I'm slowly making my way through better understanding of Stokes Theorem-type problems. I struggle with two issues, the intuition about which I'm lacking.
First, what does it mean to integrate the curl over the surface? Is this the same thing as the flux, which I usually think of as the amount of "flow" over the surface? That's confusing to me.
Second, it is usually stated that Stokes relates the surface integral of curl dot dS (over S) to the double integral of the force dot dr (over C, the contour) as in the following:
When you see a model solution, as in the following, you will see that, to calculate the right side of this equation, dS is transformed into the cross product of the partial derivatives times dA.
Why is that? dS represents the infinitesmally small surface area. Why does that become the cross of the partials times dA? Doesn't make sense to me.
Here's an example, where you can see this in action.
Best Answer
When you parametrize a surface by $\mathbf r(u,v)$, the parallelogram spanned by $\dfrac{\partial\mathbf r}{\partial u}$ and $\dfrac{\partial\mathbf r}{\partial v}$ has area $$\left\|\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}\right\|.$$ The cross product itself is a (non-normalized) normal vector and this area becomes the "fudge factor" (like the $r$ in $r\,dr\,d\theta$) that tells you how to relate area in the $uv$-plane to area on the surface. After that, it's just \begin{align*} \mathbf F\cdot\mathbf n\,dS &= \left(\mathbf F\cdot\frac{\frac{\partial\mathbf r}{\partial u}\times\frac{\partial\mathbf r}{\partial v}}{\left\|\frac{\partial\mathbf r}{\partial u}\times\frac{\partial\mathbf r}{\partial v}\right\|}\right)\left\|\frac{\partial\mathbf r}{\partial u}\times\frac{\partial\mathbf r}{\partial v}\right\|du\,dv \\ &=\mathbf F\cdot \left(\frac{\partial\mathbf r}{\partial u}\times\frac{\partial\mathbf r}{\partial v}\right) du\,dv. \end{align*}