# find length of curve of intersection

I have come to a dead end on a problem and I need someone to tell me either if I did it correctly, or how to fix it if I did not. This is Stewart Calculus 7th edition, problem 13.3.12.

Here is the problem definition:
"Find, correct to four decimal places, the length of the curve of intersection of the cylinder $4x^2 + y^2 = 4$ and the plane $x + y + z = 2$."

The answer I got is 13.5191, but I am asking for people here to review this because I reached a seemingly unsolvable integral and I had to resort to using my calculator.

Here is my work: $$x^2 +(1/4)y^2 = 1$$

\begin{align*}x &= \cos t \\ y &= 2 \sin t \\ z &= 2 - \cos t - 2\sin t \end{align*} for $0 \le t \le 2\pi$. $$r'(t) = (-\sin t, 2\cos t, (\sin t - 2 \cos t))$$

Then $$L = \int_a^b |r'(t)| dt$$

I then squared each component of $r'(t)$, summed the components, and took the square root to define the integral. I simplified twice to get:
$$\int_0^{2\pi} \sqrt{2\sin^2 t + 8\cos^2 t - 4\sin t \cos t} dt$$

For the interval 0 to $2\pi$, the calculator gives 13.51908915

Can anyone show me the solution to this?

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You are correct. The integral can be simplified a bit to $$\int_0^{2\pi} \sqrt{3 \cos(2t) - 2 \sin(2t) + 5}\ dt = \int_0^{2\pi} \sqrt{\sqrt{13} \cos(s) + 5}\ ds$$ This can be expressed in terms of elliptic integral or hypergeometric functions, e.g. it is $2\;\pi \sqrt{5} \;{\mbox{$_2$F$_1$}(-1/4,1/4;\,1;\,{ {13}/{25}})}$, but I doubt that there is an elementary solution.