The viewing portion of the rectangular glass window in a fish tank is 6969 inches wide and runs from 1.51.5 inchesinches below the water's surface to 46.546.5 inches below the surface. find the fluid force against this portion of the window. the weight-density of seawater is 64 lb divided by ft cubed .

The first thing you should know in this case is the definition of the pressure:
 P = d * g * h
 where
 P = pressure.
 d = density.
 g = gravity
 Then, the force is defined as the integral of the pressure by the area differential:
 F = int (PdA)
 The window area differential is:
 dA = w * dh
 where
 w: width
 dh: height differential
 Substituting the expression of the pressure and the area we have:
 F = int (d * g * h) (w * dh)
 where w is 69 in.
 F = d * g * w * int (hdh)
 from 1.5 to 46.5 in
 F = d * g * w * (h ^ 2) / 2
 from 1.5 to 46.5 in
 F = d * g * w (((46.5 ^ 2) - (1.5 ^ 2)) / 2])
 F = (64) * (32,16) * (69/12) * 46.5 / 12) ^ 2) - ((1.5 / 12) ^ 2)) / 2)
 F = 88761.6 lb * feet * (1 / s ^ 2)
 
 the fluid force against this portion of the window is 
 F = 88761.6 lb * feet * (1 / s ^ 2)