How to calculate the area of an asteroid using Green's theorem

Question

I came across this question in my revision: Use Green's theorem to calculate the area of an asteroid defined by $x=\cos^3{t}$ and $y=\sin^3{t}$ where $0\leqslant t \leqslant 2\pi$ . The question gives a hint by saying that the area of the asteroid is $\iint \,dx\,dy$. I interpreted this tip to be that $$\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} = 1$$ but then got stuck from there. I know the final answer is meant to be $\frac{3\pi a^2}{8}$. Would really appreciate a tip or a correction of the first step? For reference, Greens theorem: $$\oint_C (P\,dx+Q\,dy)=\iint_R (\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})\,dx\,dy.$$

Answer 1

You can take $$ Q = -\frac{1}{2}x, \qquad P = \frac{1}{2}y, $$ and then the area is $$ \iint_R dx \, dy = \frac{1}{2}\int_C (-y \, dx + x \, dy). $$ Parametrise this and do the calculation and you should get the right answer.

This formula can actually be rephrased in polar coordinates to be more transparent: $$ -y \, dx + x \, dy = -r\sin{\theta} (\cos{\theta} \, dr -r\sin{\theta} \, d\theta ) + r\cos{\theta}(\sin{\theta}\, dr + r\cos{\theta} \, d\theta) = r^2 \, d\theta, $$ the area of a sector of the circle of radius $r$ with angle $d\theta$.