Cylindrical coordinates where $z$ axis is not axis of symmetry.

Question

I'm a little bit uncertain of how to set up the limits of integration when the axis of symmetry of the region is not centered at $z$ (this is for cylindrical coordinates). The region is bounded by $(x-1)^2+y^2=1$ and $x^2+y^2+z^2=4$. This is my attempt:

Let $x=r\cos\theta$ and $y=r\sin\theta$.

We are bounded on $z$ by $\pm\sqrt{4-r^2}$. We take $0\leq r\leq 2\cos\theta$ and $-\pi/2\leq \theta \leq \pi/2$ to account for the fact that the cylinder is not centered with the $z$ axis (shift on the $x$ axis). The volume is given by $$ V = \int\limits_{-\pi/2}^{\pi/2} \int\limits_0^{2\cos\theta}\int\limits_{-\sqrt{4-r^2}}^{\sqrt{4-r^2}} dz\,(r\,dr)\,d\theta \ . $$

Answer 1

With all due respect, the OP's solution is correct. The circle $(x-1)^2+y^2=1$ sits inside the circle $x^2+y^2=4$, so the region in question projects onto the entire disk $\{(x-1)^2+y^2\le 1\}$. Thinking about $z$ cross-sections is awkward and to be discouraged.

Answer 2

I prefer to visualize the cross sections in $z$. Draw a picture of various cross-sections in $z$: there is an intersection region for each $z$

You have to find the points where the cross-sections intersect:

$$4-z^2=4 \cos^2{\theta} \implies \sin{\theta} = \pm \frac{z}{2} \ .$$

For $\theta \in [-\arcsin{(z/2)},\arcsin{(z/2)}]$, the cross-sectional area integral at $z$ looks like

$$\int_{-\arcsin{(z/2)}}^{\arcsin{(z/2)}} d\theta \: \int_0^\sqrt{4-z^2} dr \, r = (4 -z^2)\arcsin\left(\frac{z}{2}\right) \ .$$

For $\theta \in [-\pi/2,-\arcsin{(z/2)}] \cup [\arcsin{(z/2)},\pi/2]$, this area is

$$2 \int_{\arcsin{(z/2)}}^{\pi/2} d\theta \: \int_0^{2 \cos{\theta}} dr \, r = \pi - 2 \arcsin\left(\frac{z}{2}\right) -z \sqrt{1-\frac{z^2}{4}} \ . $$

To get the volume, add the above two expressions and integrate over $z \in [-2,2]$; the expression looks odd, but we are really dealing with $|z|$ :

$$V = 2 \int_{0}^2 \: \left(\pi + (2-z^2) \arcsin\left( \frac{z}{2}\right) -z \, \sqrt{1-\frac{z^2}{4}}\right)\,dz = \frac{16 \pi}{3} - \frac{64}{9} \ .$$