Answer:
[tex]w=2.216*10^{6} J[/tex]
Step-by-step explanation:
Data:
r=4m
h=10m
w=? (3 m of water out top of the tank)
g=9.8m/s2
б=1000kg/m3
We must find the work necessary to move 3m of water from the top of the tank, so there is a differential (dy) that represents a small portion of volume for which it is integrated for the total volume to move then it would be the weight for the distance that is the volume to move
dw=weight*(10-y)dy, weight=б*V, , ⇒ weight=б*Vdy*g ⇒ dw=16πg(10-y)dy ⇒
[tex]w=16\alpha \pi g(10y-\frac{y^{2}}{2})[/tex] evaluated between a and b, but a=10 and b=7 (last 3m of the full tank), finally
[tex]w=16\alpha \pi g((100-50)-(70-24.5))\\w=16\alpha \pi g(4.5)=72\alpha \pi g\\w=2.126*10^{6}J[/tex]
