I'm starting to understand Stokes theorem, but I still don't understand why are two methods around it. For example:
Evaluate $\int_{C}^{} F\ dr$, where $F(x,y,z)=(5y,-5x,3)$ and $C$ is the trace of $ x^2+y^2=4$ in the plane $ z=1$.
Now I can solve this by $2$ ways.
One is by doing $C(t)=(2\cos(t),2\sin(t),1)$, and $$\begin{align} \int_{C} F dr &=\int_{0}^{2\pi}F(C(t))C'(t)dt \\ &=\int_{0}^{2\pi}(10\sin(t),-10\cos(t),3) (-2\sin(t),2\cos(t),0) \\ &=\int_{0}^{2\pi}-20dt=-40\pi \end{align}$$
The other way is $\int_{C}^{}Fdr=\iint_C(\operatorname{curl}{F}) \cdot n dS.$ After some calculus I get $\operatorname{curl}{F}=(0,0,-10)$ and $ n=(0,0,1).$
$$\iint_C(\operatorname{curl}{F}) \cdot n dS =\iint_C(0,0,-10) \cdot (0,0,1) dS =\iint_C-20 dS=-40\pi$$
Obviously I get the same result. But my question is WHEN should I use one method or another.
It is the very essence of Stokes' theorem that two seemingly disparate integrals – one of them along a cycle, the other over a surface – have the same value.
If you are told to compute some line integral $\int_\gamma {\bf F}\cdot d{\bf r}$ along some curve $\gamma$ beginning at ${\bf p}$ and ending at ${\bf q}$ then there is no escape from actually computing this integral.
If this curve $\gamma$ is a closed curve that bounds a surface $S$ completely embedded in the domain of definition of the vector field ${\bf F}$, i.e., $\gamma=\partial S$, and only then, you can ask yourself whether it might be more advantageous to compute the surface integral $\int_S {\rm curl}({\bf F})\cdot{\bf n}\>{\rm d}\omega$ in place of the line integral $\int_\gamma {\bf F}\cdot d{\bf r}$.
Of course Stokes' theorem is of upmost importance in mathematical physics, especially electrodynamics, not so much for computational reasons, but for the insight it gives into the interplay of the various "space-time quantities" involved there.