15. Polar Coordinates

e. Arc Length of a Polar Graph

In the chapter on Arc Length, we learned that the arc length of a parametric curve, \(\vec{r}(t)=(x(t),y(t))\), from \(t=a\) to \(t=b\) is: \[ L=\int_a^b \sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\,dt \] We also know that any polar curve, \(r=r(\theta)\), can be regarded as the parametric curve \[ \vec{r}(\theta)=(x(\theta),y(\theta)) =(r(\theta)\cos\theta,r(\theta)\sin\theta) \] where the parameter \(t\) has been replaced by a parameter \(\theta\). Note on \(r(\theta)\) vs. \(\vec r(\theta)\)

Consequently:

The arc length of a polar curve, \(r=r(\theta)\), from \(\theta=\alpha\) to \(\theta=\beta\) is \[ L=\int_\alpha^\beta \sqrt{\,r(\theta)^2+\left(\dfrac{dr}{d\theta}\right)^2}\,d\theta \] Proof

You may have noticed that the cardioids have been especially easy to work with. Here, however, we encounter some complications. We first set up the integral: Since \(r=1+\cos\theta\) we have \(\dfrac{dr}{d\theta}=-\sin\theta\). Since the cardioid is swept out once between \(\theta=0\) and \(\theta=2\pi\), the arc length is: \[\begin{aligned} L &=\int_0^{2\pi} \sqrt{(1+\cos\theta)^2+\sin^2\theta }\,d\theta \\ &=\int_0^{2\pi} \sqrt{1+2\cos\theta+\cos^2\theta+\sin^2\theta}\,d\theta \\ &=\int_0^{2\pi} \sqrt{2+2\cos\theta}\,d\theta =\sqrt{2}\int_0^{2\pi} \sqrt{1+\cos\theta}\,d\theta \end{aligned}\] As with all arclength problems, there is some trick which allows us to compute the integral.

One trick is to multiply by \(\dfrac{\sqrt{1-\cos\theta}}{\sqrt{1-\cos\theta}}\) \[\begin{aligned} L &=\sqrt{2}\int_0^{2\pi} \sqrt{1+\cos\theta} \dfrac{\sqrt{1-\cos\theta}}{\sqrt{1-\cos\theta}}\,d\theta \\ &=\sqrt{2}\int_0^{2\pi} \dfrac{\sqrt{1-\cos^2\theta}}{\sqrt{1-\cos\theta}}\,d\theta =\sqrt{2}\int_0^{2\pi} \dfrac{\sqrt{\sin^2\theta}}{\sqrt{1-\cos\theta}}\,d\theta \end{aligned}\] We now hit a second problem: \(\sqrt{\sin^2\theta}=|\sin\theta|\), not \(\sin\theta\). So either we need to split up the integral according to whether \(\sin\theta \gt 0\) or \(\sin\theta \lt 0\) or we need to take advantage of the symmetry of the cardioid by finding the length of half the cardioid and doubling it. We do the latter: The upper half of the cardioid corresponds to \(0 \le \theta \le \pi\). On this interval, \(\sin\theta \ge 0\), so that \(\sqrt{\sin^2\theta}=\sin\theta\). Consequently, the arc length is: \[ L=2\sqrt{2}\int_0^\pi \dfrac{\sin\theta}{\sqrt{1-\cos\theta}}\,d\theta \] Now we can integrate by making the substitution \(u=1-\cos\theta\), so that \(du=\sin\theta\,d\theta\). Consequently; \[\begin{aligned} L &=2\sqrt{2}\int_0^2 \dfrac{du}{\sqrt{u}} =2\sqrt{2}\left[2u^{1/2}\rule{0pt}{10pt}\right]_0^2 \\ &=2\sqrt{2}\left[2\sqrt{2}\right]=8 \end{aligned}\]

A second trick is to apply the identity \(\cos^2A=\dfrac{1+\cos(2A)}{2}\) with \(A=\dfrac{\theta}{2}\). Then \(2\cos^2\dfrac{\theta}{2}=1+\cos\theta\). So: \[\begin{aligned} L&=\sqrt{2}\int_0^{2\pi} \sqrt{1+\cos\theta}\,d\theta =\sqrt{2}\int_0^{2\pi} \sqrt{2\cos^2\dfrac{\theta}{2}}\,d\theta \\ &=2\int_0^{2\pi} \sqrt{\cos^2\dfrac{\theta}{2}}\,d\theta \end{aligned}\] Again we need to compute the arclength for half of the curve and double it. For \(0 \le \theta \le \pi\), we have \(0 \le \dfrac{\theta}{2} \le \dfrac{\pi}{2}\), and \(\cos\dfrac{\theta}{2} \ge 0\). So \(\sqrt{\cos^2\dfrac{\theta}{2}}=\cos\dfrac{\theta}{2}\). Consequently, the arc length is: \[ L=4\int_0^\pi \cos\dfrac{\theta}{2}\,d\theta =4\left[2\sin\dfrac{\theta}{2}\right]_0^\pi =8\sin\dfrac{\pi}{2}=8 \]