find the area (a) that is bounded by the curve (r = 3cos(\theta)) and lies in the sector (-\frac{pi}{4}<\thet…

Explanation:

Step1: Recall area formula in polar coordinates

The area formula for a polar - curve $r = f(\theta)$ between $\theta=\alpha$ and $\theta = \beta$ is $A=\frac{1}{2}\int_{\alpha}^{\beta}r^{2}d\theta$. Here, $r = 3\cos(\theta)$, $\alpha=-\frac{\pi}{4}$, and $\beta=\frac{\pi}{3}$.

Step2: Substitute $r$ into the formula

We get $A=\frac{1}{2}\int_{-\frac{\pi}{4}}^{\frac{\pi}{3}}(3\cos(\theta))^{2}d\theta=\frac{9}{2}\int_{-\frac{\pi}{4}}^{\frac{\pi}{3}}\cos^{2}(\theta)d\theta$.

Step3: Use the double - angle formula $\cos^{2}(\theta)=\frac{1 + \cos(2\theta)}{2}$

So, $A=\frac{9}{2}\int_{-\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{1+\cos(2\theta)}{2}d\theta=\frac{9}{4}\int_{-\frac{\pi}{4}}^{\frac{\pi}{3}}(1 + \cos(2\theta))d\theta$.

Step4: Integrate term - by - term

$\int(1+\cos(2\theta))d\theta=\int 1d\theta+\int\cos(2\theta)d\theta=\theta+\frac{1}{2}\sin(2\theta)+C$.

Step5: Evaluate the definite integral

$A=\frac{9}{4}\left[\theta+\frac{1}{2}\sin(2\theta)\right]_{-\frac{\pi}{4}}^{\frac{\pi}{3}}$. $A=\frac{9}{4}\left[\left(\frac{\pi}{3}+\frac{1}{2}\sin\left(\frac{2\pi}{3}\right)\right)-\left(-\frac{\pi}{4}+\frac{1}{2}\sin\left(-\frac{\pi}{2}\right)\right)\right]$. $A=\frac{9}{4}\left[\frac{\pi}{3}+\frac{\sqrt{3}}{4}+\frac{\pi}{4}-\left(-\frac{1}{2}\right)\right]$. $A=\frac{9}{4}\left[\frac{4\pi + 3\pi}{12}+\frac{\sqrt{3}}{4}+\frac{1}{2}\right]$. $A=\frac{9}{4}\left[\frac{7\pi}{12}+\frac{\sqrt{3}}{4}+\frac{1}{2}\right]$. $A=\frac{21\pi}{16}+\frac{9\sqrt{3}}{16}+\frac{9}{8}$.

Answer:

$\frac{21\pi + 9\sqrt{3}+18}{16}$