# 10.5 | Area and Arc Length in Polar Coordinates

Let $$f$$ and $$g$$ be continuous on $$[\alpha, \beta]$$, where $$0 \leq g(\theta) \leq f(\theta)$$ and $$0 \leq \beta – \alpha < 2 \pi$$. Then the area $$A$$ of the region bounded by the graphs of $$r = g(\theta), r = f(\theta), \theta = \alpha,$$ and $$\theta = \beta$$ is given by $$A = \frac{1}{2} \int_{\alpha}^{\beta} [f(\theta)]^2 - [g(\theta)]^2 \; d \theta.$$
Let $$f$$ be a function with a continuous derivative on an interval $$[\alpha, \beta]$$. If the graph $$C$$ or $$r = f(\theta)$$ is traced exactly once as $$\theta$$ increases from $$\alpha$$ to $$\beta$$, then the length $$L$$ of $$C$$ is given by
\begin{aligned} L & = \int_\alpha^\beta \sqrt{[f'(\theta)]^2 + [f(\theta)]^2} \; d \theta \\ & = \int_\alpha^\beta \sqrt{\left(\frac{dr}{d\theta}\right)^2 + r^2} \; d \theta \\ \end{aligned}
Let $$f$$ be a function with a continuous derivative on an interval $$[\alpha, \beta]$$. If the graph $$C$$ or $$r = f(\theta)$$ is traced exactly once as $$\theta$$ increases from $$\alpha$$ to $$\beta$$, then the area of the surface obtained by revolving $$C$$ about the polar axis is given by
$$S = 2 \pi \int_\alpha^\beta r \sin \theta \sqrt{\left(\frac{dr}{d\theta}\right)^2 + r^2} \; d \theta$$
and the area of the surface obtained by revolving $$C$$ about the line $$\theta = \pi/2$$ is given by
$$S = 2 \pi \int_\alpha^\beta r \cos \theta \sqrt{\left(\frac{dr}{d\theta}\right)^2 + r^2} \; d \theta$$
Find the area of the region inside the graph of $$r = 10 \sin \theta$$ and outside the graph of $$r = 2$$.

\begin{aligned} A &= \int_{\arcsin(1/5)}^{\pi-\arcsin(1/5)} \frac{1}{2}\left([f(\theta)]^2-[g(\theta)]^2 \right) d \theta \\ &= \int_{\arcsin(1/5)}^{\pi-\arcsin(1/5)} \frac{1}{2}\left([10 \sin \theta]^2-[2]^2 \right) d \theta \\ &= \int_{\arcsin(1/5)}^{\pi-\arcsin(1/5)} 50 \sin^2 \theta-2 d \theta \\ &= \int_{\arcsin(1/5)}^{\pi-\arcsin(1/5)} 50 \left(\frac{1- \cos(2 \theta)}{2} \right)-2 d \theta \\ &= \int_{\arcsin(1/5)}^{\pi-\arcsin(1/5)} 23-25\cos(2 \theta) d \theta \\ &= \left(23 \theta -25\frac{\sin(2 \theta)}{2}\right)\Big|_{\arcsin(1/5)}^{\pi-\arcsin(1/5)} \\ &= \left(23 \theta -25\frac{\sin(2 \theta)}{2}\right)\Big|_{\arcsin(1/5)}^{\pi-\arcsin(1/5)} \\ & = 4 \sqrt{6}+23 \pi -46 \sin ^{-1}\left(\frac{1}{5}\right) \\ & \approx 72.7921 \end{aligned}

Find the area of the region inside the graph of $$r = 2+2\cos \theta$$ and outside the graph of $$r = 3$$.

\begin{aligned} A &= \int_{\alpha}^{\beta} \frac{1}{2}\left([f(\theta)]^2-[g(\theta)]^2 \right) d \theta \\ &= \int_{-\pi/3}^{\pi/3} \frac{1}{2}\left([2(1+\cos \theta)]^2-[3]^2 \right) d \theta \\ &= 2\int_{0}^{\pi/3} \frac{1}{2}\left([2(1+\cos \theta)]^2-[3]^2 \right) d \theta \\ &= \int_{0}^{\pi/3} \left([2(1+\cos \theta)]^2-[3]^2 \right) d \theta \\ &= \int_{0}^{\pi/3} 4+8 \cos \theta + 4 \cos^2 \theta-9 d \theta \\ &= \int_{0}^{\pi/3} 8 \cos \theta + 4 \left(\frac{1+\cos(2 \theta)}{2}\right)-5 d \theta \\ &= \int_{0}^{\pi/3} 8 \cos \theta + 2+2\cos(2 \theta)-5 d \theta \\ &= \int_{0}^{\pi/3} 8 \cos \theta +2 \cos(2 \theta)-3 d \theta \\ &= \left( 8 \sin \theta +\sin(2 \theta)-3 \theta \right) \Big|_{0}^{\pi/3} \\ &= \frac{9 \sqrt{3}}{2} – \pi \end{aligned}

Find the area of the region which is inside the polar curve $$r = 4\cos \theta$$ and outside the curve $$r = 3-2 \cos \theta$$.

Find the area enclosed by the polar curve
$$r=5e^{0.6 \theta}$$
on the interval $$0 \leq \theta \leq 1$$.
and the straight line segment between its ends.