Answer:
The amount of workdone by in emptying the tank by pumping the water over the top edge is [tex]\mathbf{ 1169.99 \pi \ ft-lbs}[/tex]
Explanation:
Given that the tank is 6 feet across the top and 5 feet high.
Using the similar triangles.
[tex]\dfrac{3}{5} = \dfrac{r}{y}[/tex]
5r = 3y
[tex]r = \dfrac{3}{5} y[/tex]
Thus; each disc is a circle with area
A = [tex]\pi ( \frac{3}{5} y)^2[/tex]
The weight of each disc is ;
[tex]m = \rho_{water}A[/tex]
= 62.4 × [tex]\pi ( \frac{3}{5} y)^2[/tex]
= [tex]\dfrac{561.6}{25} \pi y^2[/tex]
The distance pumped is 5-y
Thus; the workdone in pumping the tank by pumping the water over the top edge is :
[tex]W = \dfrac{561.6 \pi}{25} \int\limits^5_0 {(5-y)} \, y^2dy[/tex]
[tex]W = \dfrac{561.6 \pi}{25} \int\limits^5_0 {(5y^2-y^3)} dy[/tex]
[tex]W = \dfrac{561.6 \pi}{25}[52.083][/tex]
[tex]\mathbf{W = 1169.99 \pi \ ft-lbs}[/tex]
The amount of workdone by in emptying the tank by pumping the water over the top edge is [tex]\mathbf{ 1169.99 \pi \ ft-lbs}[/tex]
