When three cylinders intersect

The intersection of the three cylinders form eight sections, four above the $xy$ plane and four below the $xy$ plane. Finding the volume of one section gives $\frac{1}{8}$ of the answer. Therefore, I am going to calculate the volume of one section where $z>0$ and multiply the answer by 8. I am only going to consider the cylinder $x^{2}+z^{2}=1$ ; thus, the equation we need is $z=\sqrt{1-x^{2}}$ . Using Cylindrical coordinates, let $x=rcos(\theta)$ , where the radius goes from $0 \ to\ 1$ and the angle $\theta$ from $\frac{-\pi}{4}$ to $\frac{\pi}{4}$ . then the volume is defined by:

$8$ $\large\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}$ $\large\int_{0}^{1}$ $\sqrt{1-r^{2}cos^{2}(\theta)}$ $rdrd\theta$ =

$\frac{8}{3}$ $\large\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}$ $1-\frac{|sin(\theta)^{3}|}{cos^{2}(\theta)}d\theta$ =

$\frac{16}{3}$ $\large\int_{0}^{\frac{\pi}{4}}$ $1-\frac{sin(\theta)^{3}}{cos^{2}(\theta)}d\theta$ =

= $\frac{16}{3}$ $\large[ tan(\theta)-cos(\theta)-\frac{1}{cos(\theta)}]_{0}^{\frac{\pi}{4}}$ = $16 - 8\sqrt{2}$ = $\boxed{8(2- \sqrt{2})}$