# Math Help - Expansion of Green's/Stoke's theorem, find Area of smooth closed plane curve

1. ## Expansion of Green's/Stoke's theorem, find Area of smooth closed plane curve

 I was given this question in class and I assume it is a spin off of Green's theorem for finding the area of a closed curve λ in 2D but expanded to 3D I believe. Anyways I am pretty confused about it so if anyone could help I would appreciate it,. Question: Let λ be a simple closed smooth space curve that lies in a plane with unit normal vector n = (a, b, c) and has positive orientation with respect to the normal vector n of the plane. Show that the plane area enclosed by λ is 1/2∫_λ (bz−cy)dx+(cx−az)dy+(ay−bx)dz

2. ## Re: Expansion of Green's/Stoke's theorem, find Area of smooth closed plane curve

Note that the volume element on the plane is $\sigma=i(n)dx\wedge dy\wedge dz$, where $i$ is the interior product. And
$\sigma=i(n)dx\wedge dy\wedge dz$
= $i(a \frac{\partial}{\partial x} + b \frac{\partial}{\partial y} + c \frac{\partial}{\partial z})dx\wedge dy\wedge dz$
= $a dy \wedge dz + b dz \wedge dx + c dx \wedge dy$
Let $\omega=\frac{1}{2}[(bz-cy)dx+(cx-az)dy+(ay-bx)dz]$ And it is easy to verify that
$d\omega=\sigma$.
This completes the proof according to Stokes' theorem.

3. ## Re: Expansion of Green's/Stoke's theorem, find Area of smooth closed plane curve

I am just confused about i and the ^ symbols and their meaning... In class we discussed them briefly but we did not really learn about them...

4. ## Re: Expansion of Green's/Stoke's theorem, find Area of smooth closed plane curve

Those are differential forms with which we can express Stokes' theorem easily. Without that you'll have to express Stokes' theorem by Gauss/Green theorems. So if you didn't learn that you can re-write the above proof with the classic language. The proof consists of:
1. express the volume element of the plane in terms of its unit normal vector.
2. use any kinds of Stokes' theorem to turn the surface integral to the line integral.
Then you'll get the result.