Sine on a Circle I'm walking a quarter mile circular walking track. The width of the track is 8 feet across. If I walk from one side of the track to the other, walking a sine wave that has a 20 foot period, how much further am I walking when I make a complete circuit?
 A: We have that we traverse along the sine wave from the point $x=0\:\text{ft}$ to the point $x=\frac{1}{4}\:\text{mi}= 1320\:\text{ft}$. We note that the equation for the sine wave is given by:
$$y(x) = \sin\left(\frac{\pi x}{10\:\text{ft}}\right)$$
We note that we have the squared length element defined by:
$$\:\mathrm{d}s^{2}=\mathrm{d}x^{2}+\mathrm{d}y^{2} = \mathrm{d}x^{2}\left(1+\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}\right)$$
Thus we have length element:
$$\:\mathrm{d}s=\sqrt{1+\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}}\:\mathrm{d}x$$
Therefore, integrating over the region $x \in [0\:\text{ft},1320\:\text{ft}]$ we get the total length travelled by our walker:
$$s=\int_{S}\:\mathrm{d}s=\int_{0}^{1320}\sqrt{1+\frac{\pi^{2}}{100}\cos^{2}\left(\frac{\pi x}{10}\right)}\:\mathrm{d}x\approx 1351.99\:\text{ft}$$
And therefore, we have that the extra distance walked is:
$$\Delta s = s-1320\:\text{ft}=31.9905\:\text{ft}$$
A: If I understood you right,
$ b=8/2 = 4 ; 2 \pi a = 5280 /4 ;$
$ r(\theta) =  a + b \cos(\theta).  $  
Evaluate $ \int _0^{2\pi} \sqrt{r^{'2} + r^2}  d \theta $ for a full circuit or a part of it.
