Example: Evaluate $$\int_0^2\int_0^{4-x^2}\int_0^x\frac{\sin{2z}}{4-z} \text{dy dz dx}$$
I'm a little confused on how to proceed after taking the integral with respect to $y$. Is this a problem that requires switching the order of integration? If so, how would I proceed in changing the bounds?
Thanks!
The innermost integral with respect to $dy$ is easy enough, isn't it? So let's focus on the two outside integrals: $\displaystyle \int_0^2\int_0^{4-x^2}\left[\cdots\right]\,dz\,dx$. Evaluating this double integral would be simplified if we switch the order of integration between $z$ and $x$.
To see how that can be done, you should sketch a picture of the region that this double integral defines in the $xz$-plane: this is the region defined by the conditions $0\le x\le2$ and $0\le z\le4-x^2$. (And if I were you, I would graph it on the usual Cartesian plane, where you simply label the vertical axis as $z$ rather than $y$). Then looking at that picture, describe how you would integrate over the same region in the order of $dx\,dz$. Hint: you should get something like $\displaystyle \int_0^4\int_0^{\cdots}\left[\cdots\right]\,dx\,dz$.
After that, put the triple integral back together, and it should become something manageable.