Show that the area bounded by a simple closed curve is given by \(\frac{1}{2} \oint{xdy-ydx} \).

**Solution**

We use Green's Theorem, and let \(P=-y\) , \(Q= x\). Thus

\[\oint { Pdx+Qdy } =\iint { \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} \right) dxdy }\]

\[\oint { -ydx+xdy } =\iint { \left(\frac{\partial (x)}{\partial x}-\frac{\partial (-y)}{\partial y} \right)dxdy }\]

\[\oint { xdy-ydx } = 2\iint {dxdy }\]

\[\oint { xdy-ydx } = 2\iint {A }\]

\[A = \frac{1}{2}\oint { xdy-ydx }\]

Check out my other notes at Proof, Disproof, and Derivation

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TopNewestThis very question was asked in my semester examination which I could happily answer using the process described here.

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