No π in the jar — Arthur Lupton’s solution


Looking at the bottom from the drinker’s perspective, we see a semicircle of liquid.

Divide up this semicircle into vertical stripes each of thickness δx.
The central stripe is of length r, and a stripe x from the centre has length y given by
      y2   =   r2 – x2

Each stripe is one side of a triangular slice (thin prism) of liquid. The central such triangle is right-angled with its hypotenuse along the surface of the liquid, and the other adjacent side along the glass − a line that will be vertical when the glass is upright.
The volume of the triangle:
      δV   =   ½ h r δx

All the triangles are similar, and so each has a height proportional to y :
            =   h y / r

So the volume of each triangular slice is given by:
      δV   =   ½ ( h y / r ) * y δx
           δV   =   ½   h
r 		  y2 δx   =   ½   h
r 		( r2 – x2 )   δx


Integrating across the semi-circle we have:
           V   =   ½   h
r 		  &int-r+r ( r2 – x2 )   dx   =   ½   h
r 		  [ ( r2 x – x3 / 3 ) ]-r+r
                             =   ½   h
r 		  ( r3 – r3 / 3 + r3 – r3 / 3 )   =   23 r2 h