find the area of the region that lies inside the first curve and outside the second curve.\nr = 5…

Explanation:

Step1: Find intersection points

Set (5 - 5\sin\theta=5), then (- 5\sin\theta=0), so (\sin\theta = 0), (\theta=0) or (\theta=\pi).

Step2: Use the formula for the area in polar coordinates

The formula for the area (A=\frac{1}{2}\int_{\alpha}^{\beta}(r_{1}^{2}-r_{2}^{2})d\theta), where (r_{1}=5 - 5\sin\theta), (r_{2} = 5), (\alpha = 0), (\beta=\pi). [ \begin{align*} A&=\frac{1}{2}\int_{0}^{\pi}[(5 - 5\sin\theta)^{2}-5^{2}]d\theta\ &=\frac{1}{2}\int_{0}^{\pi}(25-50\sin\theta + 25\sin^{2}\theta-25)d\theta\ &=\frac{1}{2}\int_{0}^{\pi}(-50\sin\theta + 25\sin^{2}\theta)d\theta \end{align*} ] Use (\sin^{2}\theta=\frac{1-\cos(2\theta)}{2}) [ \begin{align*} A&=\frac{1}{2}\int_{0}^{\pi}(-50\sin\theta+\frac{25(1 - \cos(2\theta))}{2})d\theta\ &=\frac{1}{2}\left[50\cos\theta+\frac{25}{2}(\theta-\frac{\sin(2\theta)}{2})\right]_{0}^{\pi}\ &=\frac{1}{2}\left[(50\cos\pi+\frac{25}{2}(\pi-\frac{\sin(2\pi)}{2}))-(50\cos0+\frac{25}{2}(0-\frac{\sin(0)}{2}))\right]\ &=\frac{1}{2}\left[(- 50+\frac{25\pi}{2})-(50)\right]\ &=\frac{1}{2}\left(-100+\frac{25\pi}{2}\right)\ &=\frac{25\pi}{4}-50 \end{align*} ]

Answer:

(\frac{25\pi}{4}-50)