Verifying Green's Theorem on a simple closed curve involving area

Question

Given a simple closed curve $C$ which bounds a region $D$ and on which Green's theorem applies, I'm trying to show that $$ \text{Area}(D) = \int\limits_{C}x \ dy = - \int\limits_{C} y \ dx. \tag{1}$$ My attempt is as follows ; first assume $(1)$ is true. Since the contour integral is linear, \begin{gather} \int\limits_{C} x \ dy + \int\limits_{C} y \ dx = 0 \implies \int\limits_{C} x\ dy + y \ dx =0. \end{gather} Converting the last integral using Green's theorem and show it is equal to $0$ should imply that the initial assumption is correct. \begin{gather} \int\limits_{C} x \ dy +y \ dx = \int\limits_{C} Q \ dy + P \ dx \implies \iint\limits_{D} \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \ dA = \iint\limits_{D}(1-1)\ dA = 0. \end{gather} Therefore the initial assumption $(1)$ is valid. Can someone tell me if I went wrong somewhere as this reasoning seems quite simple , I feel like I must've missed some steps or made a wrong assumption.

Answer 1

Apply Green's Theorem by taking $Q=x, P=0.$ You get first expression in the required identity and for the second, $Q=0, P=-y$. Use the fact $\iint_{D} dA $= Area(D).