Volume of the solid bounded by $z = 4-x^2$, $y+z=4$, $y=0$ and $z=0$. If I am seeing this problem correctly, when $z=0$, $x = \pm 2$, so $-2 \le x \le 2$.
The $y$ coordinate varies from $0$ to $4$, because when $z=0, y=4$ (the plane $y+z=4$ with $z=0$). So $0 \le y \le 4$.
The $z$ coordinate varies from the plane $z=0$ to the plane $z=4-y$.
Then $0 \le z \le 4-y$.
So the integrals are:
$\displaystyle \int_{-2}^{2} \int_{0}^{4} \int_{0}^{4-y} dzdydx$
Is this correct?
 A: Unfortunately, that is not correct: the solid that you have described is bounded by the planes $x = -2,$ $x = 2,$ $y = 0,$ $y = 4,$ and $y + z = 4.$ Evidently, this is not the same as the given region.
Unfortunately, the given solid is not $z$-simple, i.e., the volume in terms of $dz \, dy \, dx$ and $dz \, dx \, dy$ involves more than one integral; however, the solid is $y$-simple, so we can evaluate the volume by $$\int_{-2}^2 \int_0^{4 - x^2} \int_0^{4 - z} 1 \, dy \, dz \, dx.$$
Of course, if you insist upon integrating with respect to $dz \, dy \, dx,$ then you must use three integrals. $$\int_{-2}^0 \int_0^{x^2} \int_0^{4 - x^2} 1 \, dz \, dy \, dx + \int_{-2}^2 \int_{x^2}^4 \int_0^{4 - y} 1 \, dz \, dy \, dx + \int_0^2 \int_0^{x^2} \int_0^{4 - x^2} 1 \, dz \, dy \, dx$$
For this type of problem, it is best to graph the surfaces to determine the integral. If you wish to see if the solid is $z$-simple, project into the $xy$-plane, and look at the resulting regions. Here, you will notice three different regions: the first region corresponds to $0 \leq y \leq x^2;$ the second gives $x^2 \leq y \leq 4;$ and the third gives $0 \leq y \leq x^2.$ So, the solid is not $z$-simple. But if we project into the $xz$-plane, there is only one region (where $0 \leq z \leq 4 - x^2$), so this solid is $y$-simple. Likewise, the solid is $x$-simple because the projection into the $yz$-plane gives the region $0 \leq z \leq 4 - y.$
A: Note that the $z=f(x,y)$ function is simply the intersection between the parabolic cylinder $z<4-x^2$ and the plane $z<4-y$, and thus the (x,y) domain is split in the intersection of both: in $y=x^2$.
Hence, the integration has two members, one for $y<x^2$ in which $z<4-x^2$:
$$
\int_{-2}^{2}\int_{0}^{x^2}\int_{0}^{4-x^2} \text{dzdydx} = \frac{128}{15}
$$
and for $y>x^2$ in which $z<4-y$:
$$
\int_{-2}^{2}\int_{x^2}^{4}\int_{0}^{4-y} \text{dzdydx} = \frac{256}{15}
$$
The answer is $\frac{128}{15}+\frac{256}{15}=\frac{128}{5}$
