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demonelite123
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A cylindrical reservoir of diameter 4 feet and height 6 feet is half-full of water weighing w pounds per cubic foot. Find the work done in emptying the water over the top.
i put the cylinder on the x-y axes, so y represents height and x represents radius of the cylinder. so the volume of a slice of water ΔV = πx^2 (Δy), and since x = 2, ΔV = 4π(Δy).
a slice of the water at height y is lifted (6 - y) feet so therefore the work,
W = w*4π(Δy)*(6 - y)
so the integral looks like 4πw ∫ (6-y) dy.
the only problem i have is how to decide the limits of integration. since you are trying to empty the water over the top, the water at the very bottom should travel to the top right? so the limits should be from 0 to 6 i think. but my book says the limits are from 0 to 3 because the reservoir is only half full. can somebody please explain this to me?
i put the cylinder on the x-y axes, so y represents height and x represents radius of the cylinder. so the volume of a slice of water ΔV = πx^2 (Δy), and since x = 2, ΔV = 4π(Δy).
a slice of the water at height y is lifted (6 - y) feet so therefore the work,
W = w*4π(Δy)*(6 - y)
so the integral looks like 4πw ∫ (6-y) dy.
the only problem i have is how to decide the limits of integration. since you are trying to empty the water over the top, the water at the very bottom should travel to the top right? so the limits should be from 0 to 6 i think. but my book says the limits are from 0 to 3 because the reservoir is only half full. can somebody please explain this to me?