The Question:
Let $F(x,y,z) = (y+\sin(x),z^2+\cos(y),x^3)$ and $\mathcal{C}$ is the curve parametrized by $r(t)=(\sin(t),\cos(t),2\sin(2t))$ with $t$ ranging in $[0,2\pi[$. Use the curl theorem to find $\displaystyle\int_\mathcal{C}F\cdot ds$.
My approach:
I know, by the curl thm. that $\displaystyle\int_{\partial S}F\cdot dr =\int_{ S}\operatorname{curl}(F)\cdot d\Sigma$
I calculated the curl for $F$ and the result was $\operatorname{curl}(F)(x,y,z)=(-2z,-3x^2,-1)$.
The boundary of the surface which the theorem talks about is the curve generated by the parametrization I was given, this means I need to find a surface $S$ such that $\partial S = \mathcal{C}$. Here lies my problem.
Once I've found the surface $S$, I need to parametrize it with $\alpha:D\subseteq \mathbb{R}^2\to \mathbb{R}^3$. With this I get $\displaystyle\int_{ S}\operatorname{curl}(F)\cdot d\Sigma = \iint_D\text{curl}(F)(\alpha(u,v))\cdot (\alpha_u\times\alpha_v) dA\,$ which is a double integral that I can calculate.
My problem:
I can't find the surface whose boundary corresponds to $r([0,2\pi[)$. I've already plotted the curve in Mathematica and it looks like the border of a pringle potato.
My gut tells me that the surface I'm supposed to use is "the potato" but I can't get my head onto it. I feel like there are many surfaces which I can't imagine that have the curve $\mathcal{C}\,$ as a boundary.
Still, if it were the "potato" I am not able to find the parametrization for it. This is where I need help.
$\textbf{Concretely}$ I need to find the parametrization of the "pringle potato" whose boundary is my curve $\mathcal{C}$ and an explanation of why is that the correct surface to think about.
Two suggestions: You could take the cone on this curve, just sticking in an $r$ parameter in front of the whole thing ($0\le r\le 1$). But what you had in mind is to use that portion of the the "potato chip surface" $z=4xy$ lying inside the curve. (How did I get this? I used the double-angle formula to write $z=2\sin(2t)=4\sin t\cos t = 4xy$.) I haven't tried doing either surface integral to see which is more accessible; I leave those to you. :)