How to find the volume enclosed by intersection of three orthogonal cylinders? If I have three cylinders $$x^2 + y^2 =1 $$ $$ x^2 + z^2 =1  $$ $$y^2 + z^2 =1$$ and I need to find the volume contained in their intersection. I know the figure I will get is a steinmetz solid looking like this. Now I was looking for the solution and I found this thread Intersection of Three Cylinders of equal radius. In this thread Mr. John Hughes explains beautifully about the figure. Here is the image . What I want is to compute the volume of this solid.   
I can infer from my research and little of intuition that the region PCB belongs to $x^2 + y^2 =1$ , region PAC belongs to $y^2 + z^2 =1$ and region PAB belongs to $x^2 + z^2 =1$. If I try to find the volume I would set an integral like this $$\iiint_E dx~ dy~ dz $$ where $E$ is the whole region ABC along with P.  
Now, the problem is the bounds of the integrals. I'm having no idea of how to get the bound. I can (with little confidence) say that $y$ is bound by the curve PCB so I may write $$0 \leq y \leq +\sqrt{1-x^2}$$ and $$ 0 \leq x \leq 1$$ . I'm stuck here, I need help with as much elaboration possible from basics.  
I know this type of question has been asked multiple times and each time a different think has been asked, so I'm going with the tradition. Please help me in cartesian coordinates only and with this diagram, if possible. Curvilinear coordinates's bounds would also work but please add a through explanation to it.
Thank you. I hope someone will surely help me through this.  
I apologize if I'm asking a conventional thing, I apologize if I'm too demanding , I apologize if I'm wrong in my very essence of the concept.
 A: It is enough to consider the areas of the sections for $z\in(0,1)$. If $z\in\left(0,\frac{1}{\sqrt{2}}\right)$ we have that $x^2+y^2\leq 1$ and $|x|,|y|\leq\sqrt{1-z^2}$ define a section which is the intersection between a square and a circle, whose area is 
$$\pi-8\int_{\sqrt{1-z^2}}^1\sqrt{1-x^2}\,dx = \pi+4z\sqrt{1-z^2}-4\arcsin(z).$$
If $z\geq \frac{1}{\sqrt{2}}$, the $z$-section is just a square with side length $2\sqrt{1-z^2}$, whose area is $4(1-z^2)$.
It follows that the enclosed volume equals
$$ 2\left[\int_{0}^{1/\sqrt{2}}\pi+4z\sqrt{1-z^2}-4\arcsin(z)\,dz+\int_{1/\sqrt{2}}^{1}4(1-z^2)\,dz\right]=\color{red}{16-8\sqrt{2}}.$$
A simpler decomposition is the following: a cube with edge length $\sqrt{2}$ and six "curved square pyramids" having volume $\int_{1/\sqrt{2}}^{1}4(1-z^2)\,dz$. It follows that the outcome is a linear combination of $1$ and $\sqrt{2}$ with rational coefficients, even without performing the explicit computation.  Increasing the number of cylinders (oriented like the lines joining the center of a regular polyhedron with the centers of its faces) leads to more accurate algebraic approximations of $\frac{4\pi}{3}$, which is the volume of the unit sphere.
