A conical container of radius 10 ft and height of 40 ft is filled to a height of

Question

A conical container of radius 10 ft and height of 40 ft is filled to a height of 38 ft of a liquid weighing 62.4 Ib/ft^3. How much work will it take to pump the contents to the rim? How much work will it take to pump the liquid to a level of 2 ft above the cone's rim?

Explanation / Answer

1) The boundaries intersect where y + 6 = x = y^2. The upper point is on the axis, and the rest of the figure below, with y+6 > y^2. The volume of a thin shell at r = 3 - y is dV = [(y+6) - y^2] 2pi r dr = [(y+6) - y^2] 2pi (3 - y) -dy Integrate from small r (big y) to big r (small y). 2) Infinity. 3) Pro forma (1), but with the roles of x and y reversed. 4) The length ds of an incrementally small segment is (ds)^2 = (dx)^2 + (dy)^2 ds = dx sqrt(1 + (dy/dx)^2) = dx sqrt(1 + (3(1/6)x^2 - 2(2x)-2)^2) Integrate ds between the limits. 5) Pro forma (2), with boundaries y = 0, y=10, y/(x-4) = 10/(3-4). 6) Force F = kx, work dW = -Fdx (in case you avoided physics). Solve for k, integrate dW. W = 1/2 k (x2^2 - x1^2) 7) The bounding curve is x^2 + (y-11)^2 = 11^2. The horizontal radius of the tank at a height y is r = x = sqrt(11^2 - (y-11)^2) and the incremental volume at that height is dV = pi r^2 dy = pi(11^2 - (y-11)^2) The weight of that water is 62.4 dV lb and the work to raise it is y(62.4 dV) ft-lb Integrate. 8) Gauge pressure in the liquid at a depth h below the surface is 60h lb/ft^2. The boundary of the cross-section is x^2 + (y-10/2)^2 = (10/2)^2 = x^2 + h^2, so the width of the tank at depth h is 2 sqrt(5^2 - h^2) and the area of one end of the tank is dA = 2 sqrt(5^2 - h^2) dh The force on that area is dF = PdA = 60h *2 sqrt(5^2 - h^2) dh Integrate.

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