Given:
Height of tank = 8 ft
and we need to pump fuel weighing 52 lb/ [tex]ft^{3}[/tex] to a height of 13 ft above the tank top
Solution:
Total height = 8+13 =21 ft
pumping dist = 21 - y
Area of cross-section = [tex]\pi r^{2}[/tex] = [tex]\pi 4^{2}[/tex] =16[tex]\pi[/tex] [tex]ft^{2}[/tex]
Now,
Work done required = [tex]\int_{0}^{8} 52\times 16\pi (21 - y)dy[/tex]
= [tex]832\pi \int_{0}^{8} (21 - y)dy[/tex]
= 832[tex]\pi([/tex][tex][ 21y ]_{0}^{8} - [\frac{y^{2}}{2}]_{0}^{8}\\[/tex])
= 113152[tex]\pi[/tex] = 355477 ft-lb
Therefore work required to pump the fuel is 355477 ft-lb