# Help with understanding calculus problem

This is from Stewart Calculus Metric Version 9E, Chapter 4 Problem Plus #12 :

A circular disk of radius $$r$$ is used in an evaporator and is rotated in a vertical plane. If it is to be partially submerged in the liquid so as to maximize the exposed wetted area of the disk, show that the center of the disk should be positioned at height $$\frac{r}{\sqrt{1+\pi^{2}}}$$ above the surface of the liquid.

The solution is :

I literally have no idea what 'rotated in a vertical plane' means and what that picture shows..

• The disc of radius $r$ is spinning about its centre. Some of it is in the water and as it spins the central part out to a distance $x$ stays dry while the ring (annulus) between $x$ and $r$ gets wet. The white part of the disc shown, beyond the distance $x$, is the exposed wetted area of the disc I think. That is the area to maximize. If $x$ is too big you narrow the ring while if $x$ is too small you submerge too much of the disc.
– Paul
Commented Oct 8, 2022 at 13:44
• Apparently "rotated in a vertical plane" is supposed to mean "mounted in a vertical plane and then spun continually around its axis". I didn't read it that way either, for what it's worth. Commented Oct 8, 2022 at 15:25

The dry disk (yellow) is lowered partially into the liquid (blue) and then rotated about its axis.

Parts of the disc that touched the liquid become wet (green).

Hopefully you can now see how the textbook diagram parameterizes the wet/dry portions of the disc.

The disk "is rotated in a vertical plane" simply means that it revolves (either clockwise or anti-clockwise), so we don't have to think about volumes. This rotation (in 2-D) is the mechanism that allows to pull liquid above its surface.

The "wetted circular region" is the annulus whose area is the area of the circle with radius $$r$$ minus the area of the inner circle with radius $$x$$. It is all wetted because the disk rotates.

The unexposed (unexposed above the liquid surface, i.e., submerged) wetted region is the area in gray in your figure and it is computed as follows: first, for simplicity, rotate the figure 90 degrees anti-clockwise. We have a half moon, of which half above the horizontal axis and half below. Let's compute the area of the part above. The horizontal interval of the area goes from $$x$$ to $$r$$, to these are the integration extremes. The height at the leftmost point is found using Pythagoras' theorem, where hypotenuse = $$r$$ and one catheter is $$x$$ (we compute the other catheter). Proceeding towards right, the height continues to be computed the same way and it is decreasing since $$x$$ is increasing, until reaching 0 when $$x = r$$. Multiply the result by two to get the total area including that below the horizontal axis.

The rest shoud be clear from the solution you posted.