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Work: Spherical Problem

  1. May 12, 2008 #1
    [SOLVED] Work: Spherical Problem

    1. The problem statement, all variables and given/known data
    A tank is full of water. Find the work W required to pump the water out of the spout. (Use 9.8 for g and 3.14 for [​IMG]. Round your answer to three significant digits.) W=_________

    r=3
    h=1
    [​IMG]

    3. The attempt at a solution
    To set up this problem, I started by taking the area of one slice of water to be pi r[tex]^{2}[/tex] multiplied by an infitesimally small height [tex]\Delta y[/tex] to get volume. Then multiply this by the density of water; 1000 kg/m[tex]^{3}[/tex] to get the volume of one slice.

    Now I want R as a function of, lets say y. We know the area of a circle is r[tex]^{2}[/tex]2+y[tex]^{2}[/tex]=3[tex]^{2}[/tex]. So r=sqrt(9-y[tex]^{2}[/tex]). Also, the distance for any slice from the top is 7-y.

    After this I tried integrating from zero to six of the function 1000*pi*(9-y[tex]^{2}[/tex])*(7-y)*y. This came out to be negative....in fact -36000*pi..

    Apparently, that is wrong. Where did I go wrong?

    Thanks.
     
  2. jcsd
  3. May 12, 2008 #2

    dx

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    Your expression for R as a function of y is wrong.
     
  4. May 12, 2008 #3
    The above equation was really meant to be r[tex]^{2}[/tex]+y[tex]^{2}[/tex]=3[tex]^{2}[/tex]. I just noticed the typo. Or, are you saying that's wrong?
     
  5. May 12, 2008 #4

    HallsofIvy

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    Be careful about what y represents. [itex]r= \sqrt{9- y^2}[/itex] is correct if is the height above the center of the sphere. But in that case, integrating from 0 to 6 is wrong. Integrating from the bottom of the tank to the top takes y from -3 to 3.
     
    Last edited: May 12, 2008
  6. May 12, 2008 #5
    I integrated and got 2469600*pi as my answer. My homework checker told me I was wrong. I even rounded to the proper amount of sig figs(3).

    Heres what I have:
    dA=(9-y[tex]^{2}[/tex])pi
    (Multiplied by infitesimally thickness dy)
    dV=(9-y[tex]^{2}[/tex])pi dy
    (Since d=mv, I multiplied density of water (1000kg/m[tex]^{3}[/tex])
    dM=1000(9-y[tex]^{2}[/tex])pi dy
    (multiplied by gravity 9.8 m/s[tex]^{2}[/tex])
    dF=9800(9-y[tex]^{2}[/tex])pi dy
    (Now multiplied by distance (7-y))
    dW=9800(9-y[tex]^{2}[/tex])(7-y) dy


    Now I integrated dW with the limits from -3 to 3 and got the above result. I'm lost. :confused: Thanks for the help Ivy.
     
  7. May 12, 2008 #6

    dx

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    The distance to the top would be 7 - y if you chose y from the bottom. It appears that you have chosen y as the distance from the center.
     
  8. May 12, 2008 #7
    Great! 4-y worked.

    I can finally put this problem to rest. Thanks for the help everyone!
     
    Last edited: May 12, 2008
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