wiki says the following two formulas are equivalent.
${\mathbf {S}}=\int d{\mathbf {S}}$
${\displaystyle \mathbf {S} ={\frac {1}{2}}\oint _{\partial S}{\vec {r}}\times d{\vec {r}}}$
I am learning exterior derivative and the generalized stokes' theorem ${\displaystyle \int _{\partial \Omega }\omega =\int _{\Omega }d\omega \,}$, but when I applied $d(r \wedge dr) = dr \wedge dr - r \wedge ddr= dr \wedge dr + 0 = 0 \Rightarrow S = 0$
where did I do wrong?
Any help is appreciated.
From Stokes's Theorem along with the vector identity $\nabla \times (\hat x_i\times\vec A )=\hat x_i \nabla \cdot \vec A-(\hat x_i\cdot \nabla)\vec A $, we have
$$\begin{align} \hat x_i\cdot \oint_{\partial S}\vec r\times d\vec r&=\oint_{\partial S}(\hat x_i\times \vec r)\cdot d\vec r\\\\ &=\int_S \hat n\cdot \nabla\times (\hat x_i\times\vec r)\,dS\\\\ &=\int_S \hat n\cdot (2\hat x_i)\,dS\\\\ &=2\hat x_i\cdot \int_S \hat n\,dS \end{align}$$
And we are done!