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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.

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You should use the area formula $$Area=\int_a^b\frac{1}{2}r(\theta)^2d\theta$$

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  • $\begingroup$ Was that derived from the area equation? I can't exactly see how. $\endgroup$ – ultrainstinct Mar 1 '16 at 3:37
  • $\begingroup$ $Area=\int\int rdrd\theta$ by substitution from the Cartesian formula, and then evaluating the inside integral gives my formula $\endgroup$ – Stella Biderman Mar 1 '16 at 3:42
  • $\begingroup$ I already know how to use double integrals to calculate area. I wanted to use the formula for the area of a region enclosed by a simple closed curve. In this case that is one petal of the curve. $\endgroup$ – ultrainstinct Mar 1 '16 at 3:44
  • $\begingroup$ @Panphobia then you'll just have to evaluate that messy integral $\endgroup$ – Stella Biderman Mar 1 '16 at 3:45
  • $\begingroup$ 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. $\endgroup$ – ultrainstinct Mar 3 '16 at 16:39

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