Find the area of the Rose's petal. 
If a Rose leaf is described by the equation $r = \sin 3\theta$, find the area of one petal.

 A: A sketch is useful here, but the only important observation is that $r=0$ when $\theta=0$, and again at $\frac{\pi}{3}$. These are your limits for one petal.  
Since the area of a polar curve between the rays $\theta=a$ and $\theta=b$ is given by $\int_{a}^{b}\frac{1}{2}r^{2}d\theta$, we have
$$A=\int_{0}^{\pi/3}\frac{1}{2}\sin^{2}(3\theta)d\theta=\frac{1}{2}\int_{0}^{\pi/3}\frac{1-\cos(6\theta)}{2}d\theta$$
$$=\frac{1}{4}\left[\theta-\frac{\sin(6\theta)}{2}\right]^{\pi/3}_{0}=\frac{1}{4}\left(\frac{\pi}{3}-\frac{1}{2}\sin\left(\frac{6\pi}{3}\right)\right)=\frac{\pi}{12}$$
A: $\newcommand{\+}{^{\dagger}}%
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$\ds{\Theta:\mathbb{R}\setminus\braces{0} \to \mathbb{R}}$ is the
$Heaviside\ Step\ Function$. $\ds{\Theta\pars{x} = \left\lbrace\begin{array}{rcl}
\ds{0} & \mbox{if} & \ds{x < 0}
\\
\ds{1} & \mbox{if} & \ds{x > 0}
\end{array}\right.}$
$\ds{}$
With $x = r\cos\pars{\theta}$, $y = r\sin\pars{\theta}$:
\begin{align}
\color{#0000ff}{\large{\cal A}}
&\equiv
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\Theta\pars{\theta}
\Theta\pars{\pi - 3\theta}\Theta\pars{\sin\pars{3\theta} - r}
\,\dd x\,\dd y
=
\int_{0}^{\pi/3}\dd\theta\int_{0}^{\infty}\dd r\,r\,
\Theta\pars{\sin\pars{3\theta} - r}
\\[3mm]&=
\int_{0}^{\pi/3}\dd\theta\int_{0}^{\sin\pars{3\theta}}r\,\dd r
=
\int_{0}^{\pi/3}\half\,\sin^{2}\pars{3\theta}\,\dd\theta
=
\half\int_{0}^{\pi/3}{1 - \cos\pars{6\theta} \over 2}\,\dd\theta
\\[3mm]&=
{1 \over 4}\bracks{{\pi \over 3} - {\sin\pars{2\pi} \over 6}}
=\color{#0000ff}{\large{\pi \over 12}}
\end{align}
