find the area of the shaded region. r = 6 + 5 cos(θ) r = sin(2θ) need help? read it submit answer

Explanation:

Step1: Recall area - formula in polar coordinates

The area $A$ of a polar - curve $r = f(\theta)$ from $\theta=a$ to $\theta = b$ is given by $A=\frac{1}{2}\int_{a}^{b}r^{2}d\theta$.

Step2: Determine the area of the outer - curve

The outer - curve is $r_1 = 6 + 5\cos\theta$. The area $A_1$ of the region bounded by $r_1$ is $A_1=\frac{1}{2}\int_{0}^{2\pi}(6 + 5\cos\theta)^{2}d\theta$. Expand $(6 + 5\cos\theta)^{2}=36+60\cos\theta + 25\cos^{2}\theta$. We know that $\cos^{2}\theta=\frac{1+\cos(2\theta)}{2}$. So, $A_1=\frac{1}{2}\int_{0}^{2\pi}(36 + 60\cos\theta+25\times\frac{1 + \cos(2\theta)}{2})d\theta$. $=\frac{1}{2}\int_{0}^{2\pi}(36+60\cos\theta+\frac{25}{2}+\frac{25\cos(2\theta)}{2})d\theta$. $=\frac{1}{2}\int_{0}^{2\pi}(\frac{72 + 25}{2}+60\cos\theta+\frac{25\cos(2\theta)}{2})d\theta$. $=\frac{1}{2}\int_{0}^{2\pi}(\frac{97}{2}+60\cos\theta+\frac{25\cos(2\theta)}{2})d\theta$. Integrating term - by - term: $\int_{0}^{2\pi}\frac{97}{2}d\theta=\frac{97}{2}\theta\big|{0}^{2\pi}=97\pi$, $\int{0}^{2\pi}60\cos\theta d\theta=60\sin\theta\big|{0}^{2\pi}=0$, $\int{0}^{2\pi}\frac{25\cos(2\theta)}{2}d\theta=\frac{25}{4}\sin(2\theta)\big|_{0}^{2\pi}=0$. So, $A_1=\frac{1}{2}(97\pi)= \frac{97\pi}{2}$.

Step3: Determine the area of the inner - curve

The inner - curve is $r_2=\sin(2\theta)$. The area $A_2$ of the region bounded by $r_2$ is $A_2=\frac{1}{2}\int_{0}^{2\pi}\sin^{2}(2\theta)d\theta$. Since $\sin^{2}(2\theta)=\frac{1-\cos(4\theta)}{2}$, then $A_2=\frac{1}{2}\int_{0}^{2\pi}\frac{1-\cos(4\theta)}{2}d\theta$. $=\frac{1}{4}\int_{0}^{2\pi}(1-\cos(4\theta))d\theta$. Integrating term - by - term: $\int_{0}^{2\pi}1d\theta=\theta\big|{0}^{2\pi}=2\pi$, $\int{0}^{2\pi}\cos(4\theta)d\theta=\frac{1}{4}\sin(4\theta)\big|_{0}^{2\pi}=0$. So, $A_2=\frac{1}{4}(2\pi)=\frac{\pi}{2}$.

Step4: Calculate the area of the shaded region

The area of the shaded region $A = A_1 - A_2$. $A=\frac{97\pi}{2}-\frac{\pi}{2}=48\pi$.

Answer:

$48\pi$