let r be the region inside the polar curve r = 4 sin(6θ) and outside the polar curve r = 1 as shown in the…

Answer:

11.781

Explanation:

Step1: Recall area formula in polar - coordinates

The area between two polar curves $r_1(\theta)$ and $r_2(\theta)$ is given by $A=\frac{1}{2}\int_{\alpha}^{\beta}(r_1^{2}(\theta)-r_2^{2}(\theta))d\theta$, where $r_1(\theta)\geq r_2(\theta)$. Here, $r_1 = 4\sin(6\theta)$ and $r_2 = 1$.

Step2: Find the intersection points

Set $4\sin(6\theta)=1$, then $\sin(6\theta)=\frac{1}{4}$. Solving for $\theta$, we get $6\theta=\arcsin(\frac{1}{4}) + 2k\pi$ or $6\theta=\pi-\arcsin(\frac{1}{4})+2k\pi$, $k\in\mathbb{Z}$. For one - petal, we can find the appropriate $\alpha$ and $\beta$. Since the curve $r = 4\sin(6\theta)$ is symmetric, we can find the area of one petal and multiply by 12 (the number of petals). For one petal, we find the limits of integration by solving $\sin(6\theta)=\frac{1}{4}$. Let $\alpha=\frac{1}{6}\arcsin(\frac{1}{4})$ and $\beta=\frac{1}{6}(\pi - \arcsin(\frac{1}{4}))$.

Step3: Calculate the area of one petal

$A_1=\frac{1}{2}\int_{\alpha}^{\beta}((4\sin(6\theta))^{2}-1^{2})d\theta=\frac{1}{2}\int_{\alpha}^{\beta}(16\sin^{2}(6\theta)-1)d\theta$. Using the identity $\sin^{2}x=\frac{1 - \cos(2x)}{2}$, we have $\sin^{2}(6\theta)=\frac{1-\cos(12\theta)}{2}$. So $A_1=\frac{1}{2}\int_{\alpha}^{\beta}(16\times\frac{1 - \cos(12\theta)}{2}-1)d\theta=\frac{1}{2}\int_{\alpha}^{\beta}(8 - 8\cos(12\theta)-1)d\theta=\frac{1}{2}\int_{\alpha}^{\beta}(7 - 8\cos(12\theta))d\theta$. Integrating term - by - term: $\int(7 - 8\cos(12\theta))d\theta=7\theta-\frac{8}{12}\sin(12\theta)+C$. Evaluating the definite integral: $A_1=\frac{1}{2}\left[7\theta-\frac{2}{3}\sin(12\theta)\right]_{\frac{1}{6}\arcsin(\frac{1}{4})}^{\frac{1}{6}(\pi - \arcsin(\frac{1}{4}))}$.

Step4: Calculate the total area

The total area $A = 12A_1$. Using a calculator to evaluate the definite integral and multiplying by 12, we get $A\approx11.781$.