Chapter 5, Problem 8P

To determine

To show:

The center of the disk should be positioned at a height $\frac{r}{\sqrt{1+{\pi }^2}}$ above the surface of the liquid

Explanation of Solution

Given information:

A circular disk of radius $r$ is used as an evaporator and is rotated in a vertical plane.

Proof:

The disc equation is $x^2+{\left(y-a\right)}^2=r^2$ . Find the value of $a$ , for which the wetted area is maximal. The wetted area is

  $\begin{array}{l}A\left(a\right)=\pi r^2-\pi a^2-\text{Area under the water}\\ A\left(a\right)\text{ =}\pi {\text{r}}^2-\pi a^2-2{\displaystyle \underset{0}{\overset{\sqrt{r^2-a^2}}{\int }}\left(\sqrt{r^2-x^2}-a\right)}\\ A\left(a\right)=\pi r^2-\pi a^2-2{\displaystyle \underset{0}{\overset{\sqrt{r^2-a^2}}{\int }}\sqrt{r^2-x^2}}dx+2a\sqrt{r^2-a^2}\end{array}$

Differentiate the function by $a$ and equate to zero,

  $\begin{array}{l}A\text{'}\left(a\right)=-2\pi a-2\sqrt{r^2-{\left(\sqrt{r^2-a^2}\right)}^2}\cdot \left(\sqrt{r^2-a^2}\right)+2\sqrt{r^2-a^2}+2a\frac{-a}{\sqrt{r^2-a^2}}\\ A\text{'}\left(a\right)=-2\pi a-2a\cdot \frac{-a}{\sqrt{r^2-a^2}}+2\sqrt{r^2-a^2}-\frac{2a^2}{\sqrt{r^2-a^2}}\\ A\text{'}\left(a\right)=-2\pi a+2\sqrt{r^2-a^2}=0\end{array}$

Solve this equation to get,

  $a=\frac{1}{\sqrt{1+{\pi }^2}}$

Since for $a=\frac{1}{\sqrt{1+{\pi }^2}}$ then,

  $A"\left(a\right)=-2\pi -\frac{2a}{\sqrt{r^2-a^2}}<0$

It follows that this is the point of maximum.

Finally, the maximal wetted area is obtained for $a=\frac{1}{\sqrt{1+{\pi }^2}}$