Common volume of three cylinders with unequal radii I would like to solve it please for the case where the radii can be a similar size - so the case where this statement is NOT true: $$ \mathbf r_1^2 \mathbf \geq \mathbf r_2^2 \mathbf + \mathbf r_3^2  $$ 
How do you solve for the common volume of 3 cylinders with unequal radii? (If you could please include the integral needed - I think it might need a cartesian equation system? Or if you have any good idea of what direction or things I could read up on to please learn to solve this. Thank you so much for your help!
$$x^2 + y^2 = r_1^2$$
$$x^2 + z^2 = r_2^2$$
$$y^2 + z^2 = r_3^2$$        where $r_1 \neq r_2 \neq r_3$ 
I can find that the common volume for equal radii using triple integration with circular co-ordinates to get the answer below (following this)
$$V_c = 8\cdot(2-\sqrt 2)\cdot r^3$$
Essentially I want to get to a place where I can find the equation for the common volume of three cylinders with different radii and at different angles.
So if anyone has an idea of how to calculate for cylinders at different angles other than 90 that would be great too! Thank you!

Let $a<b<c$ with $c<\sqrt{a^2+b^2}$ be the radii of the three cylinders
$$x^2+y^2\leq a^2,\qquad x^2+z^2\leq b^2,\qquad y^2+z^2\leq c^2\ ,$$
and denote by $S$ the part of the intersection lying in the positive octant. The set $S$ is standing on the quarter disc
$$S':\qquad x\geq0,\quad y\geq 0, \quad x^2+y^2\leq a^2$$
in the $(x,y)$-plane. The upper boundary of $S$ is determined by the two other cylinders. Therefore the height $z(x,y)$ of $S$ at the point $(x,y)\in S'$ satisfies
$$z^2(x,y)=\min\{b^2-x^2, \ c^2-y^2\}\ .$$
The two entries in the $\min$ are equal when $y^2=c^2-b^2+x^2$. It follows that
$$z(x,y)=\left\{\eqalign{\sqrt{b^2-x^2}\qquad&(y\leq\sqrt{c^2-b^2+x^2})\cr
\sqrt{c^2-y^2}\qquad&(y\geq\sqrt{c^2-b^2+x^2})\cr}\right.\ .$$ 
We now have to calculate
$${\rm vol}(S)=\int_{S'}z(x,y)\>{\rm d}(x,y)\ .\tag{1}$$
The  hyperbolic arc $y=\sqrt{c^2-b^2+x^2}$ intersects the quarter circle $x^2+y^2=a^2$ at the point
$$P=\left(\sqrt{a^2+b^2-c^2\over2}, \ \sqrt{a^2+c^2-b^2\over2}\right)=:(u,v)\ .$$
The hyperbolic arc and the lines $x=u$,  $y=v$ through $P$ partition the quarter disc $S'$ into four  sections. From $(1)$ we then obtain
$$\eqalign{{\rm vol}(S)&=\int_0^u \sqrt{c^2-b^2+x^2}\>\sqrt{b^2-x^2}\>dx+\int_u^a\sqrt{a^2-x^2}\>\sqrt{b^2-x^2}\>dx \cr&\qquad + \int_{\sqrt{c^2-b^2}}^v \sqrt{y^2-(c^2-b^2)}\>\sqrt{c^2-y^2}\>dy+\int_v^a\sqrt{a^2-y^2}\>\sqrt{c^2-y^2}\>dy\ .\cr}$$

