The deligne dam on the cayley river is built so that the wall facing the water is shaped like the region above the curve y=0.3x2 and below the line y=200. (here, distances are measured in meters.) the water level is 38 meters below the top of the dam. find the force (in newtons) exerted on the dam by water pressure. (water has a density of 1000kg/m3, and the acceleration of gravity is 9.8m/sec2. )

 F = 1.65 10⁸ N

Explanation:

In this pressure problem we have to use the definition of pressure

             P = dF / dA

            dF = P dA

we already have the expression for force and the pressure in a liquid is

            P = Po + rho g (H-y)

Where Po is the atmospheric pressure acting on both sides of the dam, whereby its contribution is canceled and (H-y) is the distance from the surface

Let's look for an expression for the area differential

            A = xy

            dA = dx dy

            y = 0.3 x²

            x = √(y/0.3)

Let's build our equation with these expressions and integrate between the initial limit where the height is measured from the bottom of the dam y = 0, x = 0 to the upper limit, let's call it H = 200m, x = RA y / 0.3 and F = 0

           ∫ dF = ∫ (ρ g (H-y) dx dy

           -F = ρ g [∫∫ H dy dx - ∫∫ ydy dx]

           F = ρ g [∫ H x dy - ∫ x2√2 / 2 dy

Let's evaluate between the limits of integration

           F = ρ g [∫ (H (√y /√0.3) dy - ∫ y/0.3 1/2 dy

           F = ρ g (H /√0.3 ∫ √y dy - 1 /0.6 ∫ y dy)

Let's do the second integral

           F = ρ g (H/√0.3 ) 2/3 - 1/0.6  y2 / 2)

           F = ρ g (2H/3√0.3 )  - 1/1.2  y2 )

We evaluate at the limits

          Y = 0

          Y = 38 m

          F = 1000 9.8 (2 200/3√3  √38³  - 1/1.2 38²)

          F = 9800 (18032 - 1203)

          F = 1.65 10⁸ N