Volume of Intersection of cylinders (different radii)

Solution 1:

It is not as simple as you may think, and you shall run into trouble when coming to the second integral. The value of the innermost integral is not $2\sqrt{a^2-x^2}$, it is $2\sqrt{a^2-x^2}$ when $|x|\leq a$ and $0$ when $x\geq a$. This has to be taken into account when doing the next integral, with respect to $x$. There will be cases to consider. Choosing another ordering of the variables, letting the variable $x$ being the outermost variable, will avoid running into cases. But wait.

Here is how I would go at it: I'm assuming $a\leq b$. Looking at the figure we can immediately see that planes $x={\rm const.}$ intersect the body $B$ whose volume we want to compute in rectangles $R_x$. It remains to find the area of $R_x$ and then to integrate over $x$. When $|x|>a$ then $R_x=\emptyset$. When $|x|\leq a$ then $$R_x=\bigl\{(y,z)\>\bigm|\>|y|\leq \sqrt{b^2-x^2}, \ |z|\leq\sqrt{a^2-x^2}\bigr\}\ .$$ It follows that $${\rm area}(R_x)=4\sqrt{(b^2-x^2)(a^2-x^2)}\qquad(-a\leq x\leq a)\ .$$ Therefore we obtain $${\rm vol}(B)=8\int_0^a \sqrt{(b^2-x^2)(a^2-x^2)}\>dx\ .$$ I'm afraid that this is an elliptic integral when $b>a$.