To determine
To show:
The center of the disk should be positioned at height r1+π2 above the surface of the liquid
Answer
Height above water surface is r1+π2
Explanation
1) Concept:
Exposed wetted area = area of the disk - area of arc OAB + area of triangle OAB - area of circle with center O and radius rcosθ
2) Calculation:

Exposed wetted area= area of the disk - area of arc OAB + area of triangle OAB - area of circle with center O and radius rcosθ
That is,
A=πr2-r2θ+0.5r2sin2θ-πrcosθ2
To maximize this area, let us find critical points of A. So differentiate it with respect to θ and equate it to zero.
dAdθ=0-r21+0.5r22cos2θ-πr2(-)sin2θ
Simplify to get
dAdθ=-r2+r2cos2θ+πr2sin2θ
Set dAdθ=0
-r2+r2cos2θ+πr2sin2θ=0
Factor out r2 from the first two terms.
-r2(1-cos2θ)+πr2sin2θ=0
Simplify.
-r22sin2θ+πr2sin2θ=0
Factor out r2.
r2-2sin2θ+2πsinθcosθ=0
Factor out 2sinθ.
2r2sinθ-sinθ+πcosθ=0
Here,r2 cannot be zero, and θ cannot be zero and
So, -sinθ+πcosθ=0
Add sinθ to both the sides.
-sinθ+sinθ+πcosθ=0+sinθ
Simplify.
πcosθ=sinθ
Divide by cosθ, and simplify.
tanθ=π Using tan2x+1=sec2x
We get
cosθ=11+π2
This means height of centre above water surface rcosθ=r1+π2
Therefore, the height of centre above water surface is r1+π2
Conclusion:
The height of centre above water surface is r1+π2