Hopefully you can figure out how to do this for $y$ as well.
A: When you have $\displaystyle\int\left( \int\left(\int \cdots \,dz\right) dy \right) dx,$ with $z$ on the inside, then you're saying $z$ goes from something to something, with $x$ and $y$ fixed. That means the bounds on the inner integral must depend on $x$ and $y.$ Likewise the bounds on the second integral, with respect to $y,$ will in general depend on $x.$  Just saying $z$ goes from $0$ to $1$ and using those as the bounds will work only if either the integral with respect to $z$ is the outermost one or else there is no dependence on $x$ and $y.$
You will have $z$ going from $1$ up to $1-x^2$ or $1-y^2,$ whichever is less.
And $1-x^2\le 1-y^2$ if $y^2 \le x^2,$ i.e. if $-|x|\le y\le |x|.$
By geometric symmetry, the integral over the region $x\ge0$ and $-x\le y\le x$ is $1/4$ times the entire integral that you seek.  So you can write
$$
4 \int_0^1 \left( \int_{-x}^x \left( \int_0^{1-x^2} \cdots\cdots \, dz \right) \, dy \right) \, dx.
$$
