>If a Rose leaf is described by the equation $r =1+ 3\sin 2\theta$, find the area of one petal.

Question

If a Rose leaf is described by the equation $r = 1+3\sin 2\theta$, find the area of one petal.

I know the formula, but don't know how to find the bounds.

Answer 1

The curve has symmetries w.r.t. the axes with polar equations $\;\theta=\pm\dfrac\pi4$:

The origin is obtained when $$\sin 2\theta=-\frac13\iff 2\theta\equiv\begin{cases}-\arcsin\frac13 \\ \pi+ \arcsin\frac13 \end{cases}\mod 2\pi\iff\theta \equiv\begin{cases}-\frac12\arcsin\frac13 \\ \frac\pi2+\frac12 \arcsin\frac13 \end{cases}\mod \pi$$

Taking the symmetries into account, the area of the big leaf is $$2\int_{-\frac12\arcsin\frac13}^{\frac\pi 4}\mspace-42mu(1+3\sin 2\theta)^2\,\mathrm d \theta$$ You can compute this integral expanding the square and linearising: \begin{align} (1+3\sin 2\theta)^2&=1+6\sin 2\theta+9\sin^22\theta=1+6\sin 2\theta+9\,\frac{1-\cos4\theta}2\\ &=\frac12(11+6\sin 2\theta-9\cos4\theta) \end{align}, so the integral is (denoting $\alpha=-\frac12\arcsin\frac13$) $$\int_\alpha^{\frac\pi 4}(11+6\sin 2\theta-9\cos4\theta)\,\mathrm d \theta = 11\theta-3\cos 2\theta-\frac94\sin4\theta\Biggm\vert_{\,\alpha}^{\tfrac\pi 4}$$ If I'm not mistaken, you should find $$\frac{11}4\Bigl(\pi+2\arcsin\frac13\Bigr)+\sqrt{6^{\vphantom{l}}}.$$