# Question #4cf0a

May 25, 2017

$18 \sqrt{3} - \frac{2}{3} \pi$

#### Explanation:

First we need to solve for the intersections between the graphs.

$17 \sin \theta = 9 - \sin \theta$
$17 \sin \theta + \sin \theta = 9$
$18 \sin \theta = 9$
$\sin \theta = \frac{1}{2}$
$\theta = {\sin}^{-} 1 \left(\frac{1}{2}\right)$
$\theta = \frac{\pi}{6}$

Now, the thing with trig functions is that they are periodic, so we need to find the next point that has $\sin \theta = \frac{1}{2}$ from $\theta = \frac{\pi}{6}$. In this case, $\sin \left(\frac{5 \pi}{6}\right) = \frac{1}{2}$.

So now we have the interval in which we have the area above ${y}_{1} = 9 - \sin \theta$ and the area below ${y}_{2} = 17 \sin \theta$ (Yellow area, Figure 1).

Figure 1

Now, we can integrate the two areas.

${A}_{1} = {\int}_{\frac{\pi}{6}}^{\frac{5}{6} \pi} 9 - \sin \theta \setminus \quad d \theta$

${A}_{2} = {\int}_{\frac{\pi}{6}}^{\frac{5}{6} \pi} 17 \sin \theta \setminus \quad d \theta$

$A = {A}_{2} - {A}_{1}$

Now, just a side note that the blue shaded area is just ${A}_{1} + {A}_{2}$

Let's integrate ${A}_{1}$ and ignore the constant

$\int \setminus \quad 9 - \sin \theta \setminus \quad d \theta = \int \setminus \quad 9 \setminus \quad d \theta - \int \setminus \quad \sin \theta \setminus \quad d \theta = x + \cos \theta$

So

${\int}_{\frac{\pi}{6}}^{\frac{5}{6} \pi} 9 - \sin \theta \setminus \quad d \theta = x + \cos \theta \setminus \quad {|}_{\frac{\pi}{6}}^{\frac{5}{6} \pi}$
$= \cos \left(\frac{5}{6} \pi\right) - \cos \left(\frac{\pi}{6}\right) + \frac{5}{6} \pi - \frac{\pi}{6}$
$= - \sqrt{3} - \frac{2}{3} \pi$

Let's integrate ${A}_{2}$ and ignore the constant

$\int \setminus \quad 17 \sin \theta \setminus \quad d \theta$
$= - 17 \cos \theta$

${\int}_{\frac{\pi}{6}}^{\frac{5}{6} \pi} 17 \sin \theta \setminus \quad d \theta = - 17 \cos \theta \setminus \quad {|}_{\frac{\pi}{6}}^{\frac{5}{6} \pi} = 17 \sqrt{3}$

So, the area is $18 \sqrt{3} - \frac{2}{3} \pi$