Can this be done with Stokes' Theorem? (Exercise) Problem description:
Let $\gamma$ denote the curve of intersection of the two surfaces $z=x^2+y^2$ and $z=1+2x$. Calculate the line integral $W=\oint_\gamma \boldsymbol{F}\cdot d\boldsymbol{r}$, where $\boldsymbol{F}=(0,x,-y)$ and $d\boldsymbol{r}=(dx,dy,dz)$.
Now, by easy substiution of the first equation into the second we get the realtionship between $x$ and $y$ as $(x-1)^2 + y^2=2$. Parametrization gives $x=1+\sqrt{2}\cos{t}\quad$ and $\quad y=\sqrt{2}\sin{t}$. Lastly the complete parametrization of the curve $\gamma$ is $\boldsymbol{r}=(1+\sqrt{2}\cos{t},\sqrt{2}\sin{t},3+2\sqrt{2}\cos{t})$, where I have inserted the new expression for $x$ into the second equation.
Well, from here we can can calculate $W$ over $0\leq t \leq 2\pi$ and get the answer of $6\pi$ relatively quickly. But I am wondering if Stokes' Theorem could be used here? My textbook implies it could be done. The result is then
$W=\frac{3}{\sqrt{5}}\iint_S dS$.
How do I know what the area of S is, that I am guessing is an ellipse? It shoud evidently become $2\sqrt{5}\pi$, so maybe the axises are $2$ and $\sqrt{5}$, respectively? But how do I reach this result otherwise?
 A: If you have an closed path integral, you can use stokes theorem.
$\oint_\gamma F\cdot \ d\gamma = \iint \nabla \times F\cdot dS$ 
Where $S$ is the elliptical disc on the plane bounded by the paraboloid.
$\nabla \times F = (-1,0,1)$
$dS = (-\frac {\partial z}{\partial x},-\frac {\partial z}{\partial y}, 1)\ dA= (-2,0,1)\ dA$
$A$ is the projection of $S$ onto the $xy$ plane (which is a circle)
$\iint 3\ dA = 3 A$
$6\pi$
A: If you have an closed path integral, you can use stokes theorem.
$\oint_\gamma F\cdot \ d\gamma = \iint \nabla \times F\cdot dS$ 
Where $S$ is the elliptical disc in the plane bounded by the paraboloid.
$\nabla \times F = (-1,0,1)$
$dS = (-\frac {\partial z}{\partial x},\frac {\partial z}{\partial y}, 1)\ dA= (-2,0,1)\ dA$
$\iint 3\ dA = 3 A$
A is the area of the ellipse projected onto the $xy$ plane (which is a circle).
$6\pi$
Regarding, 
$W=\frac{3}{\sqrt{5}}\iint_S dS$
$S$ is an ellipse with semi-major axis equal to $\sqrt {10}$, and semi-minor axis equal to $\sqrt 2$
$(\frac {3}{\sqrt 5})(\pi )(\sqrt {10})({\sqrt 2}) = 6\pi$
