find the area of the region enclosed by one loop of the curve.\nr = 6 + 12 sin(θ) (inner loop)\n24(π…

Explanation:

Step1: Find the limits of integration

Set (r = 0), so (6+12\sin\theta=0). Then (\sin\theta=-\frac{1}{2}). The solutions for (\theta) in ([0,2\pi]) are (\theta=\frac{7\pi}{6}) and (\theta=\frac{11\pi}{6}).

Step2: Use the polar - area formula

The formula for the area (A) of a polar curve (r = f(\theta)) is (A=\frac{1}{2}\int_{\alpha}^{\beta}r^{2}d\theta). Here, (r = 6 + 12\sin\theta), (\alpha=\frac{7\pi}{6}), and (\beta=\frac{11\pi}{6}). So (A=\frac{1}{2}\int_{\frac{7\pi}{6}}^{\frac{11\pi}{6}}(6 + 12\sin\theta)^{2}d\theta).

Expand ((6 + 12\sin\theta)^{2}=36+144\sin\theta + 144\sin^{2}\theta).

Use the identity (\sin^{2}\theta=\frac{1 - \cos(2\theta)}{2}). Then (36+144\sin\theta + 144\sin^{2}\theta=36+144\sin\theta+72(1 - \cos(2\theta))=108+144\sin\theta-72\cos(2\theta)).

Step3: Integrate term - by - term

(\frac{1}{2}\int_{\frac{7\pi}{6}}^{\frac{11\pi}{6}}(108 + 144\sin\theta-72\cos(2\theta))d\theta=\frac{1}{2}\left[108\theta-144\cos\theta - 36\sin(2\theta)\right]_{\frac{7\pi}{6}}^{\frac{11\pi}{6}})

First, evaluate at (\theta=\frac{11\pi}{6}): (108\times\frac{11\pi}{6}-144\cos\frac{11\pi}{6}-36\sin\left(2\times\frac{11\pi}{6}\right)=198\pi-144\times\frac{\sqrt{3}}{2}-36\times\left(-\frac{\sqrt{3}}{2}\right)=198\pi - 72\sqrt{3}+ 18\sqrt{3}=198\pi-54\sqrt{3})

Then evaluate at (\theta=\frac{7\pi}{6}): (108\times\frac{7\pi}{6}-144\cos\frac{7\pi}{6}-36\sin\left(2\times\frac{7\pi}{6}\right)=126\pi-144\times\left(-\frac{\sqrt{3}}{2}\right)-36\times\left(-\frac{\sqrt{3}}{2}\right)=126\pi + 72\sqrt{3}+18\sqrt{3}=126\pi + 90\sqrt{3})

Subtract: (\frac{1}{2}\left[(198\pi-54\sqrt{3})-(126\pi + 90\sqrt{3})\right]=\frac{1}{2}(72\pi-144\sqrt{3}) = 36\pi-72\sqrt{3}=36\left(\pi - 2\sqrt{3}\right))

Answer:

(36\left(\pi - 2\sqrt{3}\right))