I want to derive a formula for the area of the intersection of two rigth cylinders with different radii. To get an idea I attached a sketch. 
My idea is to determine the borders of $x$ and $z$ in dependence of $y$.
The borders of $x$ are: $-\sqrt{b^2-y^2} \leq x \leq \sqrt{b^2-y^2} $
And the borders of z are $-\sqrt{a^2-x^2}\leq z \leq \sqrt{a^2-x^2}$
Thus, the resulting integral should be $\int_{-b}^b \int_{-\sqrt{b^2-y^2}}^\sqrt{b^2-y^2} \int_{-\sqrt{a^2-x^2}}^\sqrt{a^2-x^2} ~\mathrm{d}z~\mathrm{d}x~\mathrm{d}y$
Is this correct?
Thank you!
Fabian
Best Answer
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 $b>a$.