use a graphing utility to graph the polar equation. inner loop of $r = 10 - 15sin(\theta)$ find the area of…

Explanation:

Step1: Find the limits of integration for the inner - loop

Set (r = 10-15\sin\theta=0). Then (15\sin\theta = 10), so (\sin\theta=\frac{2}{3}). The two values of (\theta) for the inner - loop are (\theta_1=\arcsin(\frac{2}{3})) and (\theta_2=\pi - \arcsin(\frac{2}{3})).

Step2: Recall the formula for the area in polar coordinates

The area (A) of a polar curve (r = f(\theta)) is given by (A=\frac{1}{2}\int_{\alpha}^{\beta}r^{2}d\theta), where (\alpha) and (\beta) are the limits of integration. Here, (r = 10 - 15\sin\theta), so (A=\frac{1}{2}\int_{\arcsin(\frac{2}{3})}^{\pi-\arcsin(\frac{2}{3})}(10 - 15\sin\theta)^{2}d\theta).

Step3: Expand the integrand

Expand ((10 - 15\sin\theta)^{2}=100-300\sin\theta + 225\sin^{2}\theta). We know that (\sin^{2}\theta=\frac{1 - \cos(2\theta)}{2}). So the integrand becomes (100-300\sin\theta+225\times\frac{1 - \cos(2\theta)}{2}=100-300\sin\theta+\frac{225}{2}-\frac{225\cos(2\theta)}{2}=\frac{425}{2}-300\sin\theta-\frac{225\cos(2\theta)}{2}).

Step4: Integrate term - by - term

(\int(\frac{425}{2}-300\sin\theta-\frac{225\cos(2\theta)}{2})d\theta=\frac{425}{2}\theta + 300\cos\theta-\frac{225}{4}\sin(2\theta)+C).

Step5: Evaluate the definite integral

[ \begin{align*} A&=\frac{1}{2}\left[\frac{425}{2}\theta + 300\cos\theta-\frac{225}{4}\sin(2\theta)\right]_{\arcsin(\frac{2}{3})}^{\pi-\arcsin(\frac{2}{3})}\ \end{align*} ] (\cos(\arcsin(\frac{2}{3}))=\frac{\sqrt{5}}{3}) and (\sin(2\arcsin(\frac{2}{3})) = 2\times\frac{2}{3}\times\frac{\sqrt{5}}{3}=\frac{4\sqrt{5}}{9}). [ \begin{align*} A&=\frac{1}{2}\left[\left(\frac{425}{2}(\pi-\arcsin(\frac{2}{3}))+300\cos(\pi - \arcsin(\frac{2}{3}))-\frac{225}{4}\sin(2(\pi - \arcsin(\frac{2}{3})))\right)-\left(\frac{425}{2}\arcsin(\frac{2}{3})+300\cos(\arcsin(\frac{2}{3}))-\frac{225}{4}\sin(2\arcsin(\frac{2}{3}))\right)\right]\ &=\frac{1}{2}\left[\frac{425\pi}{2}-425\arcsin(\frac{2}{3})- 300\cos(\arcsin(\frac{2}{3}))+\frac{225}{4}\sin(2\arcsin(\frac{2}{3}))-\frac{425}{2}\arcsin(\frac{2}{3})-300\cos(\arcsin(\frac{2}{3}))+\frac{225}{4}\sin(2\arcsin(\frac{2}{3}))\right]\ &=\frac{1}{2}\left[\frac{425\pi}{2}-425\arcsin(\frac{2}{3})-300\times\frac{\sqrt{5}}{3}+\frac{225}{4}\times\frac{4\sqrt{5}}{9}-\frac{425}{2}\arcsin(\frac{2}{3})-300\times\frac{\sqrt{5}}{3}+\frac{225}{4}\times\frac{4\sqrt{5}}{9}\right]\ &=\frac{1}{2}\left[\frac{425\pi}{2}-425\arcsin(\frac{2}{3}) - 100\sqrt{5}+ 25\sqrt{5}-\frac{425}{2}\arcsin(\frac{2}{3})-100\sqrt{5}+25\sqrt{5}\right]\ &=\frac{1}{2}\left[\frac{425\pi}{2}- \frac{1275}{2}\arcsin(\frac{2}{3})-150\sqrt{5}\right]\ &\approx 11.0817 \end{align*} ]

Answer:

(11.0817)