# Triple integral over a three dimensional region

The integral $$\int_{-2}^2\int_{x^2}^4\int_0^{1-y/4} 1\ dz\ dy\ dx$$is a triple integral over a three dimensional region E, Sketch E, and rewrite this integral with $y$ as a first variable and $z$ second variable and $x$ as a third variable.

What I have done so far:

$0\le z \le{1-y/4}$

$x^2\le y \le+4$

$-2 \le x\le+2$

Now, to switch the order of integration, we will be integrating in the $yz$ plane first then integrating over $x$. So, we need to think about what $yz$ slices of our region look like. Key observation: they are triangles:
I've labelled all the key points of interest; we first fix $x$, which gives us a triangular cross section. Then, we fix $z$ and vary $y$ (the horizontal line). Its first point of contact with the solid is at $y=x^2$. It leaves the solid when $y=4-4z$ (i.e. when it hits the surface $z=1-y/4$). These are your $y$ bounds. Finally we must vary $z$ within the triangle - it starts at $0$, and ends at the "top" vertex of the triangle. There, $y=x^2$, so we have $z=1-x^2/4$. Thus we have:
$$\int_{x=-2}^{x=2}\int_{z=0}^{z=1-x^2/4}\int_{y=x^2}^{y=4-4z}1dydzdx$$