Volume of a Cylinder With a Cylinder Cutout

Question

I was considering a cylinder with a cylinder cut out through the middle of it (on the curved side, parallel to the base) inspired by a doughnut themed coffee mug I saw. Would there be any way to obtain a general formula for the volume of such a cylinder? The volume of the overall cylinder is simple of course ($\pi r^2h$), and most of the cutout can be calculated easily enough. However, the ends where the cutout meet the edge of the cylinder are giving me trouble. The only way I can think of to solve it is using a triple integral, and that seems very messy. Would there be any better way to do this? Here are some pictures to show the situation for added clarity.

The Entire Cylinder

The Simple Part of the Cutout

The Complicated Part of the Cutout

Answer 1

Let's calculate the entire small radius cylinder. Easy for me to set it up that way. Let's call $x$ the horizontal direction perpendicular to both cylindrical axes, $y$ the vertical direction (along the axis of the big cylinder), and $z$ the direction along the axis of the small cylinder. Also, call the radius of the small cylinder $r$, and of the big cylinder $R$. Not divide the volume of the small cylinder into rectangles perpendicular to the $x$ axis. You can see that the width in the $y$ direction is $2\sqrt{r^2-x^2}$, and the height in the $z$ direction is $2\sqrt{R^2-x^2}$. The limits of integration are from $-r$ to $r$: $$V=8\int_0^r\sqrt{r^2-x^2}\sqrt{R^2-x^2}dx$$ Make the substitution $x=r\sin\theta$, $dx=r\cos\theta d\theta$. Can you take it from here?