# Cylinder cut out of a sphere. (volume).

1.) A cylinder $Z=${$(x,y,z)\in R^3|x^2+y^2\leq \rho^2$}$~~$ ($0<\rho<R$) is cut out of a sphere $K\in R^3$ with radius $R>0$ centered around the origin. Find the volume of the rest $R_Z:=K\backslash Z$.

2.) Let $\Omega$ be a domain, which is restricted by the paraboloid $z=x^2+y^2$ and the cone $z=\sqrt{x^2+y^2}$. Compute the moment of inertia of $\Omega$ which is given by the integral

$I_z=\int_\Omega(x^2+y^2)d(x,y,z)$ with mass density $\rho =1$ in respect to the rotation around the z-axis.

For 1.):

Okay, my thinking was that I only need the volume of the cylinder, since I assume that the volume of the sphere is just $\frac{4\pi r^3}{3}$ and the rest would be $V_{R_z}=\frac{4\pi r^3}{3}-v_z$. But I don't know how to get the volume of the cylinder. Well, at least I don't know if my approach for it is correct. Here is my idea anyway:

I looked up the general formula for the volume of a cylinder, which is:

$\int_0^{2\pi}\int_0^a\int_0^hrdzdrd\theta$, where $a=\rho$ and h the height of the cylinder. So I was guessing that I only need to find a more explicit value for h. Since the distance between the origin and one "vertex" (not really a vertex of the cylinder, but one of the rectangle in the sketch) is R and with an angle $\phi$ between R and the z-axis one could get the height as $2Rcos(\phi)=h$ and then setting this as the upper boundary for z. Might actually be just $Rcos(\phi)$ since the cylinder is centered in the origin and only goes up to $\frac{h}{2}$. But I don't know if this is the right approach since I'm integrating in respect to $\phi$.

2.): I have no idea how to solve this one, since this whole thing is enclosed by two objects and I don't know how to set the boundaries for the integral or what coordinate system to use.

Any ideas?

• By my reading of the question, in (1) the cylinder drills out a hole all the way through the sphere, it doesn't leave the "caps". You might like to set up your integral on the sphere in coordinates (it's quite possible to do this with spherical coordinates, too). May 25, 2015 at 8:41
• You are right, the caps are gone as well, noticed it while reading it once more. Makes more sense. But I still don't know how to set up the integral. May 25, 2015 at 8:48

$1.$ You can find the height of the cylinder by using Pythagorean Theorem. Draw the radius from the center to the edge of the cylinder and notice that $\text{height}=2\sqrt{R^2-\rho^2}$. Now you can subtract this cylinder and the two spherical caps to find the volume. I will not derive the formula for spherical cap volume here. You can look here: http://planetmath.org/volumeofsphericalcapandsphericalsector

$2.$ We will convert to cylindrical coordinates. The volume we want is this: http://www.wolframalpha.com/input/?i=Plot%5B%7Bz%3D%3D%7Cr%7C%2Cz%3D%3Dr%5E2%7D%2C+%7Br%2C-1%2C1%7D%5D

In cylindrical coordinates the integral becomes:

$$\int^{2\pi}_0\int^1_0\ \int ^{r}_{r^2} r^2\,r\,dz\,dr\, d\phi$$

I assume you can handle the rest.

• A question about the integral boundaries. I can't fully see how you got to $r^2$ and $r$ for the boundaries of $dz$. May 25, 2015 at 14:37
• Look at the wolframalpha sketch. There you need the small area in between. The equations of those curves are z=r^2 (the lower one) and z=|r| (the upper one) May 25, 2015 at 15:02
• You are right. I had 0 to r in mind, but it's actually the enclosed area, so it's $r^2$ to $r$. May 25, 2015 at 15:04

Hint The subset of $K$ whose volume you are computing can be described in cylindrical coordinates as $$K \cap \{(r, \theta, z)_c : r > \rho \}.$$

(It is unfortunate here that the symbol, $\rho$, which in this setting is often reserved for the [co]latitudinal spherical coordinate is used in even one other way here, let alone two.)