What is wrong with my reasoning regarding finding volumes by integration? The problem from the book is (this is Calculus 2 stuff):

Find the volume common to two spheres, each with radius $r$, if the center of each sphere lies on the surface of the other sphere.

I put the center of one sphere at the origin, so its equation is $x^2 + y^2 + z^2 = r^2$.  I put the center of the other sphere on the $x$-axis at $r$, so its equation is $(x-r)^2 + y^2 + z^2 = r^2$.
By looking at the solid down the $y$- or $z$-axis it looks like a football.  By looking at it down the $x$-axis, it looks like a circle.  So, the spheres meet along a plane as can be confirmed by setting the two equations equal to each other and simplifying until you get $x = r/2$.
So, my strategy is to integrate down the $x$-axis from 0 to $r/2$, getting the volume of the cap of one of the spheres and just doubling it, since the solid is symmetric.  In other words, I want to take circular cross-sections along the $x$-axis, use the formula for the area of a circle to find their areas, and add them all up.
The problem with this is that I need to find an equation for $r$ in terms of $x$, and it has to be quadratic rather than linear, otherwise I'll end up with the volume of a cone rather than a sphere.  But when I solve for, say, $y^2$ in one equation, plug it into the other one, and solve for $r$, I get something like $r = \sqrt{2 x^2}$, which is linear.

If you restrict yourself to the $xy$ plane you have two intersecting circles.  Now you want $y$ as a function of $x$ for the circle centered at $x=r$, so you can solve the equation for $y$, getting $y=\sqrt{r^2-x^2}$.  Your confusion may have come from reusing $r$, once for the radius of the spheres and once for the radius perpendicular to the $x$ axis out to the right hand sphere.
