How much work would it take to pump oil from the tank in Example 5 to the level of the top of the tank if the tank were completely full?

In Example 5

In Figure 6.39

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EXAMPLE 5 The conical tank in Figure 6.39 is filled to within 2 ft of the top with olive oil weighing 57 lb/ft³. How much work does it take to pump the oil to the rim of the tank? We imagine the oil divided into thin slabs by planes perpendicular to the y-axis at the points of a partition of the interval [0,8]. The typical slab between the planes at y and y + Ay has a volume of about 2 AV = (radius) (thickness) = (y)² ay = y² ay ft. The force F(y) required to lift this slab is equal to its weight, F(y) = 57 AV = 57my2 Ay lb. The distance through which F(y) must act to lift this slab to the level of the rim of the cone is about (10-y) ft, so the work done lifting the slab is about AW = 57″ (10-y)y² Ay ft-lb. Assuming there are n slabs associated with the partition of [0, 8], and that y = y denotes the plane associated with the kth slab of thickness Ay, we can approximate the work done lifting all of the slabs with the Riemann sum W = 57 (10-²Ay; ft-lb. The work of pumping the oil to the rim is the limit of these sums as the norm of the parti- tion goes to zero and the number of slabs tends to infinity: W = lim T - $ 57″ (10 – №vi Ay — * 577 (10 − y)y² dy = - 577 = Weight (weight per unit volume) x volume =: 8 (10y² - y²³) dy 57 103 1478 Jo 30,561 ft-lb.