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

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: $$\oint_C (P\,dx+Q\,dy)=\iint_R (\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})\,dx\,dy.$$

• This curve is called an astroid. – lhf Apr 25 '17 at 1:37

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 $d\theta$.

• but wait, $\frac{\partial Q}{\partial X}$ and the same for P would be zero? that wouldn't satisfy the initial tip? – Sam Somers Apr 24 '17 at 23:50
• Hang on, I've got them the wrong way round... – Chappers Apr 24 '17 at 23:50
• Perfect, just wanted to check! – Sam Somers Apr 24 '17 at 23:51
• Also, could I use any values for Q and P that satisfy the initial condition or is it just these? If it is just these, why and how did you pick them? Ease? – Sam Somers Apr 24 '17 at 23:52
• This has nice symmetry, and an interpretation in polar coordinates. You can choose anything so that the appropriate derivatives add up to $1$. – Chappers Apr 24 '17 at 23:56