find the area of the region that lies inside the first curve and outside the second curve.\nr = 3 cos(θ), r…

Explanation:

Step1: Find intersection points

Set (3\cos\theta = 1+\cos\theta). [ \begin{align*} 3\cos\theta-\cos\theta&=1\ 2\cos\theta&=1\ \cos\theta&=\frac{1}{2}\ \theta&=\pm\frac{\pi}{3} \end{align*} ]

Step2: Use polar - area formula (A=\frac{1}{2}\int_{\alpha}^{\beta}(r_{1}^{2}-r_{2}^{2})d\theta)

Here (r_{1} = 3\cos\theta), (r_{2}=1 + \cos\theta), (\alpha=-\frac{\pi}{3}), (\beta=\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}}(8\cos^{2}\theta-2\cos\theta - 1)d\theta \end{align*} ] Use (\cos^{2}\theta=\frac{1+\cos(2\theta)}{2}) [ \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*} ]

Step3: Integrate term - by - term

[ \begin{align*} \int(3 + 4\cos(2\theta)-2\cos\theta)d\theta&=3\theta+2\sin(2\theta)-2\sin\theta+C \end{align*} ] Evaluate from (-\frac{\pi}{3}) to (\frac{\pi}{3}) [ \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(\frac{2\pi}{3}\right)-2\sin\left(\frac{\pi}{3}\right)\right)-\left(3\times\left(-\frac{\pi}{3}\right)+2\sin\left(-\frac{2\pi}{3}\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)