If you have an upside-down cone with top diameter 2m and a height of 2m full of water, then how much work is needed to pump out all the water? (Single variable calculus).

My idea was something like this:
ΔV = pi * x^2
Δm = ΔV * p, (p = density of water ~ 1000 kg/m^3)
F = g (gravity)
d(x) = 2 - y = 2 - 2x, (y = 2x since height/radius = 2 and d(x) is the distance we need to move the Δm)

This should result in something like:
Integrate[ p*g*pi*x^2 ( 2 -2x ), {x,0, 1} ]
= p * g * pi / 6
But the right answer should be: p * g * pi / 3.
What mistake did I make? Thanks!

Question

If you have an upside-down cone with top diameter 2m and a height of 2m full of water, then how much work is needed to pump out all the water? (Single variable calculus).

My idea was something like this:
ΔV = pi * x^2
Δm = ΔV * p, (p = density of water ~ 1000 kg/m^3)
F = g (gravity) 
d(x) = 2 - y = 2 - 2x, (y = 2x since height/radius = 2 and d(x) is the distance we need to move the Δm)

This should result in something like:
Integrate[ p*g*pi*x^2 ( 2 -2x ), {x,0, 1} ]
= p * g * pi / 6 
But the right answer should be: p * g * pi / 3.
What mistake did I make? Thanks!

anonymous · Answer

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