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Calculus 2- Hooke's Law Spring problem

  1. May 25, 2013 #1
    Calculus 2- Water pump problem

    1. The problem statement, all variables and given/known data

    The lower half of a sphere with radius 2m is filled with water. Find the work needed to pump the water to a point 1m above the centre of the sphere.




    2. Relevant equations

    3. The attempt at a solution
    I know you have to cut the problem into small pieces, and you have to label a "x" anywhere you want but you have to solve the problem based on which part you label the x (see attachment). But I have no idea where to label the x and how to get the integral with that information. And also what is the limits I put in the definite integral? I know it has to do with the x you choose but how to know which limit to use?
     

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    Last edited: May 25, 2013
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  3. May 25, 2013 #2

    vela

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    You forgot the attachment.
     
  4. May 25, 2013 #3

    Simon Bridge

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    If you didn't have to worry about which part you'd label "x", then how would you go about it?
    Wouldn't you define a coordinate system first - exploiting the symmetry of the situation?

    Note: it would take an infinite work to pump all the water into a point - since water does not readily compress that far.

    Perhaps you just need to lift the water to a height 1m above the initial level of the water?
     
  5. May 25, 2013 #4

    Sorry forgot, there it is.
     
  6. May 25, 2013 #5

    Simon Bridge

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    Cool - looks like x is decided for you.
    Isn't it the position of the disk with thickness dx in the diagram?

    I also don't see what this has to do with Hook's Law.

    Anyway - it sounds to me that you are trying to apply an equation you do not fully understand.

    Start by defining a coordinate system. You can put the origin anywhere you like - so use your knowledge of geometry and the symmetry of the system to put it where it makes the math easier.

    Notice that a disk of thickness dx at height x has mass dm (which depends on the density of water and the volume of the disk)...

    The amount of work you need to lift a mass dm to height h from height x is dW. dW=... (you finish)
     
    Last edited: May 25, 2013
  7. May 25, 2013 #6
    My BAD!

    my bad I got the title wrong.

    Well here is another attachment...

    So the volume of the disk is [itex]\pi[/itex](4-x^2)dx ([itex]\pi[/itex]r[itex]^{2}[/itex]h)
    dw = [itex]\pi[/itex](4-x^2)dx x 1000 x 9.8 (density of water and gravitational acceleration) x dx

    Wait...work is force times distance, what would be the distance be? The distance from dx to the 1m mark above the bowl? so x+1? So the Work (W) would be dW * (1+x)?

    And the integral would be that as well? and the limits of the definite integral would be 0 to 2?
     

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    Last edited: May 25, 2013
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