How do I change the order of integration for this parabolic wedge triple integral? 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?
 A: 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 $
A: Good known way is draw figure in $\Bbb{R}^3$, to view its bounds. This gives  intuition to follow then formal way: we have
$$A =\left\lbrace \begin{array}{l}
-1 \leqslant x \leqslant 1 \\
x^2 \leqslant y \leqslant 1 \\
0 \leqslant z \leqslant y
\end{array}
\right\rbrace  =\left\lbrace \begin{array}{l}
0 \leqslant z \leqslant 1 \\
z \leqslant y \leqslant 1 \\
-\sqrt{y} \leqslant x \leqslant \sqrt{y}
\end{array}
\right\rbrace = B = \left\lbrace \begin{array}{l}
0 \leqslant y \leqslant 1 \\
0 \leqslant z \leqslant y \\
-\sqrt{y} \leqslant x \leqslant \sqrt{y}
\end{array}
\right\rbrace $$
To prove $A=B$ formally we need to show definition of sets equality, but to obtain right side we need to solve inequalities.
Addition.
In integrals it is
$$\int\limits_{-1}^{1}\int\limits_{x^2}^{1}\int\limits_{0}^{y}f(x,y,z)dzdydx= \int\limits_{-1}^{1}dx\int\limits_{x^2}^{1}dy\int\limits_{0}^{y}f(x,y,z)dz=\\
=\int\limits_{0}^{1}\int\limits_{z}^{1}\int\limits_{-\sqrt{y}}^{\sqrt{y}}f(x,y,z)dxdydz = \int\limits_{0}^{1}dz\int\limits_{z}^{1}dy\int\limits_{-\sqrt{y}}^{\sqrt{y}}f(x,y,z)dx=\\
=\int\limits_{0}^{1}\int\limits_{0}^{y}\int\limits_{-\sqrt{y}}^{\sqrt{y}}f(x,y,z)dxdzdy = \int\limits_{0}^{1}dy\int\limits_{0}^{y}dz\int\limits_{-\sqrt{y}}^{\sqrt{y}}f(x,y,z)dx$$
Above integrals and set $B$ is obtained when considering $y,z$ as independent variables. If we consider $x,z$ as independent variables, then, of course, we will have
$$B=\left\lbrace \begin{array}{l}
-1 \leqslant x \leqslant 1 \\
0 \leqslant z \leqslant x^2 \\
 x^2 \leqslant y \leqslant 1
\end{array}
\right\rbrace \bigcup \left\lbrace \begin{array}{l}
-1 \leqslant x \leqslant 1 \\
x^2 \leqslant z \leqslant 1 \\
 z \leqslant y \leqslant 1
\end{array}
\right\rbrace = \\
=C=\left\lbrace \begin{array}{l}
0 \leqslant z \leqslant 1 \\
-1 \leqslant x \leqslant -\sqrt{z} \\
 x^2 \leqslant y \leqslant 1
\end{array}
\right\rbrace \bigcup \left\lbrace \begin{array}{l}
0 \leqslant z \leqslant 1 \\
\sqrt{z} \leqslant x \leqslant 1 \\
 x^2 \leqslant y \leqslant 1
\end{array}
\right\rbrace \bigcup \left\lbrace \begin{array}{l}
0 \leqslant z \leqslant 1 \\
-\sqrt{z} \leqslant x \leqslant \sqrt{z} \\
 z \leqslant y \leqslant 1
\end{array}
\right\rbrace$$
Integral for $C$:
$$\int\limits_{0}^{1}\int\limits_{-1}^{-\sqrt{z}}\int\limits_{x^2}^{1}f(x,y,z)dydxdz + \int\limits_{0}^{1}\int\limits_{\sqrt{z}}^{1}\int\limits_{x^2}^{1}f(x,y,z)dydxdz +\int\limits_{0}^{1}\int\limits_{-\sqrt{z}}^{\sqrt{z}}\int\limits_{z}^{1}f(x,y,z)dydxdz $$
All written integrals are $\frac{4}{5}$, when $f=1$.
