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Math Help - Moment of inertia

  1. #1
    s7b
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    Moment of inertia

    Can someone explain how to set this up;

    Find the moment of inertia about the z-axis of a this shell of constant density cut from the cone 4x^2+4y^2-z^2=0, z is greater than 0 by the circular cylinder x^2+y^2=2x


    The formula for moment of inertia is:
    I=double surface integral of (x^2+y^2)density ds

    -How do you know which is the surface?
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  2. #2
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    Quote Originally Posted by s7b View Post
    Can someone explain how to set this up;

    Find the moment of inertia about the z-axis of a this shell of constant density cut from the cone 4x^2+4y^2-z^2=0, z is greater than 0 by the circular cylinder x^2+y^2=2x


    The formula for moment of inertia is:
    I=double surface integral of (x^2+y^2)density ds

    -How do you know which is the surface?
    the problem is telling you that the surface is the cone z=2 \sqrt{x^2 + y^2} and the region of integration is D=\{(x,y): \ x^2 + y^2 \leq 2x \}.

    you'll eventually need to use polar coordinates. note that then the limits of integration will be: \frac{-\pi}{2} \leq \theta \leq \frac{\pi}{2}, \ \ 0 \leq r \leq 2 \cos \theta, \ .
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  3. #3
    s7b
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    Do you just use the density for example just as "b"
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  4. #4
    s7b
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    Also when you find ds,
    how the heck do you find (dz/dx)^2
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