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?
Best Answer
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.
The fact that Stewart asks for an answer "to four decimal places" indicates that you are not expected to find an exact expression.