Example: The conical tank from y = 2x is filled to within 2 ft of the top with the olive oil weighing 57 lb/ft^3. How much work does it take to pump the oil to the rim of the tank.

Delta V = Pi (radius)^2 (thickness) = Pi(1/2y)^2 delta y = pi/4 y^2delta y ft^3

F(y) = 57 delta V = 57 Pi/4 y^2 delta y lb.

Delta W = 57pi/4 (10-y) y^2

W = Integral from 0 to 8 [ 57pi/4 (10-y)y^2dy ~ 30,561 ft-lb

Now the problem: Supposed that, instead of being full , the tank in Example 5 is only half full. How much work does it take to pump the remaining oil to a level 4 feet above the top of the tank? Answer: 60,042 ft-lb.

ok my process

For problem 18

Since it's 1/2 the volume capacity...I took half of the volume of Example 3 and the integral changes and kept everything else the same....

What's wrong?

V = 1/2[(pi)(1/2y)^2(delta y)] = 1/2 (Pi/4y^2delta y) ft^3

F(y) = (57 lb/ft^3) 1/2[pi/4 y^2 delta y] = 57/2 [pi/4 y^2 delta y]

W = 57/2 [integral from 0 to 6 pi/4 y^2 (10-y) dy = 57pi/8 [integral from 0 to 6 y^2(10-y))

= 57pi/8 [integral from 0 to 6 10y^2-y^3] = 57pi/8[10y^3/3-y^4/4] from 0 to 6

= 57pi/8 [10(6)^3/3 - (6)^4/4] = 8,864 ft lb