Answer:
Explanation:
From the given information:
The diameter of the pool = 24 ft
The radius will be = 24 ft/2 = 12 ft
The volume of water V = πr²(Δx)
V = π× 12²×(Δx)
V = 144π(Δx)
Le's assume water weighs 62.5 lb/ft³;
Then:
the Force (F) will be:
= 144π(Δx) * 62.5
= 9000πΔx lb
Also, the side of cylindrical water = 5 ft while its depth = 4ft
As such, each slide of water d = 5 - x, and the region is between 0 and 4.
∴
The required work is:
[tex]W = \int^4_0 (5-x) 9000 \ \pi dx \\ \\ W = 9000 \int^4_0 (5-x) \ dx \\ \\ W = 9000 \pi \Big [5x - \dfrac{x^2}{2} \Big]^4_0 \\ \\ W = 9000 \pi \Big [5*4- \dfrac{4^2}{2} \Big] \\ \\ W = 9000 \pi \Big [20-8 \Big] \\ \\ W = 9000 \pi \Big [12 \Big] \\ \\ W = 9000 \pi (12) \\ \\ \mathbf{W = 108000 \pi \ ft.lb}[/tex]
