Prove that the area of a region is $\int _{C} x dy = -\int _C ydx$ using Green's Theorem If $\mathbf{c}$ is a simple closed plane curve whose image bounds a region R, and which is traversed counterclockwise, then the area of R is $\int _{c} x dy = -\int _c ydx$, where x and y are the coordinates of the plane. Prove this using Green's Theorem.
Attempt:
Area = $\int\int_R 1dA$.
If we take some P(x,y) and Q(x,y) such that ($\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$) = 1, then $\int\int_R (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y})$ dA.
By Green's Theorem $\int\int_R (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y})$ dA = $\int_c Pdx + Qdy$.
Now since ($\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$) = 1 there are different possibilities for P(x,y) and Q(x,y), but I don't know how to find all of the different possibilities. 
Any help is appreciated.
 A: When you are given problems with two variables but one equation, an easy way to "solve" for one is by eliminating one variable and seeing what you get for the other. In this case, if $Q$ was zero, then you'd have
$$ 1 = -\iint_R \frac{\partial P}{\partial y} dA = \int_C P\,dx $$
and
$$ 1 = -\frac{\partial P}{\partial y}. $$
From this, what is an obvious choice of $P$ based on what you're supposed to end up with? Repeat this for $Q$. Note that I've more or less reverse engineered the problem, but if I just told you what to pick for $P$ and $Q$, you'd be left wondering why other than knowing that it works.
A: (i) Show that the integral $\int_c xdy$ represents the area of the region R. 
If we let P(x,y) = 0 and Q(x,y) = x. 
Then we get $\int_c xdy$ = $\int_c Pdx + Qdy$ = $\int\int_R (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) dxdy$ = $\int\int_R1dA$ = A.
(ii) Show that the integral -$\int_c ydx$ represent the area of the region R.
If we let P(x,y) = -y and Q(x,y) = 0.
Then $-\int_c ydx$ = $\int_c Pdx + Qdy = \int\int_R (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) dxdy = \int\int_R 1dA = A$.
A: If you're allowed it use the generalized Stokes' theorem for the 1-form $w=x \, dy$ (or $w=-y \, dx$) noticing that its exterior derivative $dw=dx\wedge dy$  is the signed volume form (area element) on $\mathbb R^2 \supset R,$
$$\int_C w = \int_R dw = \int_R dx \wedge dy = vol(R).$$
