Double integrals: finding a volume of a solid I am trying to determine the bounds given a solid bounded by the plane $y+z=4$ and the cylinder $y=x^2$, and $xz$ and $yz$ planes in the first octant. I am not sure if I am on the right track, but to find my bounds for $y$, I set $z$ equal to zero and calculated $y=4$. This gives a region enclosed by a parabola with the upper y-bound at 4. So my outer integral (assuming type I region), I believe, is $x^2 \leq y \leq 4$. And since there is a vertical asymptote at $x=1$, I am assuming that my inner integral is bounded by $0 \leq x \leq 1$.
I know that I am going to need to integrate xz and yz as the functions, but I am not sure how to express these in terms of $z$, which is what I assume I am supposed to do since all problems I have encountered up until this point that are in 3D have done this. However, all of those problems didn't have z as a coefficient of either the x or y terms (e.g. they were in a form like $x + y + z$). Am I on the right track? And if so, how should I approach this?
 A: It is useful to sketch, or plot the volume and domain of integration to get a better picture. What you suggested will work, the only bit you are missing is the function $z$ which is merely
$$z=4-x$$
When you compute a double integral, you actually sum over a set of tall thin cylinders with bases mounted on the domain of integration and heights given by the values of $z$

It would be more convenient though to swap the order of integration and integrate first with respect to $x$ which, for every given $0\le y\le 4$ ranges from $0$ to $\sqrt{y}$. That is, you slice the parabola in stripes parallel to the $x$ axis.
Therefore,
$$V=\int_0^4 dy \int_0^{\sqrt{y}}\left(4-y\right)dx=\int_0^4\left(4y^\frac{1}{2}-y^\frac{3}{2}\right)dy=\left.\left(\frac{8}{3}y^\frac{3}{2}-\frac{2}{5}y^\frac{5}{2}\right)\right|_0^4=\frac{128}{15}$$
Now you can try to calculate using your original idea of the limits and verify, if you arrive at the same result.

A: Your first paragraph is almost spot on. I cannot make sense of your second paragraph. The mistake in the first paragraph is that you are making up the limit for $x$. The intersection between $y=x^2$ and $y=4$ (the intersection of the plane $y+z=4$ and the $x$-$y$ plane). 
Since we have to stay in the first octant, the solid is bounded below by the $x$-$y$ axis. 
The volume of your solid is
$$
\int_0^2\int_{x^2}^4\,(4-y)\,dy\,dx.
$$
Properly speaking, your integrand is $(4-y)-0$, as you integrate the difference between the "ceiling" and the "floor". 
