# Find the area of a “petal” of a polar curve using Green's Theorem

Find the area of a "petal" of the curve $r^2 = 3\sin{3\theta}$ using the parametrization $\alpha (t)$ of the equation and the formula $$\frac{1}{2}\int^b_a\begin{vmatrix} \alpha_1 & \alpha_2 \\ \alpha_1' & \alpha_2' \end{vmatrix} dt$$

So I got the parametrization of $\theta = t$ and $r = \sqrt{3\sin{3t}}$, and it's obvious that $0 \leq t \leq \frac{\pi}{3}$ because the petals intersect at the origin and this is the smallest interval that gives you a petal. Is there a better way to get the parametrization? If I plugged in this parametrization it would give a very ugly integral.

You should use the area formula $$Area=\int_a^b\frac{1}{2}r(\theta)^2d\theta$$
• $Area=\int\int rdrd\theta$ by substitution from the Cartesian formula, and then evaluating the inside integral gives my formula – Stella Biderman Mar 1 '16 at 3:42
• I figured out a better way, if you let $x= r\cos \theta$ and $y=r\sin \theta$ you can parametrize it nice and you get an easy integral. – ultrainstinct Mar 3 '16 at 16:39