Maximal Humidifier Area
Maximizing The Wet Part Of The Disk Above The Water Tank
John Snyder
May, 2011
Problem
A humidifier has a circular disk of radius r that slowly rotates. It is partially submerged in a tank of water (the disk is
perpendicular to the surface of the water). In order to have the greatest humidifying effect, the total area of the wet part
of the disk that is above the water should be maximized. How high above the surface of the water should the center of
the disk be to attain this maximum?
Solution
Ÿ Summary Of Results
The optimal height of the center of the disk above the surface of the water is:
h=
r
1 + Π2
At this height the area of the wet part of the disk that is exposed to the air will be r2 IΠ - tan-1 HΠLM.
Ÿ Methodology
Let’s begin with a visualization showing the wet part of the disk (blue) above the water tank (orange).
r
BlockB:h =
, r = 1, p1, p2>,
1+Π
2
p1 = RegionPlotAh2 < x2 + y2 < r2 && y > - h, 8x, - r, r<, 8y, - r, r<, MaxRecursion ® 6E;
p2 = Graphics@8Orange, Rectangle@8- 1.1, - 1<, 81.1, - h<D<D;
Show@p1, p2, Frame ® False, Axes ® TrueDF
2
HumidifierArea148.nb
Using integration we can find the area of the blue region when the center of the disk is at a height h above the surface
of the water.
AssumingA0 < h < r,
area = FullSimplifyAIntegrateABooleAh2 < x2 + y2 < r2 && y > - hE, 8x, - ¥, ¥<, 8y, - ¥, ¥<EEE
- h2 Π + h
- h2 + r2 + r2 Π - ArcCscB
r
2 h¹r
F
-h2 +r2
1
4
r2 J2 + Π - 2 ä Log@hD + ä LogB
r2
2
True
FN
We need only concern ourselves with the first expression as the second is equal to it at the point h = r ’
2 as we see
in the next cell.
WithB:h = r ’
- h2 Π + h
2 >,
r
- h2 + r2 + r2 Π - ArcCscB
1
F ==
2
2
-h + r
r2 2 + Π - 2 ä Log@hD + ä LogB
4
FullSimplify@ð, r > 0D &F
True
So the area of the wet portion of the disk above the water’s surface is as follows.
area = areaP1, 1, 1T
- h2 Π + h
- h2 + r2 + r2 Π - ArcCscB
r
F
- h2 + r2
Plotting this area function it is clear that there is a global maximum near the red point.
r2
F 
2
HumidifierArea148.nb
BlockB8r = 1<,
PlotBarea, 8h, 0, r<, PlotStyle ® Thick,
r
Epilog ® :Red, PointSize@LargeD, PointB:
, r2 HΠ - ArcTan@ΠDL>F>,
1 + Π2
AxesLabel ® HStyle@ð, Italic, 14, FontFamily ® "Helvectia"D & ž 8"h", "Area"<L,
PlotLabel ® Style@Row@8"r=", r<D, Italic, 14, FontFamily ® "Helvectia"D,
ImageSize ® 450FF
Taking the derivative, setting it equal to zero, and solving for h we obtain.
Assuming@r > 0,
Solve@D@area, hD Š 0 && 0 < h < r, hD  FullSimplifyD
r
::h ®
>>
2
1+Π
And this value of h gives rise to the following maximal area.
area . %  FullSimplify@ð, r > 0D &
9r2 HΠ - ArcTan@ΠDL=
3