TITLE: How do I change the order of integration for this parabolic wedge triple integral?
QUESTION [7 upvotes]: The question at hand asked me to change the order of the triple integral $$\int_{-1}^{1}\int_{x^{2}}^{1}\int_{0}^{y} f(x,y,z)dz dy dx$$
to $dy dz dx$.

It seemed straight forward at first, seeing that y is bounded by $y=z$ and $y=1$, and my region in the x-z plane would look like this:

But this region seems to say that z is bound by $z=0$ and $z=1$, and x bound by $x=-1$ and $x=1$. But putting those as my bounds would just express a rectangular block cut in half and not the solid. I suppose my initial assumption of y being bound by $y=z$ and $y=1$ doesn't allow me to take into the account the $y=x^{2}$ also binding it. How should I approach this?

REPLY [4 votes]: The given integral is $\displaystyle \int_{-1}^{1}\int_{x^{2}}^{1}\int_{0}^{y} f(x,y,z) \ dz \ dy \ dx$
As it shows in your drawing, the region is bound by a parabolic cylinder $y=x^2$ and plane $y = 1$ between $z = 0$ and $z = y$.
For order of integral $dy$ then $dz$ and $dx$ last, we note that for any given $x$,

$y$ is bound below by the parabolic cylinder, when $0 \leq z \leq x^2$.

$y$ is bound below by the plane $y = z$, when $x^2 \leq z \leq 1$.


The upper bound of $y$ is $1$ in both cases. So the integral in changed order becomes,
$\displaystyle \int_{-1}^1 \int_{0}^{x^2} \int_{x^2}^1 f(x, y, z) \ dy \ dz \ dx + 
\int_{-1}^1 \int_{x^2}^{1} \int_{z}^1 f(x, y, z) \ dy \ dz \ dx $