In this task I've got to show the following identity for the area of an enclosed object $S$:
$$\text{Area}(S) = \int\int_S \mathrm{dx\,dy} = \int_{\partial S} \langle (0,x),T\rangle\, \mathrm{ds} = \int_{\partial M} \langle (-y,0),T\rangle\, \mathrm{ds} = \dfrac{1}{2}\int_{\partial M}\langle(-y,x),T\rangle\,\mathrm{ds}$$
I kinda managed proving the start and and ending point of this consideration by choosing a vector valued Function $F = \left(\begin{array}{c}-y\\ x\end{array}\right)$
Now let $\gamma(t)$ be the parametrisation of the boundary of S, thus the curve integral along this region is given by:
$\displaystyle{\int_{\gamma(t)}F\,\mathrm{dt} = \int_{\gamma(t)}-y\,\mathrm{dx}+x\,\mathrm{dy}}$. According to Greens Theorem this is equal to:
$$\int\int_S\partial_x F_2-\partial_yF_1\,\mathrm{dx\,dy} = \int\int_S 2\,\mathrm{dx\,dy}$$
Multiplying each side by $\frac{1}{2}$ yields $${\dfrac{1}{2} \int_{\gamma(t)}-y\,\mathrm{dx}+x\,\mathrm{dy} = \dfrac{1}{2}\int_{\gamma(t)}\langle(-y,x),T\rangle}\,\mathrm{ds} = \int\int_S \,\mathrm{dx\,dy} = \text{Area(S)}$$
Now, I just don't know how to show the middle part. Particularly because I can't make a distinction between $\partial S$ and $\partial M$. Those are not the same?