Finding Volume of a Cylinder with a hole (Real Life Problem) So for my engineering coursework I need to have the masses all the parts I make.
I know the density of the material I am using but I don't know the volume of this particular solid I'm about to describe. I'm wondering if anyone maybe would be able to help me find this volume or at least suggest some method I that could help me work it out. Thank you for your help.
I have a cylinder of radius R and height h. Imagine there is a $line$ $AB$ where $A$ is the point at the centre of the circular face of the cylinder and $B$ is the point at the centre of the other circular face of the cylinder. I drill a hole of $radius$ $r$ $(R > r)$ through the curved surface of the cylinder all the way through such that the centre of of the drill bit passes through line $AB$ at $90$ degrees. 
What is the volume of this cylinder with a hole in it in terms of $R$, $h$ and $r$?
Here is a drawing to help:
 A: This is not an answer, as I don't know how to evaluate the integral in terms
of elementary functions. However,
just in case someone else does...
I am assuming that $r \le R$ and that $h$ is 'large' so that it completely
surrounds the bored hole. I am also assuming that the big cylinder has its centerline along the $y$ axis and that the drilled hole is along the $x$ axis.
If we can compute the volume of the material of the original cylinder that was removed by drilling then we can compute the volume of the cylinder with a hole.
The volume of removed material is given by the volume of
$D_{r,R} = \{ (x,y,z) | z^2+y^2 \le r^2, z^2+x^2 \le R^2 \}$. This can be written as
$D_{r,R} = \{ (x,y,z) | z \in [-r,r], |x| \le \sqrt{R^2-z^2}, |y| \le \sqrt{r^2-z^2} \}$
so we can write
$\operatorname{vol}D_{r,R} = \int_{z=-r}^r \int_{y=-{\sqrt{r^2-z^2}}}^{\sqrt{r^2-z^2}} \int_{x=-{\sqrt{R^2-z^2}}}^{\sqrt{R^2-z^2}} 1 \ dx dy dz = 8 \int_0^r \sqrt{r^2-z^2} \sqrt{R^2-z^2} dz$.
As a sanity check, if we fix $r$, then
$\lim_{R \to \infty} {D_{r,R} \over 2 R} = \pi r^2$.
However, I don't think that this has an elementary solution.
A numerical solution is straightforward, if that suffices.
Then the volume of the drilled cylinder is
$\pi R^2 h-D_{r,R}$.
