What is the function that maps circumferal distance onto circular area

Question

A unit circle with points A and B are initially located at (1,0). Point A remains at (1,0) while point B moves counter clockwise along the circumference of the unit circle until it reaches (-1,0). AB represents a stretchable blade that shaves off slices of the circle as it moves along. When B has reached (0,1), it has shaved off an area equal to the area of a quarter circle, less the area of the inscribed 45,45,90 degree triangle within. When B arrives at (-1,0), B has traveled a distance equal to one half of the circumference of the unit circle and it has shaved off one half of the area of the unit circle. What is the function that maps the distance B has traveled onto the area it has shaved off?

Answer 1

Note the area which is "sliced" or "shaved" off is proportional to the distance traveled. Also, after traveling around the entire unit circle, it has gone a distance $2\pi$ and "shaved" off the entire circle area of $\pi$. Thus, for any given distance $d$, the circle area amount "shaved" would be the ratio of the entire area to the entire circumference times $d$, i.e., $f\left(d\right) = \frac{d}{2}$.

More generally, for a circle of radius $r$, the circumference would be $2\pi r$ and the area would be $\pi r^2$, so the corresponding equation for a distance of $d$ would be an area of $f\left(d\right) = \frac{dr}{2}$.

If by "distance" you mean angle transversed, then note there is $360$ degrees in total. Thus, for $d$ degrees, the area would be $f\left(d\right) = \frac{d\pi r^2}{360}$. In terms of radians, the total is $2\pi$ so, for $d$ radians, the area would instead be $f\left(d\right) = \frac{d r^2}{2}$.