10. -/1 points details my notes scalcet9 10.4.026. find the area of the region that lies inside the first…

Explanation:

Step1: Find intersection points

Set $1 + \cos(\theta)=2 - \cos(\theta)$. Then $2\cos(\theta)=1$, so $\cos(\theta)=\frac{1}{2}$, and $\theta =-\frac{\pi}{3},\frac{\pi}{3}$.

Step2: Use area - formula for polar curves

The area $A$ between two polar curves $r = f(\theta)$ and $r = g(\theta)$ is given by $A=\frac{1}{2}\int_{\alpha}^{\beta}(f^{2}(\theta)-g^{2}(\theta))d\theta$, where $f(\theta)$ is the outer - curve and $g(\theta)$ is the inner - curve. Here, $f(\theta)=1 + \cos(\theta)$ and $g(\theta)=2 - \cos(\theta)$, and $\alpha=-\frac{\pi}{3},\beta=\frac{\pi}{3}$. [ \begin{align*} A&=\frac{1}{2}\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}((1 + \cos(\theta))^{2}-(2 - \cos(\theta))^{2})d\theta\ &=\frac{1}{2}\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}(1 + 2\cos(\theta)+\cos^{2}(\theta)-(4 - 4\cos(\theta)+\cos^{2}(\theta)))d\theta\ &=\frac{1}{2}\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}(1 + 2\cos(\theta)+\cos^{2}(\theta)-4 + 4\cos(\theta)-\cos^{2}(\theta))d\theta\ &=\frac{1}{2}\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}(6\cos(\theta)-3)d\theta \end{align*} ]

Step3: Integrate

We know that $\int\cos(\theta)d\theta=\sin(\theta)$ and $\int 3d\theta = 3\theta$. [ \begin{align*} \frac{1}{2}\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}(6\cos(\theta)-3)d\theta&=\frac{1}{2}\left[6\sin(\theta)-3\theta\right]_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\ &=\frac{1}{2}\left[\left(6\sin\left(\frac{\pi}{3}\right)-3\times\frac{\pi}{3}\right)-\left(6\sin\left(-\frac{\pi}{3}\right)-3\times\left(-\frac{\pi}{3}\right)\right)\right]\ &=\frac{1}{2}\left[\left(6\times\frac{\sqrt{3}}{2}-\pi\right)-\left(-6\times\frac{\sqrt{3}}{2}+\pi\right)\right]\ &=\frac{1}{2}\left[(3\sqrt{3}-\pi)-(- 3\sqrt{3}+\pi)\right]\ &=\frac{1}{2}(6\sqrt{3}-2\pi)\ &=3\sqrt{3}-\pi \end{align*} ]

Answer:

$3\sqrt{3}-\pi$