Volume of a solid formed by 3 cylinders I am trying to find the volume of the solid enclosed by three cylinders given by $x^2+y^2=1$, $x^2+z^2=1$, and $y^2+z^2=1$. I'm supposed to be using a triple integral, and I assume, cylindrical coordinates.
So far, I've figured out that I need to evaluate a triple integral of $dV$, which is equal to $rdzdrd\theta$. However, I am having trouble figuring out what bounds to use for $r, \theta, z$.
Any assistance or hints with this problem would be greatly appreciated!
 A: Take the cylinders as solid, so $x^2+y^2 \le 1$ and so on.  
Take for symmetry the first octant, i.e. $0 \le x,y,z$.   
Then change the coordinates to cylindrical to obtain
$$
\left\{ \matrix{
  0 \le r\,(by\,def.) \hfill \cr 
  0 \le r\cos \theta ,r\sin \theta ,z \hfill \cr 
  r^2 \left( {\cos ^2 \theta  + \sin ^2 \theta } \right) \le 1 \hfill \cr 
  z^2  + r^2 \cos ^2 \theta  \le 1 \hfill \cr 
  z^2  + r^2 \sin ^2 \theta  \le 1 \hfill \cr}  \right.
$$
and simplify to
$$
\left\{ \matrix{
  0 \le \theta  \le \pi /2 \hfill \cr 
  0 \le r \le 1 \hfill \cr 
  0 \le z \le \sqrt {1 - r^2 \cos ^2 \theta }  \hfill \cr 
  0 \le z \le \sqrt {1 - r^2 \sin ^2 \theta }  \hfill \cr}  \right.
$$
To solve for the last two bounds in $z$, again using symmetry,
just reduce the angle to $\pi /4$, and integrate with these conditions
$$
\left\{ \matrix{
  0 \le \theta  \le \pi /4 \hfill \cr 
  0 \le r \le 1 \hfill \cr 
  0 \le z \le \sqrt {1 - r^2 \cos ^2 \theta }  \hfill \cr}  \right.
$$
Then finally you shall multiply by $16$.
