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?