# 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!

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It's almost correct, except I think you need to multiply by $2$ (since we are technically doing $\sqrt{16-r^2}-(-\sqrt{16-r^2})$). –  Adriano Jul 29 '13 at 3:29
Why are we subtracting that? –  Jc E Jul 29 '13 at 3:31
The integrand is usually $z_{\text{upper}}-z_{\text{lower}}$, 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. –  Adriano Jul 29 '13 at 3:33
Oh, okay! That makes sense, thank you so much! –  Jc E Jul 29 '13 at 3:36
You're very welcome. =] –  Adriano Jul 29 '13 at 3:45

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: $$\left(\sqrt{16-r^2} \right) - \left(-\sqrt{16-r^2} \right)$$