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Find the volume of the intersection of the cylinders $$\{(x,y,z)\in \mathbb{R}^3: x^2+z^2\leq 1\} \cap \{(x,y,z)\in \mathbb{R}^3:x^2 + y^2 \leq 1\}.$$

My first approach led me into contradiction, and my second got me the right answer. I am concerned to find the conceptual errors in my first approach.

I start from the set up $$V = 2\iint_{D}\sqrt{1-x^2}dA,$$

where $dA = dx\,dy$ and $D$ is the unit disc.

First pass (polar):

$$V = 2\iint_{D}\sqrt{1-x^2}dA\\ = 2\int_{0}^{2\pi}\!\!\int_{0}^1\sqrt{1-x^2}r\,dr\,d\theta$$

in the first quadrant, we can parameterize by $x,\theta$

$$= 8\int_{0}^{\pi/2}\!\!\int_{0}^1\sqrt{1-x^2}\frac{x}{\cos{\theta}}\,dr\,d\theta\\= 8\int_{0}^{\pi/2}\!\!\int_{0}^{\cos{\theta}}\sqrt{1-x^2}\frac{x}{\cos^2{\theta}}\,dx\,d\theta\\ = -4\int_{0}^{\pi/2}\sec^2{\theta}\int_{1}^{\sin^2{\theta}}u^{1/2}du\,d\theta\\=4\int_{0}^{\pi/2}\frac{1}{\cos^2{\theta}}\int_{\sin^2{\theta}}^1 u^{1/2}du \,d\theta\\ = 4\int_{0}^{\pi/2}\frac{1}{\cos^2{\theta}} \left( 3/2 - \sin^3{\theta} \right) d\theta,$$

and I notice the first term is not a convergent integral. Why not? Where is my mistake? I know parameterizing by $x,\theta$ is sort of funky, but why is it not working?

Second Pass (Cartesian/Green's theorem): $$2\iint_{D}\sqrt{1-x^2}dx\,dy\\ \overset{\text{Green's}}{=} -2\int_{\partial D}y\sqrt{1-x^2}dx\\ = -2\int_{0}^{2\pi}\sin{\theta}|\sin{\theta}|(-\sin{\theta} d\theta)\\ = 4\int_{0}^{\pi}\sin^3{\theta}d\theta = \frac{16}{3}.$$

Thanks for your help as always.

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Eric, indeed, there is no factor of $\pi$ in this. You can actually do this problem with single-variable calculus, but if you want to use polar coordinates, you'd better write $\sqrt{1-x^2}$ in polar coordinates! The Green's Theorem idea is cute, but your answer is wrong. –  Ted Shifrin Oct 17 '13 at 16:23
    
P.S. If you're going to make a different sort of coordinate change, don't forget you need the Jacobian. :) –  Ted Shifrin Oct 17 '13 at 16:37
    
It was a single variable change of variables to my mind, though, $r = \frac{x}{\cos{\theta}}$, and I tried to throw the integrating factor in there. Still trying to find my mistake in changing $dr$ to $dx$. –  Eric Auld Oct 17 '13 at 16:39
    
Eric, I suppose you can think that way, but your limits of integration had better change. –  Ted Shifrin Oct 17 '13 at 16:42
    
Yes! Thank you Ted, that is the mistake. Also there was a computational error; now I get $\frac{16}{3}$ –  Eric Auld Oct 17 '13 at 16:43

2 Answers 2

up vote 1 down vote accepted
+50

The total volume is equal to

$$8\int_{\theta=0}^{\theta=\pi/2}\int_{x=0}^{x=\cos\theta}\sqrt{1-x^2}\ \biggl|\frac{\partial(x,y)}{\partial(x,\theta)}\biggr|\,dx\,d\theta$$

and $y=x\tan\theta$, so

$$\biggl|\frac{\partial(x,y)}{\partial(x,\theta)}\biggr|=\Biggl|\begin{bmatrix}1&0\\\tan\theta&x\sec^2\theta\end{bmatrix}\Biggr|=x\sec^2\theta\,,$$

