Dizzy Eye

In polar coordination, length of a curve given $r=f(\theta)$ from $a$ to $b$ is

$\int_{a}^{b} \sqrt{{f(\theta)}^2+{f'(\theta)}^2} d\theta$ .

In this case, $r=\theta$ . So the length of this curve from $0$ to $2\pi$ is

$\int_{0}^{2\pi} \sqrt{{\theta}^2+1} d\theta$ .

Evaluating $\int \sqrt{x^2+1} dx$ lies ahead:

Let $x=\tan{t}$ . Then $\int \sqrt{x^2+1} dx=\int \sec^3{t} dt$ .

$\sec^3{t}=\tan^2{t}\sec{t}+\sec{t}$

$\int \tan^2{t}\sec{t} dt=\int \sec^3{t} dt-\int \sec{t} dt=\tan{t}\sec{t}-\int \tan^2{t}\sec{t}-\ln|{\tan{t}+\sec{t}}|$

$\int\tan^2{t}\sec{t} dt=\dfrac{1}{2}\tan{t}\sec{t}-\dfrac{1}{2}\ln{|\tan{t}+\sec{t}}|$

So, $\int \sec^3{t}=\dfrac{1}{2}\tan{t}\sec{t}+\dfrac{1}{2}\ln{|\tan{t}+\sec{t}}|$ .

And by $\tan{t}=x$ , $\int \sqrt{x^2+1} dx=\dfrac{1}{2}x\sqrt{x^2+1}+\dfrac{1}{2}\ln{|x+\sqrt{x^2+1}}|+C$ ,

where $C$ is an intergal constant.

Finally, the length of the curve above is $\pi\sqrt{4\pi^2+1}+\dfrac{1}{2}\ln{|2\pi+\sqrt{4\pi^2+1}}|=k$ .

The value of $\lfloor k \rfloor =\boxed{21}$ .