Fractal Surface Area of Lung I'm preparing for an upcoming exam and one of the questions is the following;
"A typical human lung has a volume of approximately 5 litres, and a fractal dimension of
approximately 2.97. One way of measuring the surface area $S$ of the alveoli (the cavities
where the gas exchange takes place) would be to wrap the lungs in a sheet of paper of
some thickness $\varepsilon$ and measure how much paper is needed to completely cover the lungs.
Derive an expression relating the surface area $S$ to the thickness of the paper $\varepsilon$, and hence
show that the maximum surface area is approximately $80m^2$
, once the size of the paper
approaches that of the diameter of a lung cell (approximately $50\mu m$).
"
We have done stuff in class to do with box counting dimensions and was wondering if this question had anything to do with that? I have tried doing this myself but I am finding this very difficult to make any progress. Any suggestions, or a solution to the problem would be greatly appreciated!
 A: By the definition of fractal dimension (via "shrinking rulers", aka "how long is the coastline of britain"), the surface area of the lung with an $\epsilon$-length ruler is approximately $S = S_0 \left(\frac{1}{\epsilon}\right)^{2.97}$.
A sphere of volume $5$ liters ($0.005 m^3$) has a radius approximately $0.106m$ (almost $11cm$, which is the right order of magnitude for an average adult human) and surface area approximately $0.141m^2$
Let $\epsilon = 50 \mu m = 0.00005 m$ with $S = 80m^2$, then compute $S_0 = 80m \times (0.00005m)^{2.97} = 1.35\times 10^{-11} m^{4.97}$.
Substituting the volume of the sphere $S = 0.141m^2$ then gives $\epsilon_0 = \left(\frac{1.35\times10^{-11}m^{4.97}}{0.141m^2}\right)^\frac{1}{2.97} = 0.000422m = 0.422mm$ as the initial thickness of the paper, which is the right order of magnitude for regular paper.
But if you started with paper twice as thick (I even tried starting with paper 10cm thick...) you'd end up with an area $2^{2.97}$ times bigger than $80m^2$, so I think this is a bad question with insufficient information.
