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