Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
$$(x-1)^2+y^2=1\longrightarrow x^2-2x+1+y^2=1\longrightarrow x^2+y^2-2x=0\longrightarrow r^2-2r\cos\theta=0$$ 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? :) –  B. 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.

share|improve this answer
    
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 $|z|$ :

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

share|improve this answer
    
Thank you for the insight! –  5space Apr 20 '13 at 1:57
1  
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
1  
@5space: yes, that's right; I wrote the wrong answer in. Thanks. –  Ron Gordon Apr 20 '13 at 7:40
1  
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
1  
@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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.