areas inside polar curves\nfind the areas of the regions in exercises 2,3\n2. inside the cardioid $r =…

Explanation:

Step1: Recall area formula in polar coordinates

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

Step2: Solve for the cardioid $r = a(1+\cos\theta)$

For the cardioid $r = a(1 + \cos\theta)$, the curve is symmetric about the polar axis. We can find the area of the whole cardioid by integrating from $0$ to $2\pi$. [ \begin{align*} A&=\frac{1}{2}\int_{0}^{2\pi}[a(1 + \cos\theta)]^{2}d\theta\ &=\frac{a^{2}}{2}\int_{0}^{2\pi}(1 + 2\cos\theta+\cos^{2}\theta)d\theta\ &=\frac{a^{2}}{2}\int_{0}^{2\pi}\left(1 + 2\cos\theta+\frac{1+\cos2\theta}{2}\right)d\theta\ &=\frac{a^{2}}{2}\int_{0}^{2\pi}\left(\frac{2 + 4\cos\theta+1+\cos2\theta}{2}\right)d\theta\ &=\frac{a^{2}}{4}\int_{0}^{2\pi}(3 + 4\cos\theta+\cos2\theta)d\theta\ &=\frac{a^{2}}{4}\left[3\theta + 4\sin\theta+\frac{1}{2}\sin2\theta\right]_{0}^{2\pi}\ &=\frac{3}{2}\pi a^{2} \end{align*} ]

Step3: Solve for one - leaf of the four - leaved rose $r=\cos2\theta$

For the four - leaved rose $r = \cos2\theta$, one leaf is formed from $\theta=-\frac{\pi}{4}$ to $\theta=\frac{\pi}{4}$. [ \begin{align*} A&=\frac{1}{2}\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\cos^{2}2\theta d\theta\ &=\int_{0}^{\frac{\pi}{4}}\cos^{2}2\theta d\theta\ &=\int_{0}^{\frac{\pi}{4}}\frac{1 + \cos4\theta}{2}d\theta\ &=\frac{1}{2}\left[\theta+\frac{1}{4}\sin4\theta\right]_{0}^{\frac{\pi}{4}}\ &=\frac{\pi}{8} \end{align*} ]

Answer:

The area inside the cardioid $r = a(1+\cos\theta)$ is $\frac{3}{2}\pi a^{2}$, and the area inside one leaf of the four - leaved rose $r=\cos2\theta$ is $\frac{\pi}{8}$.