# Length of the Polar Curve

Find the length of the curve $$r=\sqrt{1 + \cos (2\theta)}, 0\le\theta\le\frac{\pi\sqrt{2}}{4}$$.

I generated the integral: $$\int_0^{\frac{\pi\sqrt{2}}{4}} {\sqrt{2 - \cos (2\theta)}}\text{d}\theta$$

Is it correct? How would I solve this integral?

The length of the curve $$=\frac{\pi}{2}$$.

We know that $$x=r\cos \theta=\sqrt{1+\cos 2\theta}\cos \theta=\sqrt{2} \cos^2 \theta$$
and $$y=r\sin \theta=\sqrt{1+\cos 2\theta}\sin \theta=\sqrt{2} \sin\theta \cos \theta$$

So $$\dfrac{dy}{dx}= \dfrac{dy}{d\theta}\cdot\dfrac{d\theta}{dx}=-\cot 2\theta$$

Therefore the required arc length is \begin{align} & \int_{{\pi\sqrt{2}}/{4}}^0 \sqrt{1+\Big(\dfrac{dy}{dx}\Big)^2}dx \\ &= \int_{{\pi\sqrt{2}}/{4}}^0 \sqrt{1+\cot^2 2\theta}\quad(-\sqrt2\sin2\theta) d\theta\\ &=-\sqrt2 \int_{{\pi\sqrt{2}}/{4}}^0 dx \\ &=\frac{\pi}{2} \end{align}

Arc length is given by $$L =\int \sqrt{r^2+r^{'2} }d \theta$$ Now $$r=\sqrt 2 \cos \theta;\,r'=-\sqrt 2 \sin \theta;\,\sqrt{r^2+r^{'2} }= \sqrt 2$$ $$L= \sqrt 2 \int_0^{\frac{\pi\sqrt{2}}{4} }d \theta =\sqrt 2\cdot {\frac{\pi\sqrt{2}}{4} }= {\frac{\pi}{2} }.$$

This is an arc of circle of diameter $$\sqrt 2$$ subtending $${\dfrac{\pi\sqrt{2}}{4} }$$ radians of angle at the origin and double that $$=\dfrac{\pi}{\sqrt{2}}$$ radians at circle center.

Note $$r=\sqrt{1 + \cos (2\theta)}=\sqrt2 \cos\theta$$, which is a circle with the radius $$\frac1{\sqrt2}$$ and the range $$0\le\theta\le\frac{\pi\sqrt{2}}{4}$$ spans a circle sector of angle $$\alpha=\frac{\pi\sqrt{2}}{2}$$. Thus, the circumference length of the sector is

$$\alpha r = \frac{\pi\sqrt{2}}{2}\cdot \frac1{\sqrt2}=\frac\pi2$$

• Shouldn't it be $\sqrt{1+ \cos (2\theta)}=\sqrt{2} \cos \theta$ for $0 \leq \theta \leq \frac{\pi \sqrt{2}}{4}$? – Dunkelheit Apr 29 '20 at 20:36
• @Dunkelheit correct thx for spotting it – Quanto Apr 29 '20 at 20:54