The cylinder is $x^2 +y^2 = 1$ and the sphere is $x^2 + y^2 + z^2 = 4$. I have to find the volume of the region outside the cylinder and inside the sphere. The triple spherical integral for this problem is (from the answer key) $$\int _0^{2\pi }\int _{\frac{\pi }{6}}^{\frac{5\pi }{6}}\int _{csc\phi }^2\:\rho ^2sin\phi \:d\rho \:d\phi \:d\theta $$
What is confusing me here is that there's some space at the endcaps of the cylinder that is not being accounted for. Why is this the case?
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turns out that I was viewing the figures in the wrong way. As Andrew D. Hwang pointed out the cylinder has infinite height. Since the cylinder ends are open, it engulfs the top and bottom end of the sphere so the endcaps are not included. This explains why the integral:
$$\int _0^{2\pi }\int _{\frac{\pi }{6}}^{\frac{5\pi }{6}}\int _{csc\phi }^2\:\rho ^2sin\phi \:d\rho \:d\phi \:d\theta $$
is correct for this problem.
On
The new version of the problem (i.e. with an infinitely long cylinder through the sphere) is best solved using cylindrical coordinates $z$, $\rho$, $\phi$.
When viewed as a function of the variable $z$, the sphere consists of circular disks of radius $R = \sqrt{4 - z^2}$ and thickness $dz$. The cylinder cuts out the central region of these disks. The radius of the hole is $1$. We see that the disk is larger than the cut-out region when $z^2 > 3$. Hence the limits of the integration over $z$ are $-\sqrt{3}$ and $+\sqrt{3}$.
The actual integration is now straightforward. We get:
$$V = \int_{-\sqrt{3}}^{+\sqrt{3}} \{\pi (4-z^2)-\pi\}dz = 4 \sqrt{3}\pi$$
Why don't you just compute the $2$ volumes?
The cilinder has volume $V_c=\pi\cdot 2\sqrt{3}$ and the sphere has volume $V_s=\frac{4}{3}\pi*2^3$.
The volume inside the sphere and outside the cilinder is $V_s-V_c=\pi(\frac{2^5}{3}-2\sqrt{3})$.