📂Geometry

Derivation of the Area Formula for a Region Enclosed by a Simple Closed Plane Curve

Formula ¹

If a simple closed curve $\alpha$ surrounds the region $R$ and rotates in a counterclockwise direction, $$ V (R) = \int_{\alpha} x dy = - \int_{\alpha} y dx $$

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• $V(R)$ represents the volume of the region $R$, or in other words, the area of $R$.

Proof

According to Green’s Theorem, let’s assume that a simple planar $C^{2}$ closed curve $\mathcal{C}$, which is piecewise smooth, encloses a bounded region $\mathcal{R}$ in a counterclockwise direction.

If the two functions defined in the region $\mathcal{R}$, $P,Q$, are differentiable within $\mathcal{R}$, $$ \int_{\mathcal{C}} (Pdx + Qdy) = \iint_{\mathcal{R}} (Q_{x} - P_{y}) dx dy $$

By Green’s Theorem, $$ \begin{align*} \int_{\alpha} x dy =& \int_{\alpha} \left( 0 dx + x dy \right) \\ =& \iint_{R} \left( {{ \partial x } \over { \partial x }} - {{ \partial 0 } \over { \partial y }} \right) dx dy \\ =& \iint_{R} {{ \partial x } \over { \partial x }} dx dy \\ =& \iint_{R} 1 dx dy \\ =& V(R) \end{align*} $$ Moreover, $$ \int_{\alpha} \left( x dy + y dx \right) = \iint_{R} \left( {{ \partial y } \over { \partial y }} - {{ \partial x } \over { \partial x }} \right) dx dy = 0 $$ Hence, we obtain the following. $$ V (R) = \int_{\alpha} x dy = - \int_{\alpha} y dx $$

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