Consider an element of height `dx` at depth `x` below the surface
Area `=2pir(AB)=2pir(dx)/(costheta)`
`AD~~BC=2r=2[b-((b-a)/h)x]`
Downward thrust is
`F=intdFsintheta=int_(0)^(h)(xrhog)(2pir)/(costheta)dxsintheta dF`
`=2pitantheta[(bh^(2))/2-((b-a)/h)(h^(3))/3]`
`=2pirhoh((b+2a)/6)(b-a)`
Method 2:
the water will apply normal force on the inclined wall and at the bottom of the container as shown in figure. Force at the bottom of the container is
`F_("bottom")=` Pressure at bottom` xx` Area
`=(rhogh)(pia^(2))`
`=rhoghpia^(2)`.........i
Weight of water in container is
`W=1/3pih(a^(2)+ab+b^(2))rhog`..........ii
From the free body diagram of the water body
`W=F_(v)+F_("bottom")`
`F_(v)=W-F_("bottom")`
`=1/3pih(a^(2)+ab+b^(2))rhog-rhoghpia^(2)`
`=1/3hrhog[a^(2)+ab+b^(2)-3a]`
`=1/3pirho g[ab+b^(2)-2a^(2)]`
`=1/3pi rhogh(b+2a)(b-a)`
