The area of a region around a curve

Question

If we are given a simple closed curve with length $L$ in the plane, and we have a fixed number $r$ such that for each point $x$ on the curve there is a related disc $D(x,r)$ with radius $r$, how can we estimate the area of the region formed by all such discs from above?

Answer 1

Using the following approach it seems that the general answer is indeed just $2rL$.

Assuming the curve has no sharp points, this problem is equivalent to the sum line segments at each point $x_0$ along the curve, each centered at $x_0$, perpendicular to the curve at that point, and of length $2r$. Observing the diagram above, it can be seen that for any radius of the circle not perpendicular to the curve, such as the green one, there must be another point, $x_1$, along the curve closer to its position on the circumference. This is true because obviously any close point along tangential line (the dark red line) in the correct direction will be closer, and since the curve is continuous, in moving closer toward $x_0$ you can get a point $x_1$ arbitrarily close to the tangential line. Another way of thinking about it is to think about $x_1$ as infinitesimally farther down the curve than $x_0$, and since the measure is infinitesimal, the slope is unchanging and it can be thought of as moving down the tangent line.

If we let $R_t$ represent the radius of curvature at a distance $t$ along the curve, we can then do the following.

$$ \begin{align} A & = \frac{\mathrm{d}\theta}{2\pi} \left( \pi(R_t+r)^2 - \pi(R_t-r)^2 \right)\\[2ex] A & = \frac{\mathrm{d}\theta}{2\pi} \left(4\pi rR_t\right)\\[2ex] A & = 2rR_t \:\mathrm{d}\theta \end{align}$$

and since $\mathrm{d}\theta = \mathrm{d}t / R_t$

$$ A = 2r \:\mathrm{d}t\\[2ex] \int_0^L 2r \:\mathrm{d}t = 2rL $$

I realize that this does not hold if the curve has any sharp points. However for any sharp points you can manually add the uncounted circle section ($\frac{\theta}{2}r^2$) as well as subtracting all the overlapping area ($r^2\cot(\frac{\theta}{2})$) for a final answer of

$$2rL + r^2 \sum_{\theta_i} \frac{\theta_i}{2} - \cot\left(\frac{\theta_i}{2}\right) $$

where $\theta_i$ is the angle change at the $i^{th}$ sharp point. The only circumstance I can see in which this would not work is if two points on the curve ever came closer than 2r away from each other and you were counting some area twice.