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Moment of Inertia of Shapes

  1. Apr 16, 2008 #1
    Ok im try to understand how to find the moment of inertia of different shapes by using direct integration. In a lot of the solutions I see they always have this one thing that I dont understand. Example:

    Determine by direct integration the moment of inertia of the shaded area
    with respect to the y axis.



    I know the formula is

    Iy = INTEGRALOF(x^2 dA)

    doesnt da = xdy

    so that makes it

    Iy =INTEGRALOF(x^2 * x *dy)
    Last edited: Apr 16, 2008
  2. jcsd
  3. Apr 16, 2008 #2


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    Homework Helper

    Hi salman213,

    It appears to me that the way they are setting up the integral is by dividing the area into infinitesimal strips of length x and height dy. Each of these strips has the form of a thin rod which is rotating about one end, and that is where the (1/3) comes from. (The moment of inertia of a thin rod rotating about one end is [itex]\frac{1}{3}ML^2[/itex].)

    The integral then sums up the moments of inertia of these strips.
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