# How to find volume of solid using double integrals?

Trying to find the volume of the solid bounded by the parabolic cylinders $z=7x^2, y=x^2$ and the planes $z=0, y=4$.

I'm not sure how to set up the range of outer integral. Here's what I have so far: $$\int \int_{x^2}^4 7x^2 \ dy \ dx$$

• Are you sure what you describe is a cylinder? I cannot see how it is – user160738 Dec 14 '15 at 6:49
• @user160738 The solid is not a cylinder but it is bounded by two cylinders and two planes. – d0rmLife Dec 14 '15 at 6:51
• It's a bit unclear as to what solid you are referring to, because $z=7x^2$, $y=x^2$ only describes a parabolic curve, so solid bounded by this curve and $z=0,y=4$ does not exist. But I suspect the one you are considering has volume $14\left(8-\frac{16}{3}\right)$ – user160738 Dec 14 '15 at 7:16
• @user160738 They are parabolic cylinders, I will edit to clarify. Also that is not the answer I'm looking for. – d0rmLife Dec 14 '15 at 7:21

In this picture the vertical axis is $z$, the arrow pointing down is $y$, and the arrow pointing left is $x$.
Trying to find the volume of the solid bounded by the cylinders $z=7x^2, y=x^2$ and the planes $z=0, y=4$.
Starting from $x^2 ≤y≤ 4$, you need to then bound $z$ and $x$. $z$ is constrained by the parabolic cylinder $z=7x^2≥0$ and $z=0$ so we want $0≤z≤7x^2$. Finally we are then free to use the maximal and minimal values of $x$. As $y≤4$ and $y=x^2$ is the constraint that affects $x$, we see that $-2≤x≤2$. Putting this together,
$$V = ∫_{x=-2}^2 ∫_{z=0}^{7x^2} ∫_{y=x^2}^4 \ dy \ dz \ dx = \frac{896}{15}$$