1. The problem statement, all variables and given/known data A solid cube, mass m, side length l, is placed in a liquid of uniform density, ρ(rho), at a depth h0 below the surface of the liquid, which is open to the air. The upper and lower faces of the cube are horizontal. Find the magnitude of force, Fs, exerted on each vertical face of the cube and express it in terms of ρ (rho), l, h, atmospheric pressure p0 and gravity g. 2. Relevant equations Pressure, p, at any point in the liquid = (p0 + ρg(h0+x)) 3. The attempt at a solution surface integral Fs = -n ∫ (p0 + ρg(h0 +x))dA where n is the unit vector normal to the surface Area integral of a rectangle with side lengths a & b written as 2 single integrals Fs = -n ∫ [ ∫ (p0 + ρg(h0 +x))dy]dx . . . . . . . . . . (first integral limits 0 & a, second limits 0 & b) Fs = -n ∫ (p0 + ρg(h0 +x)) b dx Fs = -b (p0 + ρgh0) a + ½*ρ0ga2) n Fs = -ab (p0 + ρ0g(h0 + ½a)) n remembering that ab = l*l = area of cube side = l2 so Fs = -l2 (p0 + ρ0g (h0 +½a)) n is that OK? Is the integral look right. I worry as I'm doing this at home by myself.