Force of Fluid by Integration: Total hydrostatic force acting on the slant side of trapezoidal trough

A trough of length 6 ft has for its vertical cross section an isosceles trapezoid. The upper and lower base are 6 ft and 2 ft respectively and the altitude is 2 ft. If the trough is full of water, find the force on the slant side of the trough.

This is a basic problem in Hydraulics
FH=γˉhA=62.4(1)(2×6)=748.8 lb

FV=γV=62.4×12(2)(2)(6)=748.8 lb

F=FH2+FV2=1058.96 lb
 

Solution by Integration
dF=γhdA=γ(ysin45)(6dy)

F=374.4sin45220ydy

F=1058.96 lb