Solve the previous problem by direct integration of the pressure on the quarter-circle surface (unit depth into the page). Use the coordinate system proposed in the figure. Consequently, the pressure...

Solve the previous problem by direct integration of the pressure on the quarter-circle surface (unit depth into the page). Use the coordinate system proposed in the figure. Consequently, the pressure as a function of θ becomes p = ρgh = ρg R(1 − cos θ). The results must be the same as before!