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  "body": "\n## Proof\n\nWe apply Green's theorem three times.\n\n1. The area of $D$ is $A = \\iint_D 1 \\, dA$, the [[double integral|0NKBMFC18R]] of the constant $1$ over $D$.\n2. Take $L = 0$ and $M = x$. These have continuous partial derivatives on $\\R^2$, and $\\partial M / \\partial x - \\partial L / \\partial y = 1$.\n3. By [[Green's theorem|0HR3N02317]] for the [[simple closed curve|0D032NSZKY]] $C$, the [[line integral|0YS8RX87F4]] $\\oint_C x \\, dy$ equals $\\iint_D 1 \\, dA = A$.\n4. Take $L = -y$ and $M = 0$. Then $\\partial M / \\partial x - \\partial L / \\partial y = 0 - (-1) = 1$. Green's theorem gives $-\\oint_C y \\, dx = \\oint_C (-y) \\, dx = A$.\n5. Take $L = -y/2$ and $M = x/2$. Then $\\partial M / \\partial x - \\partial L / \\partial y = \\tfrac{1}{2} + \\tfrac{1}{2} = 1$. Green's theorem gives $\\tfrac{1}{2} \\oint_C (x \\, dy - y \\, dx) = A$.\n6. Steps 3 to 5 give $A = \\oint_C x \\, dy = -\\oint_C y \\, dx = \\tfrac{1}{2} \\oint_C (x \\, dy - y \\, dx)$.\n\n## Idea\n\nGreen's theorem turns a line integral into the double integral of $\\partial M / \\partial x - \\partial L / \\partial y$. If this expression is the constant $1$, the double integral is the area. Each of the three choices of $(L, M)$ has this property.\n",
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