Find the volume of intersection between cylinders

Question

Find the volume of intersection of the cylinder
{$ x^2 + y^2 \leq 1 $} , {$ x^2 + z^2 \leq 1$}, {$ y^2 + z^2 \leq 1$}.

i am having tough time finding the volume how do i solve this kind of questions ? .

my trial : i will move to the cylinder coordinates of the xy cylinder let :

$x^2 + y^2 = r^2 $

$ z = z $

$0\leq\theta \leq 2\pi$

solving the inequalties i get :

$ 0 \leq r^2 \leq 1$

$ -\sqrt{1-\frac{r^2}{2}}\leq z \leq \sqrt{1-\frac{r^2}{2}}$

$0\leq\theta \leq 2\pi$

the integral is :

$ \int_{z=-\sqrt{1-\frac{r^2}{2}}}^{z=\sqrt{1-\frac{r^2}{2}}}\int_{r=0}^{r=1}\int_0^{2\pi} dz \ dr \ d{\theta}$ = $ \frac{4\pi}{\sqrt{2}}\frac{(2-r^2)^{\frac{3}{2}}}{-3} |_{r=0}^{r=1}$

Answer 1

$$V = 16 \int\limits_0^{\pi/4} \int\limits_0^1 s \sqrt{1 - s^2 \cos^2 \theta}\ ds\ d\theta = 8(2 - \sqrt{2}) $$

Answer 2

Perhaps surprisingly, it is not necessary to use polar coordinates, nor even multiple integration or trigonometry, to find the volume.

What we need to do is consider the volume of a solid whose cross-sections perpendicular to the $x$-axis are rectangles whose width is $x$ and whose height is $\sqrt{1-x^2} - 1/\sqrt{2}$, over the interval $x \in [0,1/\sqrt{2}]$. This is simply $$\int_{x=0}^{1/\sqrt{2}} x \left(\sqrt{1 - x^2} - 1/\sqrt{2}\right) \, dx = \left[-\frac{(1-x^2)^{3/2}}{3} - \frac{x^2}{2\sqrt{2}}\right]_{x=0}^{1/\sqrt{2}} = \frac{8 - 5 \sqrt{2}}{24}.$$ Then the desired volume is $48$ times this volume, plus the volume of the inscribed cube, which gives $$2\sqrt{2} + 2(8-5\sqrt{2}) = 8(2 - \sqrt{2}).$$ I have left the reasoning as an exercise for the reader.

An alternative computation can be set up that includes the inscribed cube in the integral, simply by changing the height of the perpendicular cross section to $\sqrt{1-x^2} - x$, giving $$\frac{V}{48} = \int_{x=0}^{1/\sqrt{2}} x \left(\sqrt{1-x^2} - x\right) \, dx.$$ Again, the details are left for the reader.