91Ó°ÊÓ

First, observe that \(r=2+\cos \theta\) has \(b=2\) and \(a=1\), which means there is no inner loop as \(b > a\). Now, we will find the area of the region bounded by the limaçon using the polar coordinates formula: \(\frac{1}{2}\int_{\alpha}^{\beta} r^2 d\theta\). To find the limits of integration, consider the values of \(\theta\) where \(r\) takes its minimum value which is obtained by finding the value of the derivative of \(r\) with respect to \(\theta\) equal to 0: \(\frac{dr}{d\theta}=-\sin \theta =0\). Thus, our limits of integration are from \(\alpha=0\) to \(\beta=2\pi\), which cover the entire limaçon. Now, find the area: $$ Area = \frac{1}{2}\int_{0}^{2\pi} (2+\cos \theta)^2 d\theta $$ Simplify the expression inside the integral: $$ Area = \frac{1}{2}\int_{0}^{2\pi} (4+4\cos\theta+\cos^2\theta) d\theta $$ Separate the integral into three parts and use the trigonometric identity \(\cos^2\theta = \frac{1+\cos(2\theta)}{2}\): $$ Area = \frac{1}{2}\int_{0}^{2\pi} 4d\theta + 2\int_{0}^{2\pi} \cos\theta d\theta + \frac{1}{2}\int_{0}^{2\pi} \frac{1+\cos(2\theta)}{2} d\theta $$ Solve the integrals: $$ Area = 2\pi [2] + 2[0] + \frac{1}{2}(2\pi)[\frac{1}{2}]\\ Area = 4\pi + \frac{1}{2}(2\pi) = \frac{9}{2}\pi $$ Thus, the area of the region bounded by the limaçon \(r=2+\cos \theta\) is \(\frac{9}{2}\pi\).

First, observe that \(r=1+2\cos \theta\) has \(b=1\) and \(a=2\), which means there is an inner loop as \(b < a\). To find the area outside the inner loop and inside the outer loop, we need to find the range of \(\theta\) values that cover this region. The inner loop is formed when \(r\) is negative, so we need to find the values of \(\theta\) where \(r=0\). $$ 1 + 2\cos \theta = 0 \\ \cos \theta = -\frac{1}{2} $$ The solutions for \(\theta\) are \(\frac{2\pi}{3}\) and \(\frac{4\pi}{3}\). The region outside the inner loop and inside the outer loop corresponds to the range \(\theta \in [\frac{2\pi}{3}, \frac{4\pi}{3}]\). Therefore, we integrate over this range as follows: $$ Area = \frac{1}{2}\int_{\frac{2\pi}{3}}^{\frac{4\pi}{3}} (1+2\cos \theta)^2 d\theta $$ Now, we can simplify and solve the integral as in part a to find the area. The value of the area will be \(2\pi\).

The area inside the inner loop corresponds to the complement of the area found in part b with respect to the complete area of the limaçon. Thus, our limits of integration are from \(\alpha=0\) to \(\beta=2\pi\). The total area of the limaçon can be found by integrating over this range: $$ TotalArea = \frac{1}{2}\int_{0}^{2\pi} (1+2\cos \theta)^2 d\theta $$ After simplifying and solving this integral, we find that the total area is \(5\pi\). To find the area inside the inner loop, subtract the area outside the inner loop and inside the outer loop from the total area: $$ Area_{innerloop} = TotalArea - Area_{outside\_innerloop} \\ Area_{innerloop} = 5\pi - 2\pi = 3\pi $$ Thus, the area of the region inside the inner loop of the limaçon \(r=1+2\cos \theta\) is \(3\pi\).