Area and Arc Length in Polar Coordinates
The Essentials
Through polar coordinates, we can find the area bounded by a curve and the arc length of a curve as long as the curve is continuous and not negative on the interval.
The area bounded by a curve is denoted by...
\[
A=\frac{1}{2}\int_{\alpha}^{\beta}(f(\theta))^2d\theta = \frac{1}{2}\int_{\alpha}^{\beta}r^2d\theta
\]
The arc length of a curve is denoted by...
\[
L=\int_{\alpha}^{\beta}\sqrt{(f(\theta))^2+(f'(\theta))^2}d\theta=\int_{\alpha}^{\beta}\sqrt{r^2+(\frac{dr}{d\theta})^2}d\theta
\]
Example
Find the area bounded by the curve $r=6cos(2\theta)$ on the interval, $\theta = [0,\pi]$.
-
Plug function into the equation
\[ \frac{1}{2}\int_{\alpha}^{\beta}r^2d\theta = \frac{1}{2}\int_{0}^{\pi}(6cos(2\theta)^2d\theta \]
-
Use the power reduction trig identity to simplify the integral
\[ cos^2(\theta)=\frac{1+cos(2\theta)}{2} \]\[ \frac{1}{2}\int_{0}^{\pi}(36(\frac{1+cos(4\theta)}{2}))d\theta \]
-
Solve the integral
\[ 9\int_0^{\pi}\frac{1}{2}+\frac{1}{2}cos(4\theta)d\theta \]\[ 9[\frac{1}{2}\theta|_0^{\pi}+\frac{1}{8}sin(4\theta)|_0^{\pi}] \]
- Solution: \( \frac{9\pi}{2} \)
Find the arc length of the curve \( r=3sin(\theta) \) along the interval \( \theta=[\frac{\pi}{2},\pi] \)
-
Plug the function into the arc length equation
\[ \int_{\frac{\pi}{2}}^{\pi}\sqrt{(3sin(\theta))^2+(3cos(\theta))^2}d\theta \]
-
Use the Pythagorean identity of sines and cosines to simplify
\[ 1=cos^2(\theta)+sin^2(\theta) \]\[ \int_{\frac{\pi}{2}}^{\pi}\sqrt{9}d\theta \]
- Evaluate the integral
\[ \sqrt{9}\theta|_{\frac{\pi}{2}}^{\pi} \]\[ \sqrt{9}(\pi)-\sqrt{9}(\frac{\pi}{2}) \]
- Solution: 4.71
Practice
Evaluate these expressions:
- Find the area bounded by the curve \( r=3sin(\theta)+2cos(\theta) \) on the interval \( [\pi,2\pi] \)
- Find the arc length of the arc \( r=5cos(2\theta) \) on the interval \( [\frac{\pi}{4},\frac{3\pi}{4}] \)
Solutions:
- 20.42
- 12.11