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September 2, 2014

September 2, 2014

Posted by **Jerry** on Wednesday, October 10, 2012 at 11:12am.

- Calculus II -
**Steve**, Wednesday, October 10, 2012 at 11:39amLet the center of the tank be at (0,0)

when the water level is at y, the radius of the water surface is

r^2 = 25-y^2

Work is ∫F(y) dy

F(y) is the weight of the water. So, since each slice of water in the tank is raised 21+y feet, and water weighs 62.4 lbs/ft^3,

W(y) = ∫[-5,5] 62.4 π (25-y^2) (21+y) dy

= 218400π

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