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Math Help - Moment of inertia

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    MHF Contributor arbolis's Avatar
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    Moment of inertia

    I'd like to know where I went wrong :
    A lattice with constant density \rho (x,y)= \rho is on the region under the curve y= \sin x from x=0 to x=\pi. Find the moments of inertia I_x and I_y.
    My attempt : I_x :
    I realize that the region is an upper semi circle with radius r=1.
    I_x=\iint _D y^2 \rho (x,y)dA=\int_0^{\pi} \int_0^{1} (\rho \sin (\theta))^2 \rho ^2 d \rho d\theta=\frac{\pi}{10}, which mean that I_x does not depend on the mass of the lattice!
    I'm guessing that my error comes from the first step, that is \iint _D y^2 \rho (x,y)dA=\int_0^{\pi} \int_0^{1} (\rho \sin (\theta))^2 \rho ^2 d \rho d\theta. Should I replace y by \sin x, so y=\sin (\cos (\theta))?
    If so, then I don't understand why I can't take y as \rho \sin (\theta)...
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  2. #2
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    Quote Originally Posted by arbolis View Post
    I'd like to know where I went wrong :
    A lattice with constant density \rho (x,y)= \rho is on the region under the curve y= \sin x from x=0 to x=\pi. Find the moments of inertia I_x and I_y.
    My attempt : I_x :
    I realize that the region is an upper semi circle with radius r=1.
    I_x=\iint _D y^2 \rho (x,y)dA=\int_0^{\pi} \int_0^{1} (\rho \sin (\theta))^2 \rho ^2 d \rho d\theta=\frac{\pi}{10}, which mean that I_x does not depend on the mass of the lattice!
    I'm guessing that my error comes from the first step, that is \iint _D y^2 \rho (x,y)dA=\int_0^{\pi} \int_0^{1} (\rho \sin (\theta))^2 \rho ^2 d \rho d\theta. Should I replace y by \sin x, so y=\sin (\cos (\theta))?
    If so, then I don't understand why I can't take y as \rho \sin (\theta)...
    You've got to be careful. You are using \rho for two different things here. A density and a changing radius.
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