91Ó°ÊÓ

To find the boundaries for the inner loop, we set \(r=0\) and solve for \(\theta\). \(0=2+4\cos\theta \Rightarrow \cos\theta=-\frac{1}{2}\). We know that cosine of \(\theta\) equals \(-\frac{1}{2}\) when \(\theta= \frac{2\pi}{3}\) and \(\theta= \frac{4\pi}{3}\), which are the limits of the inner loop.

The formula for finding the area of a polar graph is given by \(A = \frac{1}{2}\int_{\alpha}^{\beta} (r(\theta))^2 d\theta\), where \(\alpha\) and \(\beta\) are the lower and upper boundaries for \(\theta\) respectively. In our case, \(r(\theta) = 2+4\cos\theta\), so our integral becomes: \(A = \frac{1}{2} \int_{\frac{2\pi}{3}}^{\frac{4\pi}{3}} (2+4\cos\theta)^2 d\theta\).

First, expand the square: \((2+4\cos\theta)^2 = 4+16\cos\theta+16\cos^2\theta\) Now use the double angle identity for \(\cos^2\theta\): \(16\cos^2\theta = 8(2\cos^2\theta -1)=8\cos(2\theta)+8\) So, our integral becomes: \(A = \frac{1}{2} \int_{\frac{2\pi}{3}}^{\frac{4\pi}{3}} (12+32\cos\theta+8\cos(2\theta)) d\theta\) Let's integrate \(12\), \(32\cos\theta\), and \(8\cos(2\theta)\) separately: \(I_1 = \int_{\frac{2\pi}{3}}^{\frac{4\pi}{3}} 12 d\theta = 12\theta\Big|_{\frac{2\pi}{3}}^{\frac{4\pi}{3}}\) \(I_2 = \int_{\frac{2\pi}{3}}^{\frac{4\pi}{3}} 32\cos\theta d\theta = 32\sin\theta\Big|_{\frac{2\pi}{3}}^{\frac{4\pi}{3}}\) \(I_3 = \int_{\frac{2\pi}{3}}^{\frac{4\pi}{3}} 8\cos(2\theta) d\theta = 4\sin(2\theta)\Big|_{\frac{2\pi}{3}}^{\frac{4\pi}{3}}\)

Now, let's plug in the limits to evaluate the definite integrals: \(I_1=12\theta\Big|_{\frac{2\pi}{3}}^{\frac{4\pi}{3}}=12(\frac{4\pi}{3} -\frac{2\pi}{3})=8\pi\) \(I_2=32\sin\theta\Big|_{\frac{2\pi}{3}}^{\frac{4\pi}{3}} = 32(\sin(\frac{4\pi}{3})-\sin(\frac{2\pi}{3}))=0\) \(I_3 = 4\sin(2\theta)\Big|_{\frac{2\pi}{3}}^{\frac{4\pi}{3}}=4(\sin(\frac{8\pi}{3})-\sin(\frac{4\pi}{3}))=4(\sin(\frac{2\pi}{3})-\sin(\frac{4\pi}{3}))=12\) Finally, add up the integral results, and multiply by \(\frac{1}{2}\) to get the final area: \(A =\frac{1}{2}(I_1+I_2+I_3) = \frac{1}{2}(8\pi + 0 + 12) = 4\pi + 6\).

The area of the region inside the inner loop of the limaçon \(r=2+4\cos\theta\) is \(4\pi + 6\).