A tank is is half full of oil that has a density of 900kg/m3. Find the work W re

Question

A tank is is half full of oil that has a density of 900kg/m3. Find the work W required to pump the oil out of the spout. (Use 9.8 for g and 3.14 for . Round your answer to three significant digits.) W = J r=13.2 h=3.3

Explanation / Answer

Suppose that we inscribe a cylinder that is x units from the center of the sphere and has a height of ?x. Superimpose a coordinate system where the center of the sphere is the origin. Then, since the radius of the sphere is r, the equation of the sphere is: x^2 + y^2 + z^2 = r^2. At the point where the cylinder is x units from the center, the width of the sphere, y, is: 0^2 + y^2 + x^2 = 12^2 ==> y = v(144 - x^2) The square of the radius of the cylinder is then: y^2 = 144 - x^2 The volume of the cylinder is then: ?V = pr^2*h = p(144 - x^2)?x. (As a sanity check, the volume of the sphere is: ? p(144 - x^2) dx (from x=-12 to 12) = 2304p = (4/3)p(12)^3, which checks with V = (4/3)pr^3.) The weight of water is: ?F = mg = p?Vg = (900)[p(144 - x^2)](9.8) = 8820p(144 - x^2). Then, the work to move this water up as: ?W = ?Fd = 8820p(144 - x^2)(18 + x). (Note that the water moves up x to units to the center, 12 units up to the top of the sphere, and 6 units to get to the top of the spout.) Since x varies from -12 to 0 in the bottom half of the sphere, the required work is: W = ? 8820p(144 - x^2)(18 + x) dx (from x=-12 to 0) = 137,168,640p J.

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