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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

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  • $\begingroup$ 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
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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 $b>a$.

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

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