Different ways about Stokes theorem 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.
 A: 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.
A: Things like Stokes' Theorem have multiple uses. One use is completely logical: besides being very interesting and important in its own right, it can be used in the proof of other theorems. Another (lesser) use is calculational: it could make the calculation of certain integrals easier. In this use case there is no hard and fast rule. Sometimes using Stokes' theorem might not help make the integral easier. Sometimes it might be easier to do the line integral than the surface integral, or vice versa. Maybe in some cases one is easier for you than the other, but the opposite for your friend, because of your different skill sets. It's really just a matter of trying it out as a calculational tool and seeing how much it helps you.
A: The first point to make is that the whole point of Stokes's theorem is that the integral of the curl over the surface is equal to the integral of the function around the curve bounding the surface:
$$ \iint_S (\operatorname{curl}{\mathbf{F}}) \cdot \mathbf{n} \, dS = \int_C \mathbf{F} \cdot d\mathbf{r}. $$
This means that if we need to calculate the value of one of these integrals, we can use the other one: for practical reasons, it may be that one is much easier to do than the other. For example, $S$ may be a very complicated wiggly thing, but bounded by something simple like a circle. Or the curl of the vector field may vanish. Or, we may have a nasty curve with corners in it, like a square, where it is actually easier to parametrise the interior as a surface. 
This effectively boils down to the same thing as the fundamental theorem of calculus: it's just that in two dimensions, there is so much more freedom in choosing the path of integration, so the relative simplicity of both sides varies much more.
Another thing to think about is numerics: suppose I want to calculate the integral of some vector field on a surface numerically. That's about $n^2$ points I have to evaluate the function at and add up. But if I can find a field of which the original field is the curl, then I can just do $n$ evaluations on the curve around the surface.
As a physical example, we may be interested in something like the total charge inside a surface. But if we can't look inside the surface and add it up directly, we can use the divergence theorem to find the total charge by examining the electric field on the surface, by Gauss's law $ Q = \iiint_V \rho \, dV = \epsilon_0 \iint_{S} \mathbf{F} \cdot d\mathbf{S} $. (An example with Stokes's theorem itself would be the magnetic field caused by the current passing through a surface.)