hence the corresponding volume is equal to

$$\begin{align*} 8\int_{\theta=0}^{\theta=\pi/2}\sec^2\theta\int_{x=0}^{x=\cos\theta}\sqrt{1-x^2}\ x\,dx\,d\theta=&\,8\int_{\theta=0}^{\theta=\pi/2}\sec^2\theta\ \frac{-(1-x^2)^{3/2}}{3}\biggl|_{x=0}^{x=\cos\theta}\,d\theta\\ =&\,\frac83\int_{\theta=0}^{\theta=\pi/2}\sec^2\theta(1-\sin^3\theta)\,d\theta\\ =&\,\frac83\int_{\theta=0}^{\theta=\pi/2}\sec^2\theta(1-\sin^3\theta)\,d\theta\\ =&\,\frac83\biggl[\,\int_{\theta=0}^{\theta=\pi/2}\sec^2\theta-\int_{\theta=0}^{\theta=\pi/2}\sec^2\theta\sin^3\theta\,d\theta\,\biggr]\\ =&\,\frac83\bigl[\tan\theta-\sec\theta-\cos\theta\bigr]\biggl|_{\theta=0}^{\theta=\pi/2}\\ =&\,\frac83\ \lim_{\theta\to(\pi/2)^-}\,(\tan\theta-\sec\theta+2)\\ =&\,\frac{16}3\,. \end{align*}$$

Moral of the story: you cannot split the last integral of yours, because both integrals diverge as improper integrals.

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A ha...I thought it was a sign that something was wrong that one of the terms was clearly divergent. Thank you for the insight. –  Eric Auld Oct 22 '13 at 7:54
    
How are you integrating $\sec^2 {\theta} \sin^3{\theta}$? –  Eric Auld Oct 22 '13 at 10:31
    
@EricAuld If $v=\cos\theta$ then $-\sec^2\theta\sin^3\theta\,d\theta=v^{-2}\sin^2\theta\,dv=v^{-2}(1-v^2)\,dv\,.$ –  Matemáticos Chibchas Oct 22 '13 at 23:14

Let $A$ be the intersection of the cylinders. What are the possible values of $x$ for $(x,y,z)\in A$? Certainly $|x|\leq 1$ is a necessary condition (because $x^2+y^2\leq1$), and it is also sufficient, because given such $x$ you have $(x,0,0)\in A$. Therefore

$$vol(A)=\int_{x=-1}^{x=1}\cdots\ \,.$$

Now let $x\in[-1,1]$ be fixed. What are the possible values of $z$, with $(x,y,z)\in A$? You must have $|z|\leq\sqrt{1-x^2}$, and again it is also sufficient, because in that case $(x,0,z)\in A$. Therefore your volume becomes

$$vol(A)=\int_{x=-1}^{x=1}\int_{z=-\sqrt{1-x^2}}^{z=\sqrt{1-x^2}}\cdots\ \,.$$

Finally, given $x,z$ as before, what are the possible values of $y$ with $(x,y,z)\in A$? condition $|y|\leq\sqrt{1-x^2}$ is automatically necessary and sufficient (no need this time to "adjust variables"), and so we arrive to the formula

$$vol(A)==\int_{x=-1}^{x=1}\int_{z=-\sqrt{1-x^2}}^{z=\sqrt{1-x^2}}\int_{y=-\sqrt{1-x^2}}^{y=\sqrt{1-x^2}}\ 1\,dy\,dz\,dx\,.$$

The innermost iterated integral is equal to $2\sqrt{1-x^2}$; since the following iterated integral is based on the variable $z$, it follows that the next iterated integration yields $2\sqrt{1-x^2}\cdot2\sqrt{1-x^2}=4(1-x^2)$. Therefore your volume becomes $\int_{x=-1}^{x=1}4(1-x^2)\,dx=$ do it yourself!

EDIT: Sorry, I didn't read your specific question, I just read the title...

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No problem, I'm looking for the conceptual error in the first method I tried. The first method led me into a contradiction... –  Eric Auld Oct 21 '13 at 6:16

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