No π in the jar


Shaded area (a segment) = ½ r2 (2 θ) – ½ r2 sin 2 θ

Volume of typical ‘segmant’ = ( r2 θ – ½ r2 sin 2 θ ) δx

and total volume = &int0h ( r2 θ – ½ r2 sin 2 θ ) dx

           Now y
x 		= r
h 		where y = r ( 1 – cos θ )
           \ x = h
r 		r ( 1 – cos θ ) and dx = h sin θ dθ
           So V = &int0π/2 ( r2 θ – ½ r2 sin 2 θ ) h sin θ dθ
                   = r2 h &int0π/2 ( θ – ½ sin 2 θ ) sin θ dθ

           &int0π/2 θ sin θ dθ   =   [ – θ cos θ ]0π/2   –   &int0π/2 – cos θ dθ   =   1

           &int0π/2 ( – ½ sin 2 θ ) sin θ dθ   =   ¼ &int0π/2 ( cos 3 θ   –   cos θ ) dθ
                                             =   ¼ [ – 13 sin 3θ   –   sin θ ]0π/2   =   – 13

           So V = r2 h ( 1 – 13 )   =   23 r2 h