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Math Help - work needed to pump water out a tank

  1. #1
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    Post work needed to pump water out a tank

    a tank is formed by revolving the curve y=x^2 ; 0<=x<=4 ; around y-axis
    if filled with water with weight 10000 N/m^3
    find work needed to empty the tank by pumping it to the top ?

    the problem is that i dont have the height to set the integral
    is it right if i use the radius as 4?

    this way ill get V=PI r^2 dy
    F=10000*V
    W=(?) *F
    integral limits ==??

    any help or hints pls
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  2. #2
    Super Member Aryth's Avatar
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    Here's a hint for the height:

    If the farthest on the x-axis (horizontally) you can go is 4, then would it not be correct that the farthest on the y-axis (vertically) you can go is y= x^2 = 4^2 = 16 ?

    But, if the tank is full when the work begins, then the distance is only a distance x^2 (starting at x=4) from the maximum height. So what do you think the height is?

    And as a result, what does your work integral look like?
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  3. #3
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    Question thnx 4 replying

    actually my work integral is : Force*volume*hight of slab*dy ]with integral limit to cover the whole hieght

    from ur hint i got the limit of integral from 0-16 (cuz im pumping out from the top)

    *do u think its correct if i use it as follows

    w= (16*PI)(10000)(16-y)dy [integral 0-16]???

    thank you
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  4. #4
    Super Member Aryth's Avatar
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    You're right

    W = (16*10000)\pi \int_{0}^{16} [16 - y] ~dy

    Do that and you're finished.
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  5. #5
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    Thumbs up thank u

    thanks alot for ur help
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