# Find the area of a single loop in curve r=\sin(6\theta)?

## I am told the formula is $A = \frac{1}{2} \setminus {\int}_{a}^{b} {r}^{2} d \setminus \theta$, so $A = \frac{1}{2} \setminus {\int}_{a}^{b} {\left(\setminus \sin \left(6 \setminus \theta\right)\right)}^{2} d \setminus \theta$ But what are the bound values, $a$ and $b$?

May 28, 2018

The area of 1 loop of the given polar curve is $\frac{\pi}{24}$ square units.

#### Explanation:

Start by drawing the polar curve. It helps to picture it. As you can see, each loop starts and ends when $r = 0$. Thus our bounds of integration will be consecutive values of $\theta$ where $r = 0$.

$\sin \left(6 \theta\right) = 0$

$6 \theta = 0 \mathmr{and} 6 \theta = \pi$

$\theta = 0 \mathmr{and} \theta = \frac{\pi}{6}$

Thus we will be finding the value of $\frac{1}{2} {\int}_{0}^{\frac{\pi}{6}} {\sin}^{2} \left(6 x\right) \mathrm{dx}$ to find the area.

$A = \frac{1}{2} {\int}_{0}^{\frac{\pi}{6}} {\sin}^{2} \left(6 x\right) \mathrm{dx}$

Recall that $\cos \left(2 x\right) = 1 - 2 {\sin}^{2} \left(x\right)$, thus $\cos \left(12 x\right) = 1 - 2 {\sin}^{2} \left(6 x\right)$, and it follows that ${\sin}^{2} \left(6 x\right) = \frac{\cos \left(12 x\right) - 1}{- 2} = \frac{1}{2} - \cos \frac{12 x}{2}$

$\setminus \textcolor{m a r \infty n}{A = \frac{1}{2} \setminus {\int}_{0}^{\setminus \frac{\pi}{6}} \left(\frac{1}{2} - \setminus \cos \frac{12 x}{2}\right) \mathrm{dx}}$
\color(maroon)(A=1/2{:[1/2x-1/2(1/12\sin(12x))]|:}_0^(\pi/6))
A = 1/2{:[1/2x - 1/24sin(12x)]|:}_0^(pi/6)
$A = \frac{1}{2} \left(\frac{\pi}{12}\right)$
$A = \frac{\pi}{24}$

Hopefully this helps!