Work needed to pump water to a tank

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SUMMARY

The discussion focuses on calculating the work required to pump water into a spherical tank, specifically addressing the integration process involved. The user attempted to compute the work using two integrals: one for the bottom half and another for the top half of the sphere, yielding a result that differed from the provided answer key. The discrepancy arises from an error in the first integral, where the height of the water was incorrectly set at 90m instead of accounting for the tank's dimensions, which should have started at 0m for the bottom of the sphere.

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Homework Statement


I have attached the question with a diagram and answer key.


The Attempt at a Solution


What I did was found the work needed to pump the water into the bottom half of the sphere from 90-100 and then add that to the work needed to pump the water into the top half of the sphere.

integral from 0 to 10 9800pi(100-x^2)(90+x) + integral from 0 to 10 9800pi(100-x^2)(100+x)

When I plug in my numbers using wolfram I get:
http://www.wolframalpha.com/input/?i=integrate+from+0+to+10+(9800pi(100-h^2)(190+2h))

Which is slightly off compared to what the answer key gets:
http://www.wolframalpha.com/input/?i=integrate+from+-10+to+10+(9800pi(100+x)(100-x^2)

Does anyone know why they would be different? I feel like my logic is correct...
 

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Your flaw is in your first integral. When x=0 (in the first integral), you're 90m above the ground. Yet, the diameter is 10m, when it should be 0m ;)
 

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