Volume bounded by cylinders $x^2 + y^2 = r^2$ and $z^2 + y^2 = r^2$ I am having trouble expressing the titular question as iterated integrals over a given region. I have tried narrowing down the problem, and have concluded that the simplest way to approach this is to integrate over the XZ plane in the positive octant and multiply by 8, but I am having trouble identifying the bounding functions.
 A: The solid lies above the region $D$ in the $xy$-plane bounded by the circle $x^{2} + y^{2} = r^{2}$, so the volume is given by the integral $$\int\int\limits_{D} f(x,y) \ dA = \int\limits_{-r}^{r}\int\limits_{-\sqrt{r^{2}-y^{2}}}^{\sqrt{r^{2}-y^{2}}} f(x,y) \ dx dy$$
Therefore the required volume of the solid is: $$\int\limits_{-r}^{r}\int\limits_{-\sqrt{r^{2}-y^{2}}}^{\sqrt{r^{2}-y^{2}}} 2\sqrt{r^{2}-y^{2}} \ dx dy = \frac{16}{3}r^{3}$$
A: This is one of those results in calculus which were anticipated by
Archimedes. He gave a correct formula for the volume but it is not known exactly 
how Archimedes solved this problem. There is, however, a simple way to
obtain the answer without much calculus. Let me quote from late Gardner's The Unexpected Hanging and Other Mathematical Diversions (Gardner considers the case $r=1$ but this is not essential, of course):

Imagine a sphere of unit radius inside the volume common
  to the two cylinders and having as its center the point
  where the axes of the cylinders intersect. Suppose that the
  cylinders and sphere are sliced in half by a plane through
  the sphere's center and both axes of the cylinders. 
  The cross section of the volume common to the cylinders will be a square. 
  The cross section of the sphere will be a circle that fills the square.

 

Now suppose that the cylinders and sphere are sliced by a
  plane that is parallel to the previous one but that shaves off
  only a small portion of each cylinder (have a look at the picture on the left).
  This will produce parallel tracks on each cylinder,
  which intersect as before to form a square cross section of
  the volume common to both cylinders. Also as before, the
  cross section of the sphere will be a circle inside the square.
  It is not hard to see (with a little imagination and pencil
  doodling) that any plane section through the cylinders, parallel
  to the cylinders' axes, will always have the same result:
  a square cross section of the volume common to the cylinders,
  enclosing a circular cross section of the sphere.
  Think of all these plane sections as being packed together
  like the leaves of a book. Clearly, the volume of the sphere
  will be the sum of all circular cross sections, and the volume
  of the solid common to both cylinders will be the sum of all
  the square cross sections. We conclude, therefore, that the
  ratio of the volume of the sphere to the volume of the solid
  common to the cylinders is the same as the ratio of the area
  of a circle to the area of a circumscribed square. A brief calculation
  shows that the latter ratio is $\pi/4$. This allows the
  following equation, in which $x$ is the volume we seek:

$$\frac{4\pi r^3/3}{x}=\frac{\pi}{4}.$$

The $\pi$'s drop out, giving $x$ a value of $16r^3/3$. The radius in
  this case is 1, so the volume common to both cylinders is
  $16/3$. As Archimedes pointed out, it is exactly $2/3$ the volume
  of a cube that encloses the sphere; that is, a cube with
  an edge equal to the diameter of each cylinder.

