# Integrating a specific integral based on the volume of the intersection of two cylinders

Volume of Intersection of cylinders (different radii)

I have been able to obtain the formula to calculate the volume like the solution to the link above.

My formula looked like this:

$$V = \int_{-r}^{r} \sqrt{r^2 -y^2}\sqrt{R^2 - y^2}dy$$

Where $r \leq R$

if $b = \frac{R}{r}$, how can I show $V = r^3 F(b)$.

I am having trouble finding an expression where volume is equal to $r^3$ times some expression that is only dependant on b

• Hint: Make the substitution $y=rx$ in the integral. – mickep Feb 22 '16 at 19:53
• something like u = ry? – Mat Feb 22 '16 at 19:57
• Rather $u=y/r$ if you want to call the new variable $u$. – mickep Feb 22 '16 at 19:59
• I have come up with $V = 4r^3\int_{-r}^{r} \sqrt{1 - u^2}\sqrt{a^2 - u^2}du$ and don't know how to integrate this. Do I have to use substitution again? – Mat Feb 22 '16 at 20:44

If you let $u=y/r$, you get the integral $$V=\int_{-1}^1\sqrt{r^2-r^2u^2}\sqrt{R^2-r^2y^2}\,r\,du =r^3\int_{-1}^1\sqrt{1-u^2}\sqrt{R^2/r^2-u^2}\,du.$$ Since $b=R/r$, we find that $V=r^3F(b)$, with $$F(b)=\int_{-1}^1\sqrt{1-u^2}\sqrt{b^2-u^2}\,du.$$ Note that we do not need to calculate $F$ explicitly.