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moment of inertia - cylinder

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Determine the moment of inertia of the cylinder $C$ of density varies as $\\rho = \\var{d}xy$, height $\\var{h}$, and radius $\\var{a}$ about the axis passing through the centres of its circular faces.

", "advice": "

We can, naturally, consider this problem in cylindricla coordinates. 

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We may assume that the cylinder $C$ sits in the space such that the axis rotation described in the question is the $z$-axis.

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Then $C$ can be described by 

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\\[0\\leq r \\leq \\var{a} \\quad \\mbox{and } \\quad 0\\leq \\theta\\leq 2\\pi \\quad\\mbox{and}\\quad 0\\leq z\\leq h.\\]

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We also have to turn $\\rho(x,y)$ into polar coordinates. 

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\\[\\rho(x,y) = \\var{d}(r\\cos\\theta)(r\\sin\\theta) = \\var{d}\\cdot r^2\\cos\\theta\\sin\\theta.\\]

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Then, in cylindrical coordinates, the mass element becomes $dM = \\rho dV = r^2\\cos\\theta\\sin\\theta \\,r\\,d\\theta\\,dr\\, dz$. Also, observe that each mass element is at the distance of $r$ to the axis of rotation. Then the moment of inertia is 

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\\[I_z = \\int_C r^2 \\, dM = \\int_C r^2\\rho\\, dV =\\iiint_C \\var{d}\\cdot r^5 \\cos\\theta\\sin\\theta\\, d\\theta\\,dr\\, dz =\\\\[3mm] \\var{d}\\int_0^{\\var{h}}\\int_0^{2\\pi}\\int_0^{\\var{a}} r^5\\cos\\theta\\sin\\theta\\, dr\\,d\\theta \\, dz = \\var{d}\\int_0^{\\var{h}}\\, dz \\int_0^{2\\pi} \\cos\\theta\\sin\\theta\\,d\\theta \\int_0^{\\var{a}} r^5\\, dr . \\]

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Note that for the last equality we used the fact that boundaries of the integrals are constants.

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We can compute the integrals separately as follows:

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Thus we obtain 

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\\[I_z = \\var{d}\\times\\var{h}\\times 0\\times \\frac{\\var{a^6}}{6} = 0.\\]

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