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Homework Help: Second moment of area

  1. Apr 12, 2007 #1
    1. The problem statement, all variables and given/known data

    Ok, I have the normal Y(up) and x(to the right) axis. I've got the curve(starting at orgin) y^2=x. On the x axis, it goes to x =4 inches. So, that is the area i'm looking at(shaded).

    Question states this:

    Find the second moment of the shaded area with respect to
    a) x - axis
    b) y - axis

    I'm not exactly sure how to do this....but I tried.

    Oh, and I found the top of the curve, is at y=2.

    2. Relevant equations



    3. The attempt at a solution


    Ix = int(y^2 * dA)

    = int(from 0 to 2)(y^2 *(4-y^2) dy = 4.27in^4

    Is this portion correct? I'm not sure

    Thanks,
    Brad
     
  2. jcsd
  3. Apr 12, 2007 #2

    Dick

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    An integral dA is an integral over an area. So you will get a double integral over dx and dy. Look up an example.
     
  4. Apr 12, 2007 #3
    I have looked at examples....I'm fairly confident this is how they did it.

    Where do you see that I went wrong? I think my equations are correct....
     
  5. Apr 12, 2007 #4

    Dick

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    Yes, your basic equations are good. Sorry. I didn't realize you'd already done the x integration. But draw the figure and think about your limits of integration.
     
  6. Apr 12, 2007 #5
    I have the picture drawn, on my paper here.

    I was thinking, that due to integrating in terms of y, I should integrate from zero to the height of the curve.

    So, if x=4, then y = sqrt(4) = 2. This is why zero to 2 seemed logical to me.
     
  7. Apr 12, 2007 #6

    Dick

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    Why not -2 to +2 in y? Does it say only y>=0? If you've included a picture, I can't see it.
     
  8. Apr 12, 2007 #7
    no.......i didn't include a pic.....I said, in the first post(confusing, I know)......that we are starting at (0,0)....like the axis is at the begining of this curve(coming out of orgin)....I can post a pic, if you'd like...with paint.
     
  9. Apr 12, 2007 #8

    Dick

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    Don't include a pic. It's probably not necessary. But can y be negative? I know x can't.
     
  10. Apr 12, 2007 #9
    Nope.....both start at 0,0......then goes out to 4 inches on the x axis......with that curve(y^2=x)......and need to find how tall it is(on the y-axis).
     
  11. Apr 12, 2007 #10

    Dick

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    The curve y^2=x has both of the points (4,2) and (4,-2) on it. So the area enclosed between x=4 and x=0 has regions both above and below the x-axis. I'm asking if you are being specifically asked to not include the region below. If so then your answer is already correct.
     
    Last edited: Apr 12, 2007
  12. Apr 12, 2007 #11
    ok........Yes, I've only got a picture of quadrant 1 Guess that is how I should have put it, all along.....

    Ok....that is great....thank you!!!!
     
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