If C is a positively oriented simple closed contour, use Green's formula to show that the area of the region enclosed by C can be calculated as $$\frac{1}{2i}\int_C \bar{z}dz$$
I approached this problem by writing down $\int_C zdz = \int_C (u+iv)(dx+idy)$ and applying Green's formula to get $\int \int(-v_x+iu_x-u_y+iv_y)dxdy$. I also applied Green's formula to $\int_C \bar{z}dz$ to get $\int \int(v_x+iu_x-u_y+iv_y)dxdy$.
I don't see how to get from the point that I am at to $\frac{1}{2i}\int_C \bar{z}dz$.
Best Answer
We have
$$\frac{1}{2i} \int_C \bar{z}\, dz = \frac{1}{2i} \int_C (x - iy)(dx + i\, dy) = \frac{1}{2i} \int_C (x\, dx + y\, dy) + i(x\, dy - y\, dx).$$
Since $x\, dx + y\, dy$ is exact,
$$\int_C (x\, dx + y\, dy) = 0.$$
Therefore
$$\frac{1}{2i} \int_C \bar{z}\, dz = \frac{1}{2}\int_C (x\, dy - y\, dx).$$
Now if $R$ is the region enclosed by $C$, by Green's theorem, the last expression equals
$$\frac{1}{2}\iint_R \left(\frac{\partial}{\partial x} x - \frac{\partial}{\partial y}(-y)\right)\, dx\, dy = \frac{1}{2} \iint_R (1 + 1)\, dx\, dy = \text{Area}(R).$$