Show for a positively oriented simple closed contour $C$ that the area of the region $G \subset{\mathbb{C}}$ enclosed by C is given by $\frac{1}{2i}\int_C \bar{z}\text{ d}z$. Use the Gauss' integral theorem.
So my steps towards solving the problem were as follows:
$$ \int_C \bar{z}\text{ d}z = \int_C (x-iy)\text{ d}(x+iy) = \int_C x\text{ d}x + y\text{ d}y + i\int_C x\text{ d}y - y\text{ d}x. $$ So this can be written as the dot product of two vectors
$$ \int_C (x,y)\cdot(\text{d}x,\text{d}y) + i\int_C (-y,x)\cdot(\text{d}x,\text{d}y) = \int_C (x,y)\text{ d}(x,y) + i\int_C (-y,x)\text{ d}(x,y). $$ Using the Gauss' integral theorem $\int_{\partial G} \vec{x}\text{ d}\vec{s} = \int_G \vec{\nabla}\cdot\vec{v}\text{ d}A$ this leads to
$$ \int_C (x,y)\text{ d}(x,y) + i\int_C (-y,x)\text{ d}(x,y) = \int_G \vec{\nabla}\cdot(x,y)\text{ d}A + i\int_G \vec{\nabla}\cdot(-y,x)\text{ d}A = \int_G 2\text{ d}A + i\int_G 0\text{ d}A = 2\cdot\text{vol}_2(G). $$
So clearly this isn't the correct result. Where exactly is my misconception in the calculation?
I am thankful for any help.