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Homework Help: Work and Fluid Force: Calculus II

  1. Jun 19, 2010 #1
    I worked out these two problems and got an answer but was not sure about the answer...

    If you could just check it and if it is not write tell me what I did wrong or give me a step by step, it would be greatly appreciated.

    1. A tank is formed by revolving y = 3x^2, x = [0,2] around the y axis is filled to the 4 feet point with water (w = 62.4 lb/ft^3). Find the work necessary to pump the water out of the tank over the top.

    I got the integral being from a = 0 and b = 4 and integrating [12y^(1/2) - y^(3/2)]dy with w and pi as constants outside of the integral.

    Would this be correct?

    2. A plate shaped as in the figure [question 2.jpg is attached] is submerged vertically in a fluid as indicted. Find the fluid force on the plate if the fluid has weight density 62.4 lb/ft^3

    The integral I set up was the limits being -5 to -1 and the integral being (5-y)(-7/4y - 7/4) dy.

    Which is integrated out with the limits it comes out to be 665/6, and I was wondering if it was correct or not [the way I did it]

    Thank you.

    Can someone please step by step show me how to solve this if I am not correct.

    Please help! I greatly appreciate it!
     

    Attached Files:

  2. jcsd
  3. Jun 19, 2010 #2

    LCKurtz

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    No, it wouldn't be correct. Figure out the mass of a circular slab of water of thickness dy and how far it must be lifted. Then integrate that.
     
  4. Jun 19, 2010 #3

    LCKurtz

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    Why are you using 5 - y as the depth? When y = -5 that would be 10. :confused:
     
  5. Jun 19, 2010 #4
    Mass of the circular slab? Like a slice? How do we find the mass of that? Is that not just the 'w' or the 'weight density'

    ?
     
  6. Jun 19, 2010 #5
    I am using 5-y for the depth because the top of the fluid to the bottom of the plate is y-5?

    Is that not correct? I don't understand how it can be 10-y?

    Is the other part of the integral correct for that part?
     
  7. Jun 19, 2010 #6

    LCKurtz

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    I should have said "weight". But anyway, you must multiply the weight density by the volume to get the weight. Your tank has circular cross sections being formed by revolving the curve. Your element of volume would be a cylindrical shaped disk of thickness dy. What is its volume dV?
     
  8. Jun 19, 2010 #7

    LCKurtz

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    Did I say it should be 10 - y??

    Apparently you have taken your origin to be at the water level above the point, judging by your equation of the slanted line. That's fine. What is the depth when y = -1? -2? -3? -4?

    Use a formula that gives correct answers.

    Yes, the other part looks OK.
     
  9. Jun 19, 2010 #8
    We set up the integral in the dy, right?

    So we would use pi(r)^2 for the base of the circle/cylindrical disk
    Meaning pi[ 1/[2(sqrt)3] y^(-1/2)]

    Because I solved y=3x^2 for y and squared it, right?

    And we multiply that by by (12-y)?

    Would I have the weight (density X pi) / 18 root(3) outside the integral and y^(1/2)(12-y) dy inside to integrate?

    I think im more confused then I started. :[
     
  10. Jun 19, 2010 #9
    Wait...so, the answer I have is correct?

    I don't understand this post by you.

    Im sorry!
     
  11. Jun 19, 2010 #10

    LCKurtz

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    When you revolve it around the y axis the radius is measured in the x direction.
     
  12. Jun 19, 2010 #11
    I do have it in the x direction...right?

    Im not sure what you're trying to say...
     
  13. Jun 19, 2010 #12

    LCKurtz

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    No, your formula of 5 - y for the depth is not correct. What does it give when you try y = -1 or -2 or -3 or -4? Does it give the correct depth? Can you fix it?
     
  14. Jun 19, 2010 #13

    LCKurtz

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    I'm saying that the radius of your little disk is the x value not the y value of the point on your curve. Have you drawn a picture?
     
  15. Jun 19, 2010 #14

    Why would I try y = -1 or -2 or -3 or -4?


    I really appreciate your help!
     
  16. Jun 19, 2010 #15

    LCKurtz

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    Because in the coordinate system you have chosen, those are example values of y where the triangle is located.
     
  17. Jun 19, 2010 #16
    So you want me to plug in those values in where to find the exact location of the depth?
     
  18. Jun 19, 2010 #17
    So do I solve the equation for y? And then do the integral?

    I think setting up the integral is the hardest part! And im getting so confused by looking in the book and trying to figure it out and trying to do what you are telling me to!

    :[
     
  19. Jun 19, 2010 #18

    LCKurtz

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    You are using the formula 5 - y for the depth. I am telling you that it doesn't work and you can check it for yourself by trying those example numbers and comparing what you get with the actual depth. You need to use a formula that gives the correct depth.
     
  20. Jun 19, 2010 #19

    Oh. Ok, So the actual depth is 5 ft, correct?

    So if I put 0 into y it will give me the actual depth? Is that what you're saying?

    But how does that help at all?
     
  21. Jun 19, 2010 #20

    LCKurtz

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    In your choice of coordinates, the triangle sits at a level where the y values are between -5 at the bottom and and -1 at the top of the triangle, right? And the actual depth of points in the triangle range from 5 at the bottom to 1 at the top, measured in how far it is under water.

    You need a formula that for any y between -5 and -1 gives the correct depth. 5-y doesn't do it. You are making it much harder than it is.
     
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