Find the area of the Rose's petal.
$\endgroup$ 3If a Rose leaf is described by the equation $r = \sin 3\theta$, find the area of one petal.
2 Answers
$\begingroup$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
$$A=\int_{0}^{\pi/3}\frac{1}{2}\sin^{2}(3\theta)d\theta=\frac{1}{2}\int_{0}^{\pi/3}\frac{1-\cos(6\theta)}{2}d\theta$$
$$=\frac{1}{4}\left[\theta-\frac{\sin(6\theta)}{2}\right]^{\pi/3}_{0}=\frac{1}{4}\left(\frac{\pi}{3}-\frac{1}{2}\sin\left(\frac{6\pi}{3}\right)\right)=\frac{\pi}{12}$$
$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\Theta:\mathbb{R}\setminus\braces{0} \to \mathbb{R}}$ 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} $\endgroup$ 1
