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:
$$
\begin{cases}
x &= -9 \sin (2 t)-5 \sin (3 t) \\[6pt]
y & = 9 \cos (2 t)-5 \cos (3 t)
\end{cases}
$$
but obtained $87\pi$, which does not seem to be correct from the following simple comparison (the circle has a radius of $\sqrt{87}$).
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?
Best Answer
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 $1.575\pi$.)
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 $$ 5A + B = \oint_{C} - B = 10\oint_{C'} - 2\oint_{C}. $$