Question

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 is the area of the fluid used?

• A. $\frac{43}{9}-2\ln 3$
• B. $\frac{40}{9}+3\ln 2$
• C. $\frac{40}{9}-2\ln 3$
• D. $\frac{40}{9}+2\ln 3$

Answer 1

The correct option is C $\frac{40}{9}-2\ln 3$

Observe that the level surface of the fluid, in the non-rotated system, is given by the line $x+y=2c,$ for some $c>0.$
The depth of the fluid is then the distance from the point $(1,1)$ (at the bottom of the rotated graph) to the point $(c,c).$
Given, this distance is $\frac{2\sqrt{2}}{3}.$
$\Rightarrow c=\frac{5}{3}.$
Thus, the region of fluid is the area bounded by the curves $y=\frac{10}{3}-x$ and $y=\frac{1}{x}.$
Solving the two curves, we get the intersection points $(\frac{1}{3},3)$ and $(3,\frac{1}{3}).$
Hence, area of the fluid is given by
$\underset{1/3}{\overset{3}{\int }}(\frac{10}{3}-x-\frac{1}{x})dx=\frac{40}{9}-2\ln 3$