Quiz

mg=integrate (h from 0 to H) ρg(H-h)*sin(α)*dS(h) ----------- (1) It seems like there should be a "-" sign here. Because the direction is upwards, not downwards, and because I need to adjust the answer. integrate (h from 0 to H) ρgHsin(α)*dS(h) = integrate (α from 0 to arcsin(H/R)) ρgH*2π(Rcos(α))sin(α)*Rdα = πρgHR^2 integrate (α from 0 to arcsin(H/R)) sin(2α)dα = -0.5 πρgHR^2 cos(2α) dα (α from 0 to arcsin(H/R)) = -0.5 πρgHR^2 (cos(2arcsin(H/R))-1) = -0.5 πρgHR^2 (1-2(H/R)^2-1) = πρgH^3 -------------- (2) integrate (h from 0 to H) ρghsin(α)*dS(h) = integrate (α from 0 to arcsin(H/R)) ρgRsin(α)sin(α)*2π(Rcos(α))*Rdα = integrate (α from 0 to arcsin(H/R)) ρgsin(α)sin(2α)*πR^3 dα = ... Let's recall the formula: sinx siny = 1/2 (-cos(x+y)+cos(x-y)) and continue: ... = integrate (α from 0 to arcsin(H/R)) ρg((cos(α)-cos(3α))/2)*πR^3 dα = 0.5 πρgR^3 (sin(α) - (1/3)sin(3α)) (α from 0 to arcsin(H/R)) = 0.5 πρgR^3 (H/R - (1/3)sin(3arcsin(H/R))) = ... The sine of a triple angle sin(3x) = 3 sin x - 4 sin^3 x ... = 0.5 πρgR^3 *(4/3) (H/R)^3 = 2/3 πρgH^3 ----------------- (3) (1), (2), (3): m = (1/3)πρH^3 H=(3m/πρ)^(1/3)