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. enter image description here

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

  • $\begingroup$ This paper link (p.139) states that the volume is given by $\int_{-b}^b \int_{-\sqrt{b^2-y^2}}^\sqrt{b^2-y^2} \int_{-\sqrt{a^2-y^2}}^\sqrt{a^2-y^2} ~\mathrm{d}z~\mathrm{d}x~\mathrm{d}y$ $\endgroup$ – fmeyer Feb 27 '15 at 10:48
  • $\begingroup$ This post is chosen to be the target/mother for (abstract) duplicates of the particular variation of orthogonal of different radii. $\endgroup$ – Lee David Chung Lin Jan 22 at 11:55

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 $b>a$.

  • $\begingroup$ You are absoulutely rigth. Thank you very much indeed. $\endgroup$ – fmeyer Mar 1 '15 at 9:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.