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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 $$ V = \int\limits_{-\pi/2}^{\pi/2} \int\limits_0^{2\cos\theta}\int\limits_{-\sqrt{4-r^2}}^{\sqrt{4-r^2}} dz\,(r\,dr)\,d\theta \ . $$

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$$(x-1)^2+y^2=1\longrightarrow x^2-2x+1+y^2=1\longrightarrow x^2+y^2-2x=0\longrightarrow r^2-2r\cos\theta=0$$ so the range for $r$ would be $r|_{0}^{2\cos\theta}$. The range for $\theta$ looks fine and for $z$ you did it correct. So what is it your question exactly? :) –  Babak S. Apr 19 '13 at 18:57
Mostly I am uncertain about the range of $r$ when the axis of symmetry is shifted. I have not seen how to get the limits for $r$ in this case and there were no similar problems or examples in my text. Then someone said my set up is incorrect, but I could not see what was wrong. –  5space Apr 19 '13 at 19:01

2 Answers 2

up vote 1 down vote accepted

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.

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Thank you for the feedback, Dr. Shifrin! Btw, I have been enjoying your multivariable text as a reference in my course! So double thanks! –  5space Apr 20 '13 at 1:54
With all due respect, sir, I find the cross sections easier to visualize. The result below is not at all awkward in my opinion - beauty is in the eye of the beholder. The results both agree. –  Ron Gordon Apr 20 '13 at 2:29

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 $z$

3D Figure

Cross sections

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 $|z|$ :

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

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Thank you for the insight! –  5space Apr 20 '13 at 1:57
It is an interesting approach and I appreciate seeing different ways of solving the same problem. I tried something similar before realizing that the upper $r$ limit is $r=2\cos\theta$, but this is much nicer than what I was attempting. I also tried shifting the cylinder so that its axis of symmetry was along $z$ (of course shifting the sphere appropriately). But this method did not work out as nicely. –  5space Apr 20 '13 at 3:48
@5space: yes, that's right; I wrote the wrong answer in. Thanks. –  Ron Gordon Apr 20 '13 at 7:40
Well, if your goal is to reduce problems to single integration, then it's the way to go. As a small challenge, I'll throw out one of my favorite problems: Find the volume of the region bounded by the three cylinders $x^2+y^2=1$, $x^2+z^2=1$, and $y^2+z^2=1$. –  Ted Shifrin Apr 20 '13 at 13:26
@TedShifrin: here is a link to my answer to your problem. stackexchange.moderatenerd.com/2013/04/21/… –  Ron Gordon Apr 21 '13 at 14:37

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