find the area of the region that lies inside the first curve and outside the second curve.\n$r = 4…

Explanation:

Step1: Find intersection points

Set $4 - 4\sin(\theta)=4$. Then $- 4\sin(\theta)=0$, so $\sin(\theta)=0$. The solutions in the range $[0,2\pi]$ are $\theta = 0$ and $\theta=\pi$.

Step2: Use the polar - area formula

The area formula for a polar region is $A=\frac{1}{2}\int_{\alpha}^{\beta}(r_1^{2}-r_2^{2})d\theta$, where $r_1$ is the outer - curve and $r_2$ is the inner - curve. Here, $r_1 = 4 - 4\sin(\theta)$ and $r_2 = 4$, and $\alpha = 0$, $\beta=\pi$. [ \begin{align*} A&=\frac{1}{2}\int_{0}^{\pi}((4 - 4\sin(\theta))^{2}-4^{2})d\theta\ &=\frac{1}{2}\int_{0}^{\pi}(16 - 32\sin(\theta)+16\sin^{2}(\theta)-16)d\theta\ &=\frac{1}{2}\int_{0}^{\pi}(- 32\sin(\theta)+16\sin^{2}(\theta))d\theta \end{align*} ] Recall that $\sin^{2}(\theta)=\frac{1 - \cos(2\theta)}{2}$. [ \begin{align*} A&=\frac{1}{2}\int_{0}^{\pi}(-32\sin(\theta)+16\times\frac{1 - \cos(2\theta)}{2})d\theta\ &=\frac{1}{2}\int_{0}^{\pi}(-32\sin(\theta)+8 - 8\cos(2\theta))d\theta\ &=\frac{1}{2}\left[32\cos(\theta)+8\theta-4\sin(2\theta)\right]_{0}^{\pi} \end{align*} ]

Step3: Evaluate the definite integral

[ \begin{align*} &\frac{1}{2}\left[(32\cos(\pi)+8\pi-4\sin(2\pi))-(32\cos(0)+8\times0 - 4\sin(0))\right]\ &=\frac{1}{2}\left[(-32 + 8\pi-0)-(32+0 - 0)\right]\ &=\frac{1}{2}\left(-32 + 8\pi-32\right)\ &=4\pi-32 \end{align*} ] Since area cannot be negative, we take the absolute value. The area $A = 32 - 4\pi$.

Answer:

$32 - 4\pi$