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