Calculus III - Lateral Surface Area emm, I got this problem on a text book and it looked nice to done with, but I'm really unsure about my answer because it goes into a non-elementary equation to integrate, well, the problems goes like this...
"The figure below shows a piece of tin that has been cut from a circular cylinder. The base of the circular cylinder is modeled by \begin{eqnarray}x^2 +y^2 =9\end{eqnarray}At any point (x, y) on the base, the height of the object is
\begin{eqnarray}
f(x,y) = 1 + \cos\frac{\pi x}{4}\end{eqnarray}
Explain how to use a line integral to find the surface area of the piece of tin."
So, what do I do?, well, i started converting the original equation into a parametric one where the only thing to change should be the x, so it would be something like this \begin{eqnarray} f(x(t),y(t)) = 1 + \cos\frac{\pi 3\cos t}{4}\end{eqnarray} So I finished there because this gives me the limits of integration from 0 to 2(pi), and the dS=3, but as I said, I'm very untrusted on my process,the nob-elementary integral gives me problems, if anyone could help me explaining to me how to finish it or fix some interpretation problem, I'll be appreciated, thanks.

 A: Your computations are correct; $dS=3$ as you wrote, and the final integral is
$$\int_0^{2\pi}3(1+\cos(3\pi/4\cdot\cos t))\,dt$$
The integral is also truly non-elementary, as you write. We will nevertheless derive some "nice" expression for it. Split up the integral:
$$=\int_0^{2\pi}3\,dt+\int_0^{2\pi}3\cos(3\pi/4\cdot\cos t)\,dt=6\pi+3\int_0^{2\pi}\cos(3\pi/4\cdot\cos t)\,dt$$
By periodicity we can halve the interval of the second integral:
$$=6\pi+6\int_0^{\pi}\cos(3\pi/4\cdot\cos t)\,dt$$
Now we note that $\cos(3\pi/4\cdot\cos x)=\cos(3\pi/4\cdot\sin(x-\pi/2))$, and both functions have the same period of $\pi$. We may harmlessly substitute $\sin t$ for $\cos t$ (and throw in a negative sign in the outer cosine for good measure, since $\cos t$ is even);
$$=6\pi+6\int_0^{\pi}\cos(-3\pi/4\cdot\sin t)\,dt$$
We recognise the integral as the Bessel function of the first kind:
$$=6\pi+6\pi J_0\left(\frac{3\pi}4\right)=6\pi\left(1+J_0\left(\frac{3\pi}4\right)\right)$$
This is probably the simplest expression for the surface area of the tin.
A: So you obviously wrote $x$ and $y$ in terms of polar coordinates as $$x=r\cos t\\y=r\sin t$$with $r=3$. You know the height as a function of $t$ and all you need now is the formula for the area. You are almost there. $$A=\int h(t) ds$$ The remaining part is $ds$. In polar coordinates, this length is $ds=r\ dt=3dt$. Finally your area is $$A=\int_0^{2\pi}\left(1+\cos\frac{3\pi\cos t}{4}\right)3dt$$ 
However, you are asked to do a little different. You still have $ds=3dt$, so $t=\frac{s}{3}$. Also the limits for $s$ are from $0$ to $3\times 2\pi$
$$A=\int_0^{6\pi}\left(1+\cos\frac{3\pi\cos \frac s3}{4}\right)ds$$
I don't think it has a nice value, but I don't know if they really want that. They asked to show how you would do it. 
