Q:

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: $$ \int \int_{x^2}^4 7x^2 \ dy \ dx $$ 

A:

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,
$$ V = ∫_{x=-2}^2 ∫_{z=0}^{7x^2} ∫_{y=x^2}^4 \ dy \ dz \ dx = \frac{896}{15}$$

