Turn the graph of $y=\frac{1}{x}$ by $45^{\circ}$ counter-clockwise and consider the bowl-like top part of
the curve (the part above $y=0$). We let a $2D$ fluid accumulate in this $2D$ bowl until the
maximum depth of the fluid is $\frac{2\sqrt{2}}{3}$. What’s the area of the fluid used?
For convenience, we still work on the un-rotated graph. The bottom of the bowl is $(1, 1)$ and the surface of the liquid will be $x+y=k$ where $k$ is to be determined. By symmetry, the depth of the accumulated liquid must equal the distance between $(1, 1)$ and $\left(\frac{k}{2}, \frac{k}{2}\right)$. Setting this to $\frac{2\sqrt{2}}{3}$ leads to $k=\frac{10}{3}$. Therefore, the to-be-determined area is the one bounded by the curves $y=\frac{1}{x}$ and $x+y=\frac{10}{3}$. These two curves insect at $\left(\frac{1}{3}, 3\right)$ and $\left(3, \frac{1}{3}\right)$. It follows that the desired result is
$$\int_{\frac{1}{3}}^{3}\left(\left(-x+\frac{10}{3}\right) - \left(\frac{1}{x}\right)\right)dx=\left.\left(-\frac{1}{2}x^2 +\frac{10}{3}x-\ln{x}\right)\right|_{\frac{1}{3}}^{3}=\boxed{\frac{40}{9}-2\ln{3}}$$