Areas and Lengths in Polar Coordinates

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 10.4 – Areas and Lengths in Polar Coordinates

Expected Educational Results

• Objective 17–01: I can set up the integral to compute the area bounded by one or two polar curves.

• Objective 17–02: I can evaluate the integral to compute the area bounded by one or two polar curves.

• Objective 17–03: I can set up the integral to compute the arclength of a polar curve.

• Objective 17–04: I can estimate the integral for the arclength of a polar curve.

Bloom’s Taxonomy

A modern version of Bloom’s Taxonomy is included here to recognize various different levels of understanding and to encourage you to work towards higher-order understanding (those at the top of the pyramid). All Objectives, Investigations, Activities, etc. are color-coded with the level of understanding.

Areas and Lengths in Polar Coordinates

Arclength

Definition: Length of a Polar Curve

${\displaystyle L={\int }_a^b\sqrt{r^2+(\frac{dr}{d\theta }{)}^2}d\theta }$$\displaystyle L=\int_a^b{\sqrt{r^2+\bigg(\frac{dr}{d\theta}\bigg)^2}\,d\theta}$

where $r=f(\theta )$$r=f(\theta)$.

NOTE: Due to the sum of two terms under the square root in the formula for arclength, the integral is a challenge to evaluate exactly. Thus, exact answers are expected where the exact value is the correct and complete set up of the integral.

Exact Arclength

Investigation 03

Find the following lengths:

1. The length of ${\displaystyle r=2(1+\cos (\theta ))}$$\displaystyle r=2\left(1+\cos{(\theta)}\right)$ on ${\displaystyle 0\le \theta \le 2\pi }$$\displaystyle 0\leq\theta\leq 2\pi$

2. The length of ${\displaystyle r=\sin (6\theta )}$$\displaystyle r=\sin{(6\theta)}$ on ${\displaystyle 0\le \theta \le \frac{\pi }{3}}$$\displaystyle 0\leq\theta\leq \frac{\pi}{3}$

3. The length of ${\displaystyle r={\sin }^2\left(\frac{\theta }{2}\right)}$$\displaystyle r=\sin^2{\!\left(\frac{\theta}{2}\right)}$ on ${\displaystyle 0\le \theta \le \pi }$$\displaystyle 0\leq\theta\leq \pi$

4. The length of ${\displaystyle r=\tan (\theta )}$$\displaystyle r=\tan{(\theta)}$ on ${\displaystyle \frac{\pi }{6}\le \theta \le \frac{\pi }{3}}$$\displaystyle \frac{\pi}{6}\leq\theta\leq \frac{\pi}{3}$

Approximate Arclength

When asked to estimate the arclength of a curve, you need to show the correct and complete set up of the integral, then use technology to find a decimal approximation of the arclength.

Use Technology to Approximate Arclength

Mathematica

1
(* Example from Investigation 03-1 *)
2
(* Find arclength of r=2(1+cos(theta)) on [0,2pi] *)
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Integrate[Sqrt[(2(1+Cos[t]))^2+(-2Sin[t])^2], {t,0,2Pi}]//N

Warnings:

1. Be very careful with the syntax. Syntax is the set of rules on how to write computer code. Every software program has its own unique syntax. Some basic Mathematica syntax is located at: http://www.jjw3.com/TECH_Common_Functions.pdf.

2. To execute code (including comment codes), press and hold the SHIFT key and press the ENTER key.

Investigation 04

Find the approximate lengths (to 3 decimal places) of the following:

1. The approximate length of ${\displaystyle r=2(1+\cos (\theta ))}$$\displaystyle r=2\left(1+\cos{(\theta)}\right)$ on ${\displaystyle 0\le \theta \le 2\pi }$$\displaystyle 0\leq\theta\leq 2\pi$

2. The approximate length of ${\displaystyle r=\sin (6\theta )}$$\displaystyle r=\sin{(6\theta)}$ on ${\displaystyle 0\le \theta \le \frac{\pi }{3}}$$\displaystyle 0\leq\theta\leq \frac{\pi}{3}$

3. The approximate length of ${\displaystyle r={\sin }^2\left(\frac{\theta }{2}\right)}$$\displaystyle r=\sin^2{\!\left(\frac{\theta}{2}\right)}$ on ${\displaystyle 0\le \theta \le \pi }$$\displaystyle 0\leq\theta\leq \pi$

4. The approximate length of ${\displaystyle r=\tan (\theta )}$$\displaystyle r=\tan{(\theta)}$ on ${\displaystyle \frac{\pi }{6}\le \theta \le \frac{\pi }{3}}$$\displaystyle \frac{\pi}{6}\leq\theta\leq \frac{\pi}{3}$

Homework

At this time, you should be able to complete the following assignments:

• Section 10.4: # 45, 47.

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Last Modified: Tuesday, 3 November 2020 7:04 EDT