Find the net downward thrust on the inclined surface of a bucket full of water of height `h` and radii `a` and `b`.

Explanation

Consider an element of liquid of height dx at depth x below the free surface of liquid. Let r be the radius of this element.

Area of element, `Delta A = 2 pi r xx (dx)/(cos theta)`
in fig. `tan theta = (b-a)/(h) = (b-r)/(x)`
or `r=[b-(b-a)/(h)x]`
Thrust of liquid on the elementary curved surface
`dF = (x rho g) Delta A`
Downward thrust on the entire curved surface
`F = int_(0)^(h) F sin theta = (x rho g) (2 pi r)/(cos theta) dx sin theta`
`= int_(0)^(h) 2 pi rho g [b-((b-a)/(h))x] xx x dx tan theta`
`= 2 pi rho g tan theta [ (bh^2)/(2) - ((b-a).(h)h^(3))/(3)]`
` 2 pi rho g ((b-a)/(h))[(bh^2)/(2)-((b-a)h^(2))/(3)]`
`= 2 pi rho g (b-a)h(2a+b)6`.