1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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!!!!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Second moment of area
  1. Second Moment of Aria (Replies: 2)

  2. Second moment of area (Replies: 5)

Loading...