find the area of the region. interior of $r = 3cos(\theta)$

Explanation:

Step1: Recall area formula in polar coordinates

The area $A$ of a polar - curve $r = f(\theta)$ is given by $A=\frac{1}{2}\int_{\alpha}^{\beta}r^{2}d\theta$. For the circle $r = 3\cos\theta$, the range of $\theta$ for one - complete loop is from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$.

Step2: Substitute $r$ into the formula

Substitute $r = 3\cos\theta$ into the area formula: $A=\frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(3\cos\theta)^{2}d\theta$.

Step3: Simplify the integrand

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

Step4: Integrate term - by - term

$\frac{9}{4}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(1+\cos(2\theta))d\theta=\frac{9}{4}\left[\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}1d\theta+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos(2\theta)d\theta\right]$. The integral of $1$ with respect to $\theta$ is $\theta$, and $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}1d\theta=\left[\theta\right]{-\frac{\pi}{2}}^{\frac{\pi}{2}}=\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)=\pi$. For $\int{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos(2\theta)d\theta$, let $u = 2\theta$, $du=2d\theta$. When $\theta=-\frac{\pi}{2}$, $u =-\pi$; when $\theta=\frac{\pi}{2}$, $u=\pi$. $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos(2\theta)d\theta=\frac{1}{2}\int_{-\pi}^{\pi}\cos(u)du=\frac{1}{2}[\sin(u)]_{-\pi}^{\pi}=0$.

Step5: Calculate the area

$A=\frac{9}{4}(\pi + 0)=\frac{9\pi}{4}$.

Answer:

$\frac{9\pi}{4}$