Under the conditions of Green’s Theorem, the area of a region $R$ enclosed by a curve $C$ is

$$\oint_C x \, dy=-\oint_C y \, dx=\frac{1}{2}\oint_C (x \, dy - y \, dx)$$

I tried to use the result to calculate the area of region defined by the following plane curve:

$$ \begin{cases} x &= -9 \sin (2 t)-5 \sin (3 t) \\[6pt] y & = 9 \cos (2 t)-5 \cos (3 t) \end{cases} $$

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but obtained $87\pi$, which does not seem to be correct from the following simple comparison (the circle has a radius of $\sqrt{87}$).

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How should I use Green's theorem in this case? What would be a more convenient way to calculate the area enclosed by such a plane curve?

  • 2
    $\begingroup$ I believe your calculation is counting the interior "pentagon" two times - you can divide that curve into the boundary and the boundary of the interior pentagon, so integrating the former gives you the proper area, and integrating the latter gives the pentagon's area - but this method integrates both! You'll need to split the curve into segments between its self-intersections to get just the boundary. $\endgroup$ – Milo Brandt Jun 29 '15 at 1:25

Let $A$ denote the area of one of the five triangular "points" and $B$ the area of the central pentagon. Your Green's theorem integral double-counts $B$, so its value is $5A + 2B$, whereas you want to find $5A + B$.

Let $t_{0}$ denote the first root of $x(t)$ larger than $\pi$, i.e., the time at which the curve first crosses the negative $y$-axis after reaching the very top of the curve, and let $C'$ denote the quasi-triangular curve obtained by following $C$ from $\pi$ to $t_{0}$, the using the vertical segment to close up (shaded). (Very roughly, $t_{0}$ is in the vicinity of $1.575\pi$.)

The region enclosed by C'

Twice the "Green's theorem integral" over $C'$ is equal to $3A + B$.

Since $$ B = (15A + 6B) - (15A + 5B) = 3\oint_{C} - 10\oint_{C'}; $$ the desired area is $$ 5A + B = \oint_{C} - B = 10\oint_{C'} - 2\oint_{C}. $$

  • $\begingroup$ Incidentally, the "$1.575\pi$" figure isn't intended as a value for $t_{0}$, but it's close enough to use as a seed for Newton's method applied to the $x$ component of the curve. $\endgroup$ – Andrew D. Hwang Jun 29 '15 at 2:03
  • $\begingroup$ thank you ! Is it possible to obtain symbolic solution instead of approximate ones? $\endgroup$ – LCFactorization Jun 30 '15 at 8:12
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    $\begingroup$ As it turns out, yes: A bit of trig and algebra shows $t_{0}$ is a root of$$20\cos^{2} t + 18\cos t - 5 = 0;$$that is, $t_{0} = 2\pi - \arccos\frac{-9 + \sqrt{181}}{20}$. (Also, as you may have noticed, it suffices to integrate from $\pi$ to $t_{0}$; the contribution from the vertical segment is zero for all three integrands since $x \equiv 0$ on the $y$-axis.) $\endgroup$ – Andrew D. Hwang Jun 30 '15 at 10:39

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