Question

WORTH 98 POINTS! PLEASE HELP! I will mark Brainliest and give stars if all steps are shown and the integral is included

Answer 1

Both curves complete one loop over the interval $0\le\theta\le2\pi$. See the attached plot. First find when the curves intersect in this interval:

$2-2\sin\theta=3\implies\sin\theta=-\dfrac12\implies\theta=\dfrac{7\pi}6,\theta=\dfrac{11\pi}6$

From the plot we can see that $r=2-2\sin\theta$ is "larger" than $r=3$, so the area is given by

$\displaystyle\int_{7\pi/6}^{11\pi/6}((2-2\sin\theta)-3)\,\mathrm d\theta=-\int_{7\pi/6}^{11\pi/6}(1+2\sin\theta)\,\mathrm d\theta$

$\displaystyle=\int_{11\pi/6}^{7\pi/6}(1+2\sin\theta)\,\mathrm d\theta$

$=(\theta-2\cos\theta)\bigg|_{\theta=11\pi/6}^{\theta=7\pi/6}$

$=\left(\dfrac{7\pi}6-2\cos\dfrac{7\pi}6\right)-\left(\dfrac{11\pi}6-2\cos\dfrac{11\pi}6\right)$

$=2\sqrt3-\dfrac{2\pi}3$