Average distance to perimeter of a polygon? Trying to calculate heat transfer which is a function of distance of each molecule to the closest wall for various container shapes.  For example, a rectangular prism versus a cylinder.
So I think that  a 'thin' rectangular prism of volume V average distance to wall can be much less than average distance of a cylinder.  Reducing only to the cross section, assuming a rectangle of dimension $x$, $.5x$, versus a cylinder of of radius $\sqrt{\frac{x^2}{2 \pi}}$ (which is same volume I think) what is the average distance from a point in the circle to the perimeter versus the average distance of a point in the rectangle to its closest perimeter?
For a circle I have this idea that if inscribe a smaller circle inside the big circle with the same center point, such that the smaller circle contains 1/2 the volume of the outer circle, then the average distance is radius of outer circle minus radius of inner circle.  Is that correct?
For the rectangle I don't quite know  -- whether the same approach could be used to inscribe a rectangle of the same aspect ratio which contains 1/2 the area of the outer rectangle and the average distance is the length of the perpendicular connecting the inner and outer rectangle?
Is this a correct approach?
 A: The expected distance from a unit circle to a randomly chosen point inside the circle is given by
$$
\frac{1}{\pi}\int_{0}^{1}(1-r)(2\pi r)dr=\left(r^2-\frac{2}{3}r^3\right)\bigg\vert_{0}^{1}=\frac{1}{3};
$$
as a function of the area of the circle, then, the expected distance is
$$
{d}_{\text{circle}}(A)=\frac{1}{3\sqrt{\pi}}\sqrt{A} \approx 0.1881 \sqrt{A}
$$
(which holds for circles of any size).  For a square of side length $2$, the expected distance from the square to a random interior point (which can be calculated by considering a single quadrant) is
$$
\int_{0}^{1}y(2-2y)dy = \left(y^2-\frac{2}{3}y^3\right)\bigg\vert_{0}^{1}=\frac{1}{3}
$$
as well, so
$$
d_{\text{square}}(A)=\frac{1}{6}\sqrt{A} \approx 0.1667 \sqrt{A}
$$
for an arbitrary square.  Finally, for a $2\times 4$ rectangle, you need to consider the short and long sides differently.  Essentially you have two $2 \times 1$ end caps that behave like the square (so the average distance is $1/3$), and a $2\times 2$ central block for which the average distance is just $1/2$.  The two components have equal areas, so the overall average distance is the average of $1/2$ and $1/3$, or $5/12$.  Since the area of the entire rectangle is $8$, we have
$$
d_{\text{rect}}(A)=\frac{5}{24\sqrt{2}}\sqrt{A} \approx 0.1473 \sqrt{A}
$$
for any rectangle with aspect ratio $2$.
If it helps, you can think of "unrolling" each shape, while preserving its area, so that the set of points at distance $d$ from the perimeter lies along $y=d$.  For the circle and the square (and for any regular polygon), this gives a triangle with base equal to the original shape's perimeter.  For the rectangle, though, it gives a trapezoid, because the points maximally distant from the perimeter are a line segment, not a single point.
A: For a circle of radius $r$, the average distance to the circumference is $\frac 1{\pi r^2}\int_0^r (r-x)2\pi x dx$ because at radius $x$ the distance is $r-x$ and the area between $x$ and $x+ dx$ is $2 \pi x dx$.   This gives $\frac 1{\pi r^2}\int_0^r (r-x)2\pi x dx=\frac 2{ r^2}(rx^2/2-x^3/3)|_0^r=\frac 13r$
For a rectangle it is harder, as the corners round off as you look for areas with the same distance to the edge.  If the rectangle is very long and skinny, you can ignore the ends and the average distance will be one quarter the short dimension.
