Let $c$ be the curve of intersection of the plane defined by $x+y+z = a$ and the cylinder $x^{2} + y^{2} = a^{2}$ ($a > 0$). Evaluate the path integral:
$\displaystyle\oint_{c} \sqrt{a^2 + xy} \mathrm{d}s$
I have first tried to parametrize the curve of intersection with the followng (let $0 \leq t \leq 2\pi$):
$x = a\cos(t)$
$y = a\sin(t)$
$z = a - x -y = a - a\cos(t) - a\sin(t)$
Then, we can find that $|c'(t)| = \sqrt{(-a\sin(t))^2 + (a\cos(t))^2 + (a\sin(t) - a\cos(t))^2}$, to then have that:
$\displaystyle\oint_{c} \sqrt{a^{2} + xy} \mathrm{d}s = \int_{0}^{2\pi} \sqrt{a^{2}+a^{2}\cos(t)\sin(t)} |c'(t)|\mathrm{d}t = \int_{0}^{2\pi} \sqrt{a^{2}+a^{2}\cos(t)\sin(t)} \sqrt{2a^{2}-2a^{2}\cos(t)\sin(t)} \mathrm{d}t = \sqrt{2}a^{2} \int_{0}^{2\pi} \sqrt{1-\cos^{2}(t)\sin^{2}(t)} \mathrm{d}t$
I've been trying to solve the last integral, but apparently I checked if it could be solved with Wolfram Alpha, and it gives the solution in terms of the elliptic integral of the second kind. On the other hand, Stokes' Theorem cannot be used that easily as we would not have a line integral in this case.
What would be the best way to solve this problem to obtain a closed form solution? Perhaps I made a mistake in my calculations. Thank you for the help.