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moment of inertia - cylinder

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

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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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In cylindrical coordinates, the mass element is $dM = \\rho dV = \\rho r\\,d\\theta\\,dr\\, dz$. 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 r^3\\rho\\, d\\theta\\,dr\\, dz =\\\\[3mm] \\int_0^{\\var{h}}\\int_0^{2\\pi}\\int_0^{\\var{a}} r^3\\rho\\, dr\\,d\\theta \\, dz = \\rho\\int_0^{\\var{h}}\\, dz \\int_0^{2\\pi}\\,d\\theta \\int_0^{\\var{a}} r^3\\, 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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Thus we obtain 

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\\[I_z = \\rho [z]_0^{\\var{h}} [\\theta]_0^{2\\pi}\\left[ \\frac{r^4}{4}\\right]_0^{\\var{a}} = \\frac{1}{2}\\pi\\rho\\times\\var{h}\\times{a^4} = \\simplify{1/2*pi*{d}*{h}*{a}^4}.\\]

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Also note that $I_z = \\frac{1}{2}M\\var{a}^2$ where $M = \\rho V$, with $V=\\pi\\times\\var{a}^2\\times\\var{h}$. 

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