Surface Integrals and Stokes Theorem I had a question regarding closed curve and its interior parameter domain when expressed as a surface integral using Stokes Theorem.
For example, if I have the plane $2x+3 = z$ intersecting the paraboloid $x^{2} + y^{2} = z$, this would lead to certain curve that would be hard to parameterize, but stokes would allow conversion to a surface integral.  My question is what would be the parameter domain?
I tried equating the $z$ and this yields a circle, but I wasn't sure if this would be the projection of the curve onto the xy plane.  Any help would be appreciated!  Thanks.
 A: The curve is not particularly hard to parametrize -- and neither is the surface.  In fact, much the same work goes into both problems.
The plane $2x+3 = z$ intersects the paraboloid $x^2 + y^2 = z$ in points that satisfy $x^2 + y^2 = 2x + 3$.  Moving the $2x$ to the other side and completing the square, we have
$$(x-1)^2 + y^2 = 4$$
which is the equation of a circle in the $xy$-plane, centered at $(1, 0)$ and with radius $2$.  This circle can be parametrized as $$\left<x, y\right> = \left<1 + 2\cos(t), 2\sin(t)\right>, \quad 0\le t<2\pi$$
But we are not interested in the circle in the $xy$-plane; rather, we are interested in the curve that lies above it.  But this is easy to write down, now; we just parametrize the $z$-coordinate by plugging the parametrization of the $x$-coordinate into the equation of the plane, $2x + 3 = z$:
$$\left<x, y, z\right> = \left<1 + 2\cos(t), 2\sin(t), 5 + 2\cos(t)\right>, \quad 0\le t<2\pi$$
Now, what about parametrizing the surface that is the interior of the circle?  In the $xy$-plane we can write simply
$$\left<x, y\right> = \left<1 + u\cos(v), u\sin(v)\right>, \quad 0\le v<2\pi, 0\le u<2$$
and then the portion of the $xy$-plane that is contained within the paraboloid would be
$$\left<x, y, z\right> = \left<1 + u\cos(v), u\sin(v), 5 + u\cos(v)\right>, \quad 0\le v<2\pi, 0\le u<2$$
As for whether it would be easier to explicitly integrate a vector field $\textbf{F}$ around the curve, or instead to integrate $\textrm{curl } \textbf{F}$ over the surface, it really depends a lot on the specific vector field $\textbf{F}$ you are working with.
A: Stoke's theorem relates an integral over the boundary of a surface $\delta \mathcal{S}$ to an integral over the surface $\mathcal{S}$. In your instance the boundary curve can be parametrized by $\textbf{r}(t)=\langle rcos(\theta), rsin(\theta), 2rcos(\theta)+3\rangle$. Let $x=rcos(\theta), y=rsin(\theta)$ and find $z$ by plugging in the equation of the plane. $\theta$ goes from $0$ to $2\pi$
So you would use $\int_{\mathcal{C}} \textbf{F}\cdot d\textbf{r}$
So it would be easier to work with the vector line integral given some vectors field going through the surface bounded by the plane and the paraboloid!
