Finding the volume of this intersection How to find the volume of this:

The region common to the interiors of the cylinders $x^2+y^2=1$ and
  $x^2+z^2=1$ and the first octant.

I tried finding the volume via double integration and the integration gets too complicated. I want to know how to solve this via triple integration. Usually there are $x,y,z's$ in both equations and that I can solve with triple but I dont know what to do here.
I started of doing $x^2+z^2=x^2+y^2$ so $z=y$ but dont know what to do with this it doesnt look like the line of intersection is $z=y$ in the diagram given.
 A: Let $A$ be the intersection of the two solid cylinders, considered only in the first octant $x,y,z\ge 0$.
Use Fubini for the computation of the volume. It is clear that $x\in[0,1]$. Now for a fixed $x$ we are searching for all $(y,z)$ with the property
$0\le y,z\le \sqrt{1-x^2}$. This is a square with side $\sqrt{1-x^2}$, its area is $(1-x^2)$ so the needed volume is
$$
\text{Volume}(A)=\int_0^1(1-x^2)\; dx =1-\frac 13=\frac 23\ .
$$
A: Draw out the surfaces and see that $x$ varies from $0$ to $1$, $y$ varies from $0$ to $\sqrt{1-x^2}$, and $z$ varies from $0$ to $\sqrt{1-x^2}$ as well. There are $5$ other ways to order the integration but aren't as useful as this ordering (check why) and this ordering makes use of squaring the square root
$$V=\int_0^1\int_0^\sqrt{1-x^2}\int_0^\sqrt{1-x^2}dzdydx$$
This is the same as 
$$V=\int_0^1\int_0^\sqrt{1-x^2}\sqrt{1-x^2}dydx$$
Which is the area integral of the function $f(x,y)=\sqrt{1-z^2}$, the surface of the "top" cylinder. The other cylinder, $x^2+y^2=1$ serves as the boundary of the integration region. The triple and double integrals are equivalent.
When using the triple integral to find volume, the integrand is simply $1$ and integrate wrt all variables whereas with using the double integral to find volume, the integrand is the function $f(x,y)$ and then integrate wrt $x$ and $y$
As for $z=y$, it appears on the $y-z$ plane as the cylinders make contact at $(0,1,1)$ 
