<s>[INST] 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?
 [/INST] 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 $\frac{128}{15}+\frac{256}{15}=\frac{128}{5}$
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