I have found the integral that can give me the volume for the common part of two cylinders that meet at a right angle. The cylinders have radii of $r$ and $R$, respectively with $r\le R$. The integral I have found is

$V = 8\int_{0}^r \sqrt{R^2-x^2} \cdot \sqrt{r^2-x^2}$d$x$

This is a nonelementary integral. However, now I need to find a $F(a)$ such that $V = r^3F(a)$ where $a = R/r$.

I found the volume by dividing the object into infinite rectangles, hence the integral.

To find F(a) I tried an approach where I tried using the relationship between the Area of Ellipse and the area of Rectangle (ratio of $4/pi$), I know the volume of an ellipsoid is $(4/3)pi\cdot a \cdot b \cdot c$, I figured multiplying this by $4/pi$ would give me the volume of the object in a more accessible manner. However, this did not work. It just gave me a function $\dfrac{16r^2a\beta}{3}$, where $\beta$ is an unknown variable that I was not able to figure out.

I'm completely lost as to how to proceed and would appreciate any help.


Your integral is missing $\mathrm dx$.

To bring it into the form you want, express it in terms of $a=\frac Rr$ and $\xi=\frac xr$:

$$ 8\int_0^r\sqrt{R^2-x^2}\sqrt{r^2-x^2}\mathrm dx = 8r^3\int_0^1\sqrt{a^2-\xi^2}\sqrt{1-\xi^2}\mathrm d\xi\;. $$

  • $\begingroup$ This is a big help and I appreciate it. However, I need to find a function that depends exclusively on $a$ and not on r or R seperately. I think since we have $\frac{x}{r}$, this doesn't quite fit the bill. $\endgroup$ – jessicajjensen Feb 22 '16 at 4:19
  • $\begingroup$ @jessicajjensen: The result doesn't depend on $\frac xr$. The variables $x$ and $\xi$ are just integration variables. The result is exactly of the form you want, $V=r^3F(a)$, with $F(a)=8\int_0^1\sqrt{a^2-\xi^2}\sqrt{1-\xi^2}\mathrm d\xi$. $\endgroup$ – joriki Feb 22 '16 at 6:37
  • $\begingroup$ I see. I'm really sorry but I'm having trouble seeing exactly how you got here. I can see that it works but if you could list some steps that would be incredibly helpful. Thank you so much. $\endgroup$ – jessicajjensen Feb 22 '16 at 7:52
  • $\begingroup$ @jessicajjensen: $$ \int_0^r\sqrt{R^2-x^2}\sqrt{r^2-x^2}\mathrm dx=r^3\int_0^{\frac rr}\frac{\sqrt{R^2-x^2}}r\frac{\sqrt{r^2-x^2}}r\mathrm d\left(\frac xr\right)\;. $$ $\endgroup$ – joriki Feb 22 '16 at 7:54

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