A cylindrical tank of radius 3 ft length 8 ft is laid out horizontally .The tank is half full of oil that weighs 60 pounds per cubic foot.

Question

a.)determine the work done in pumping out oil to the top of the tank.
b.)determine the work required to pump out the oil to leave 4 feet above the top of the tank.

Answer 1

Draw a cross-section. If the oil is at depth y, then the width of the thin rectangle of oil is 2x, where

(3-y)^2 + x^2 = 9
x = √(6y-y^2)

So, the weight of oil for that thin sheet is

2x*8*60 dy

and the total work to lift all the oil out of the top of the tank is thus

∫[0,3] 2*8*60(6-y)√(6y-y^2) dy = 2160(4+3π) = 28998

Now our engineer friends will just want to find out how far the center of mass is lifted, times the weight of the oil.

The weight is 1/2 * 9π * 8 * 60 = 2160π lbs

The centroid of a semicircle of radius r is at a distance 4r/(3π) from the center of the circle. That means we have to lift the centroid a distance of r + 4r/3π = r(4+3π)/3π

Multiply that by the weight of oil, and you have work of

3(4+3π)/3π * 2160π = 2160(4+3π)

as above.