Length of the intersection between a sphere and a cylinder

Question

Note: while writing this question I realized what I was missing so there is no question here, but I thought it is a nice exercise to share, and I'd like to see more solutions.

I am trying to calculate the length of the curve given by $$x^2+y^2+z^2=1, x^2+y^2=x$$

The curve is the intersection between a sphere and a cylinder, and we can notice is is symmetric around the $xy$ and $xz$ planes, so I tried to calculate only the part where $y>0, z>0$ and multiply by 4.

For this part I used the parameterization: $\gamma(t)=(t, \sqrt{t-t^2}, \sqrt{1-t}), t \in [0, 1]$

And then the length should be $l=4\int_0^1 ||\gamma'(t)|| dt = 4\int_0^1 \sqrt{1+\frac{(1-2t)^2}{4(t-t^2)} +\frac{1}{4-4t}} dt = 4\int_0^1 \sqrt{\frac{t+1}{4t-4t^2}} dt$

And I'm not sure how to solve this integral but using an online calculator we can see the length is $\approx 7.64$

Different solutions, or suggestions for ways to solve this integral are welcome!

Answer 1

This again requires elliptic integral, similar to your working.

In cylindrical coordinates, $x = r \cos t, y = r \sin t, z$.

Cylinder $x^2 + y^2 = x $ or $(x-\frac{1}{2})^2 + y^2 = \big(\frac{1}{2}\big)^2$ can be written in cylindrical coordinates as
$r = \cos t, -\frac{\pi}{2} \leq t \leq \frac{\pi}{2}$.

At the intersection of sphere and cylinder,

$x = \cos^2 t, y = \cos t \sin t, z = \sqrt{1-r^2} = \sin t$

So $x'_t = -\sin2t, y'_t = \cos 2t, z'_t = \cos t$

Length of the curve = $ \displaystyle 2 \int_{-\pi/2}^{\pi/2} \sqrt{(x'_t)^2 + (y'_t)^2 + (z'_t)^2} \ dt = 4 \int_{0}^{\pi/2} \sqrt{1 + \cos^2t} \ dt \approx 7.64$

$\displaystyle \int_0^{\pi/2} \sqrt{1 + \cos^2t} \ dt $ can be written as,

$\displaystyle \sqrt2 \int_0^{\pi/2} \sqrt{1 - \frac{1}{2} \sin^2t} \ dt = \sqrt2 E(\frac{1}{\sqrt2}).$

$E(k) = \displaystyle \int_0^{\pi/2} \sqrt{1 - k^2 \sin^2t} \ dt$ is complete elliptic integral of the second kind as shared by Raffaele in the comment.