# Volume of the region outside of a cylinder and inside a sphere

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? • I don't think the answer is correct. The answer below by Martin is the best way to do it. If you want to use integral, it should be $\int _0^{2\pi }\int _{\frac{\pi }{6}}^{\frac{5\pi }{6}}\int _{1/\sin\phi}^{\sqrt{2}}\rho ^2\sin\phi \:d\rho d\phi d\theta+2\int _0^{2\pi }\int _{0}^{\frac{\pi }{6}}\int _{1/\cos\phi}^{\sqrt{2}}\rho ^2\sin\phi \:d\rho d\phi d\theta$ – KittyL Jun 28 '15 at 18:05
• @lasec0203: The cylinder in your question has infinite height, which doesn't match the figure. The sphere in your question (radius $2$) doesn't match the diagram (radius $\sqrt{2}$). The answer key integral, as written, does not give the volume outside a cylinder, but outside a cone. Could you please carefully check the problem statement, the purported answer, and the diagram to be sure they're consistent? :) – Andrew D. Hwang Jun 28 '15 at 18:22
• @AndrewD.Hwang, yes the figure is incorrect. I will change it here in a sec – lasec0203 Jun 28 '15 at 18:27
• @KittyL I think I understand it now, the endcaps are technically not outside of the cylinder because the boundary of the cylinder ends at (1, sqrt(3)) where the two figures intersect. I was viewing the cylinder as a can, closed top and bottom, which is wrong. The ends are open. – lasec0203 Jun 28 '15 at 18:28

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 $V_s-V_c=\pi(\frac{2^5}{3}-2\sqrt{3})$.

• You forget the volume of the two cap. – achille hui Jun 28 '15 at 17:54
• I don't see the flaw in what I've done. The sphere has a certain volume including the caps. The cilinder has another volume and it's inside the sphere. To compute the volume outside the cilinder and inside the sphere I've computed each volume and substracted. Could you elaborate? – Martín Forsberg Conde Jun 28 '15 at 17:56
• @achillehui yea, I though of going this route too, but when I asked my prof, he said "That actually won't work. If you do that, then you'd get $\:\frac{32\pi }{3}-2\pi \sqrt{3}$. Which is wrong. This method is ignoring the curved end caps of the cylinder, and just to be sure these endcaps are not hemispheres! So there is no geometry formula to use here instead of calculus." – lasec0203 Jun 28 '15 at 17:56
• @lasec0203: So are you looking for volumes including the caps or not? – KittyL Jun 28 '15 at 18:08
• @KittyL I'm quoting what my professor said. "Exactly. The region outside the cylinder and inside the sphere doesn't include the end caps. Just imagine shooting a hole through a sphere, and then finding the volume of what remains." this reply lost me – lasec0203 Jun 28 '15 at 18:11

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:

$$V = \int_{-\sqrt{3}}^{+\sqrt{3}} \{\pi (4-z^2)-\pi\}dz = 4 \sqrt{3}\pi$$

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. • please edit this into the original problem statement – MichaelChirico Jun 28 '15 at 20:43
• @MichaelChirico I will, but I leave the actually shapes the way I envisioned them originally. – lasec0203 Jun 28 '15 at 20:51
• I believe the limits for $\rho$ should be from $\csc\phi$ to 2 instead of 0 to $\sqrt{2}$. – user84413 Jun 29 '15 at 0:18
• @user84413, thanks for the catch. – lasec0203 Jun 29 '15 at 2:49