Double integral in polar coordinates Use polar coordinates to find the volume of the given solid inside the sphere $$x^2+y^2+z^2=16$$ and outside the cylinder $$x^2+y^2=4$$
When I try to solve the problem, I keep getting the wrong answer, so I don't know if it's an arithmetic error or if I'm setting it up incorrectly. I've been setting the integral like so: $$\int_{0}^{2\pi}\int_{2}^4\sqrt{16-r^2}rdrd\theta$$
Is that the right set up? If so then I must have made an arithmetic error, if it's not correct, could someone help explain to me why it's not that? Thanks so much!

It's almost correct. Recall that the integrand is usually of the form $z_\text{upper}−z_\text{lower}$, where each $z$ defines the lower and upper boundaries of the solid. As it is currently set up, you are treating the sphere as a hemisphere, where your lower boundary is the $xy$-plane. Hence, you need to multiply by $2$, since we are technically doing:
$$
\left(\sqrt{16-r^2} \right) - \left(-\sqrt{16-r^2} \right)
$$
