# Stokes theorem and line integral

1. Dec 12, 2009

1. The problem statement, all variables and given/known data

Prove that $$2A=\oint \vec{r}\times d\vec{r}$$

2. Relevant equations

3. The attempt at a solution

From stokes theorem we have $$\oint d\vec{r}\times \vec{r}=\int _{s}(d\vec{s}\times \nabla)\times \vec{r}= \int _{s}(2ds\frac{\partial f}{\partial x},-ds+ds\frac{\partial f}{\partial y},-2ds)$$. I'm stuck here and I don't know how to continue. Please help me.

Another method is $$\oint \vec{r}\times d\vec{r}=\oint (ydz-zdy,zdx-xdz,xdy-ydx)$$. Please help me solve it.

2. Dec 12, 2009

### Dick

Maybe they didn't state it very clearly, but that formula is only true for a planar surface, say a region in the xy plane. And I'm not sure what formulation you are using of Stokes theorem. If you are using the vector form, that means a unit normal points in the z direction, say k. To get the area you want to integrate 1*dx*dy. Now if you define the vector field F=(-y/2,x/2,0) what is curl(F)?

3. Dec 12, 2009

So in this case z=0 and then we have: $$\oint \vec{r}\times d\vec{r}=\oint (ydz-zdy,zdx-xdz,xdy-ydx)=\oint (0,0,-ydx+xdy)$$ so $$\vec{F}=(-y,x,0) and \vec{n}=(0,0,1) and \nabla \times \vec{F}=(0,0,2) and so (\nabla \times \vec{F}).\vec{n}=(0,0,2)$$ so we have $$\oint \vec{F}.d\vec{r}=\int_{s}(\nabla\times \vec{F}).\vec{n}ds=\int\int_{D}2dxdy=2A$$