# Three-Dimensional Lebesgue Measure

I've been trying to calculate the three-dimensional Lebesgue measure of $$\left\{(x,y,\theta)\in\mathbb{R}^2\times[0,\pi):\ x^2+y^2\leq 1;\; \theta\in[0,\pi);\; (x+\cos(\theta))^2+(y+\sin(\theta))^2\leq1\right\}.$$

When I was working on it, I tried to do what seemed most natural: using some kind of polar coordinates, which made it seem much more spherical, but I haven't made any further progress on what to make of it because of the last restraint. I tried thinking about this geometrically to get the intuition of where to go with it, but I haven't figured it out yet. Any suggestions?

-

Fix $\theta\in[0,\pi)$, the planar region $$R(\theta):=\{(x,y)\in\mathbb{R}^2:x^2+y^2\le 1; (x+\cos\theta)^2+(y+\sin\theta)^2\le 1\}$$ is just the intersection of two closed disks with the same radius $1$: one disk is centered at $(0,0)$ and the other is centered at $(-\cos\theta,-\sin\theta)$. Then some easy planar geometry argument will show that the area of $R(\theta)$ is $\frac{2\pi}{3}-\frac{\sqrt{3}}{2}$. Then by Fubini's theorem, the Lebesgue measure of your set is $\frac{2\pi^2}{3}-\frac{\sqrt{3}\pi}{2}$.

-
I think I get the first part and the last part (except where you say open disks, I think they might be closed). The planar geometry argument has me confused, but thank you very much for the light you've shed on the problem. At least to some extent I see it now. –  Clayton Dec 11 '12 at 6:04
@Clayton: Yes, they are closed disks. Sorry for my typo. Hope you have filled in the gaps now. –  23rd Dec 11 '12 at 6:24