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?
Best Answer
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.