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.


E 10.5 Exercises