Stokes theorem to get $\oint \vec{F}d\vec{R}$ I have the vector field 
\begin{equation*}
\vec{F}=(ye^x,x^2+e^x,z^2e^z)
\end{equation*}
and the curve $C$ that us given by 
\begin{equation*}
\vec{r}(t)=(1+\cos t, 1+\sin t, 1-\cos t-\sin t),~0\leq t\leq 2\pi. 
\end{equation*}
What is $\oint_C \vec{F}d\vec{R}$?
I know that my curl is $(0,0,2x)$, but what is my final integral? How do I use Stokes' theorem here?
 A: Surface Integral
You would need to calculate
$$
I = \int_A\limits \mbox{curl} F \cdot dA =
\int_A\limits (0, 0, 2x) \cdot dA =
2 \int_A\limits x \, dA_z
$$
where $A$ is a surface with boundary $\partial A = C$.
The open task to choose some feasible $A$ which allows for simple calculation of the last term.

So what is $C$? We had
$$
r(t) = (1+\cos t, 1+\sin t, 1-\cos t-\sin t)^t
$$
this gives
$$
r(t) \cdot (1,1,1)^t = 3 \iff 
r(t) \cdot n = \sqrt{3} \quad n = (1,1,1)^t/\sqrt{3}
$$
So the $r(t)$ endpoints form a curve $C$ which lies in a plane $E$ with normal vector $n$ and origin $\sqrt{3} n = (1,1,1)^t$:
$$
E = \{ x \mid \left(x - \sqrt{3} n\right) \cdot n = 0 \}
$$
Let $A$ be the part of $E$ that is inside $C$. From the observation that the projection of $A$ onto the $x$-$y$-plane is a circle of radius $1$ with center $(1,1)$ we assume the parametrization:


$$
A(r, t) = (1 + r \cos t, 1 + r \sin t, 1 - r \cos t - r \sin t)^t
\quad (r \in [0, 1], t \in [ 0, 2\pi ])
$$
with the partial derivatives along the parameters
$$
\partial_r A = (\cos t, \sin t, - \cos t - \sin t)^t \\
\partial_t A = (-r \sin t, r \cos t, r \sin t - r \cos t)^t
$$
then we get
$$
(\partial_r A \times \partial_t A)_x =
r (\sin t)^2 - r \cos t \sin t + r (\cos t)^2  + r \cos t \sin t =
r \\
(\partial_r A \times \partial_t A)_y =
r \cos t \sin t + r (\sin t)^2 - r \cos t \sin t + r (\cos t)^2 =
r \\
(\partial_r A \times \partial_t A)_z =
r (\cos t)^2 + r (\sin t)^2 = r
$$
this gives
$$
dA 
= \partial_r A \times \partial_t A \, dr \, dt 
= r \, (1,1,1)^t \, dr \, dt
= n \, \sqrt{3} \, dA^{xy}
$$
So we have
\begin{align}
I 
&= 2 \int\limits_A (1 + r \cos t) r dr dt \\
&= 2 \int\limits_0^1 dr \, \int\limits_0^{2\pi} dt \, (r + r^2 \cos t) \\
&= 2 \int\limits_0^1 dr \, 2\pi r \\
&= 2 \pi
\end{align}
Line Integral
With the help of a computer algebra system (link) we get
\begin{align}
I
&= \int\limits_C F \cdot dr \\ 
&= \int\limits_C (ye^x, x^2+e^x, z^2 e^z) \cdot dr \\
&= \int\limits_0^{2\pi} (ye^x, x^2+e^x, z^2 e^z) \cdot \dot{r} dt \\
&= \int\limits_0^{2\pi} (ye^x, x^2+e^x, z^2 e^z) \cdot (-\sin t, \cos t, \sin t - \cos t) dt \\
&= \int\limits_0^{2\pi} \left(-(1+\sin(t)) e^{1+\cos(t)} \sin(t) + \cos(t) ((1+\cos(t))^2 + e^{1+\cos(t)}) + (\sin(t) - \cos(t)) (1-\cos(t) - \sin( t))^2 e^{1-\cos(t) - \sin(t)}\right) dt \\
&= \left[ t+\frac{7}{4}\sin(t)+\frac{1}{12} \sin(3 t)+\sin(t) \cos(t)+(\sin(t)+1) e^{\cos(t)+1}+(\sin(2 t)+2) e^{-\sin(t)-\cos(t)+1}\right]_0^{2\pi} \\
&= 2 \pi
\end{align}
A: Note that $r(t)$ lies on the plane $x+y+z-3=0$. So $r(t)$ encloses a region on this plane, whose unit normal (compatible with the orientation of $r(t)$) is $\frac{1}{\sqrt{3}}(1, 1, 1)$. Also note that $r(t)$ is actually a circle whose projection onto the xy-plane is $(x-1)^2+(y-1)^2=1$.
By Stokes, 
\begin{align*}
\oint_C F\cdot dR&=\int\!\!\!\!\int_S (0, 0, 2x)\cdot\frac{1}{\sqrt{3}}(1, 1, 1)dS\\
&=\int\!\!\!\!\int_S \frac{2}{\sqrt{3}}xdS
\end{align*}
I will leave the rest to you.
A: You know what $F$ is, you know whar $dr$ is. Compute $F \cdot dr$ and then formulate $\int_{0}^{2\pi}F \cdot dr$ and evaluate the integral above.
EDIT
Stokes theorem 
$$\iint_{\Sigma} \nabla \times \mathbf{F} \cdot \mathrm{d}\mathbf{\Sigma} = \oint_{\partial\Sigma} \mathbf{F} \cdot \mathrm{d} \mathbf{r}.$$ 
