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!
 A: According to this [first] picture made in GeoGebra, this enclosed volume [seems to be] strictly inside the intersection of the 2 smaller cylinders. Am I right? [No! I was not!] 
This second picture shows that the third one must join the diagonal point where the other two meet, viewed along the z axis or be larger than that if it is to not interfere with the answer for the two first cylinders.


The intersection of the solid parts of [two] cylinders are on a surface called hyperbolic cylinder (Could be something else if they are not perpendicular, like part of another quadric). We find it's equation by eliminating the common variable in two equations (see picture). You will have to decide where you position your cylinders (smaller radius along x axis, the other along y). Then you need change the angle of only one of them, and find the new equation (with a slope parameter).

A: 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
$$\eqalign{{\rm vol}(S)&=\int_0^u \sqrt{c^2-b^2+x^2}\>\sqrt{b^2-x^2}\>dx+\int_u^a\sqrt{a^2-x^2}\>\sqrt{b^2-x^2}\>dx \cr&\qquad + \int_{\sqrt{c^2-b^2}}^v \sqrt{y^2-(c^2-b^2)}\>\sqrt{c^2-y^2}\>dy+\int_v^a\sqrt{a^2-y^2}\>\sqrt{c^2-y^2}\>dy\ .\cr}$$

