11. 0/1 points details my notes scalcet9 10.4.027. find the area of the region that lies inside the first…

11. 0/1 points details my notes scalcet9 10.4.027. find the area of the region that lies inside the first curve and outside the second curve. r = 3 cos(θ), r = 1 + cos(θ) enhanced feedback please try again, keeping in mind that you should first find the intersection points. then use the formula for the area between two parametric curves, which is a = ∫a to b 1/2(f(θ)² - g(θ)²) dθ, where f(θ) ≥ g(θ) ≥ 0. need help? read it watch it submit answer

11. 0/1 points details my notes scalcet9 10.4.027. find the area of the region that lies inside the first curve and outside the second curve. r = 3 cos(θ), r = 1 + cos(θ) enhanced feedback please try again, keeping in mind that you should first find the intersection points. then use the formula for the area between two parametric curves, which is a = ∫a to b 1/2(f(θ)² - g(θ)²) dθ, where f(θ) ≥ g(θ) ≥ 0. need help? read it watch it submit answer

Answer

Explanation:

Step1: Find intersection points

Set $3\cos\theta=1 + \cos\theta$. $3\cos\theta-\cos\theta=1$, so $2\cos\theta=1$, then $\cos\theta=\frac{1}{2}$. The solutions in $[0, 2\pi]$ are $\theta=\pm\frac{\pi}{3}$.

Step2: Apply area - between - curves formula

The area formula for the region between two polar curves $r = f(\theta)$ and $r = g(\theta)$ with $f(\theta)\geq g(\theta)$ is $A=\frac{1}{2}\int_{a}^{b}[f(\theta)^{2}-g(\theta)^{2}]d\theta$. Here $f(\theta)=3\cos\theta$ and $g(\theta)=1 + \cos\theta$, and $a =-\frac{\pi}{3}$, $b=\frac{\pi}{3}$. [ \begin{align*} A&=\frac{1}{2}\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}[(3\cos\theta)^{2}-(1 + \cos\theta)^{2}]d\theta\ &=\frac{1}{2}\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}(9\cos^{2}\theta-(1 + 2\cos\theta+\cos^{2}\theta))d\theta\ &=\frac{1}{2}\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}(9\cos^{2}\theta - 1-2\cos\theta-\cos^{2}\theta)d\theta\ &=\frac{1}{2}\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}(8\cos^{2}\theta-2\cos\theta - 1)d\theta \end{align*} ] Since $\cos^{2}\theta=\frac{1 + \cos(2\theta)}{2}$, we have: [ \begin{align*} A&=\frac{1}{2}\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}(8\times\frac{1+\cos(2\theta)}{2}-2\cos\theta - 1)d\theta\ &=\frac{1}{2}\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}(4 + 4\cos(2\theta)-2\cos\theta - 1)d\theta\ &=\frac{1}{2}\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}(3 + 4\cos(2\theta)-2\cos\theta)d\theta \end{align*} ] Integrating term - by - term: $\int 3d\theta=3\theta$, $\int 4\cos(2\theta)d\theta = 2\sin(2\theta)$, $\int - 2\cos\theta d\theta=-2\sin\theta$. [ \begin{align*} A&=\frac{1}{2}\left[3\theta+2\sin(2\theta)-2\sin\theta\right]_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\ &=\frac{1}{2}\left[\left(3\times\frac{\pi}{3}+2\sin\left(2\times\frac{\pi}{3}\right)-2\sin\frac{\pi}{3}\right)-\left(3\times\left(-\frac{\pi}{3}\right)+2\sin\left(2\times\left(-\frac{\pi}{3}\right)\right)-2\sin\left(-\frac{\pi}{3}\right)\right)\right]\ &=\frac{1}{2}\left[\left(\pi + 2\times\frac{\sqrt{3}}{2}-2\times\frac{\sqrt{3}}{2}\right)-\left(-\pi+2\times\left(-\frac{\sqrt{3}}{2}\right)-2\times\left(-\frac{\sqrt{3}}{2}\right)\right)\right]\ &=\frac{1}{2}(\pi-(-\pi))\ &=\pi \end{align*} ]

Answer:

$\pi$