{
  "aliases": [
    "area-by-greens-theorem"
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  "body": "\n## Statement\n\nLet $C$ be a positively oriented, piecewise smooth, simple [[closed curve|0D032NSZKY]] in $\\mathbb{R}^2$ that encloses the bounded region $D$.\nThen the area $A$ of $D$ satisfies\n$$A = \\oint_C x \\, dy = -\\oint_C y \\, dx = \\frac{1}{2} \\oint_C (x \\, dy - y \\, dx).$$\n\n## Remarks\n\nProof: apply [[Green's theorem|0HR3N02317]] with $(L, M)$ equal to $(0, x)$, $(-y, 0)$ and $(-y/2, x/2)$. In each case $\\partial M / \\partial x - \\partial L / \\partial y = 1$.\nExample: the ellipse $x = a \\cos t$, $y = b \\sin t$, $t \\in [0, 2\\pi]$, with $a, b \u003e 0$, gives $A = \\tfrac{1}{2} \\int_0^{2\\pi} a b \\, dt = \\pi a b$.\nFor a polygon with vertices $(x_1, y_1), \\dots, (x_n, y_n)$ in counterclockwise order, the formula gives $A = \\tfrac{1}{2} \\sum_{k=1}^n (x_k y_{k+1} - x_{k+1} y_k)$, with $(x_{n+1}, y_{n+1}) = (x_1, y_1)$.\n\n## Explanations\n\n### L1 human\n\nYou can find the area of a region by going once around its boundary, without looking inside.\nThis is a direct use of Green's theorem with a field whose curl is $1$.\n",
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