Area and Arc Length in Polar Coordinates: Learn It 2

Giselle Miller

Area and Arc Length in Polar Coordinates: Learn It 2

Arc Length in Polar Curves

When you need to find the length of a curve defined in polar coordinates, you can adapt the familiar arc length formula from rectangular coordinates.

Rectangular Arc Length: For a parameterized curve [latex](x(t), y(t))[/latex], arc length is [latex]L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt[/latex]

To derive the polar formula, we start with the conversion equations:

[latex]x=r\cos\theta =f\left(\theta \right)\cos\theta \:\:\text{and} \:\:y=r\sin\theta =f\left(\theta \right)\sin\theta[/latex],

Taking derivatives with respect to [latex]\theta[/latex]:

[latex]\begin{array}{c}\frac{dx}{d\theta }={f}^{\prime }\left(\theta \right)\cos\theta -f\left(\theta \right)\sin\theta \hfill \\ \frac{dy}{d\theta }={f}^{\prime }\left(\theta \right)\sin\theta +f\left(\theta \right)\cos\theta .\hfill \end{array}[/latex]

Now we substitute these into the arc length formula, replacing the parameter [latex]t[/latex] with [latex]\theta[/latex]:

[latex]\begin{align} L &= \int_{\alpha}^{\beta} \sqrt{\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2} d\theta \\ &= \int_{\alpha}^{\beta} \sqrt{\left(f'(\theta)\cos\theta - f(\theta)\sin\theta\right)^2 + \left(f'(\theta)\sin\theta + f(\theta)\cos\theta\right)^2} d\theta \end{align}[/latex]

Expanding the terms under the square root and using the identity [latex]\cos^2\theta + \sin^2\theta = 1[/latex]:

[latex]\begin{align} L &= \int_{\alpha}^{\beta} \sqrt{[f'(\theta)]^2(\cos^2\theta + \sin^2\theta) + [f(\theta)]^2(\cos^2\theta + \sin^2\theta)} d\theta \\ &= \int_{\alpha}^{\beta} \sqrt{[f'(\theta)]^2 + [f(\theta)]^2} d\theta \\ &= \int_{\alpha}^{\beta} \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} d\theta \end{align}[/latex]

This gives us the following theorem.

arc length of a curve defined by a polar function

Let [latex]f[/latex] be a function whose derivative is continuous on [latex][\alpha, \beta][/latex]. The length of the curve [latex]r = f(\theta)[/latex] from [latex]\theta = \alpha[/latex] to [latex]\theta = \beta[/latex] is:

[latex]L={\displaystyle\int }_{\alpha }^{\beta }\sqrt{{\left[f\left(\theta \right)\right]}^{2}+{\left[{f}^{\prime }\left(\theta \right)\right]}^{2}}d\theta ={\displaystyle\int }_{\alpha }^{\beta }\sqrt{{r}^{2}+{\left(\frac{dr}{d\theta }\right)}^{2}}d\theta[/latex].

Find the arc length of the cardioid [latex]r=2+2\cos\theta[/latex].

When [latex]\theta =0,r=2+2\cos0=4[/latex]. Furthermore, as [latex]\theta[/latex] goes from [latex]0[/latex] to [latex]2\pi[/latex], the cardioid is traced out exactly once. Therefore these are the limits of integration. Using [latex]f\left(\theta \right)=2+2\cos\theta[/latex], [latex]\alpha =0[/latex], and [latex]\beta =2\pi[/latex], the theorem equation becomes

[latex]\begin{array}{cc}\hfill L& ={\displaystyle\int }_{\alpha }^{\beta }\sqrt{{\left[f\left(\theta \right)\right]}^{2}+{\left[{f}^{\prime }\left(\theta \right)\right]}^{2}}d\theta \hfill \\ & ={\displaystyle\int }_{0}^{2\pi }\sqrt{{\left[2+2\cos\theta \right]}^{2}+{\left[-2\sin\theta \right]}^{2}}d\theta \hfill \\ & ={\displaystyle\int }_{0}^{2\pi}\sqrt{4+8\cos\theta +4{\cos}^{2}\theta +4{\sin}^{2}\theta }d\theta \hfill \\ & ={\displaystyle\int }_{0}^{2\pi }\sqrt{4+8\cos\theta +4\left({\cos}^{2}\theta +{\sin}^{2}\theta \right)}d\theta \hfill \\ & ={\displaystyle\int }_{0}^{2\pi }\sqrt{8+8\cos\theta }d\theta \hfill \\ & =2{\displaystyle\int }_{0}^{2\pi }\sqrt{2+2\cos\theta }d\theta .\hfill \end{array}[/latex]

Next, using the identity [latex]\cos\left(2\alpha \right)=2{\cos}^{2}\alpha -1[/latex], add 1 to both sides and multiply by 2. This gives [latex]2+2\cos\left(2\alpha \right)=4{\cos}^{2}\alpha[/latex]. Substituting [latex]\alpha =\frac{\theta}{2}[/latex] gives [latex]2+2\cos\theta =4{\cos}^{2}\left(\frac{\theta}{2}\right)[/latex], so the integral becomes

[latex]\begin{array}{cc}\hfill L& =2{\displaystyle\int }_{0}^{2\pi }\sqrt{2+2\cos\theta }d\theta \hfill \\ & =2{\displaystyle\int }_{0}^{2\pi }\sqrt{4{\cos}^{2}\left(\frac{\theta }{2}\right)}d\theta \hfill \\ & =2{\displaystyle\int }_{0}^{2\pi }2|\cos\left(\frac{\theta }{2}\right)|d\theta .\hfill \end{array}[/latex]

The absolute value is necessary because the cosine is negative for some values in its domain. To resolve this issue, change the limits from [latex]0[/latex] to [latex]\pi[/latex] and double the answer. This strategy works because cosine is positive between [latex]0[/latex] and [latex]\frac{\pi }{2}[/latex]. Thus,

[latex]\begin{array}{cc}\hfill L& =4{\displaystyle\int }_{0}^{2\pi }|\cos\left(\frac{\theta }{2}\right)|d\theta \hfill \\ & =8{\displaystyle\int }_{0}^{\pi }\cos\left(\frac{\theta }{2}\right)d\theta \hfill \\ & =8{\left(2\sin\left(\frac{\theta }{2}\right)\right)}_{0}^{\pi }\hfill \\ & =16.\hfill \end{array}[/latex]

Watch the following video to see the worked solution to the example above.

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