= Axial moment of inertia of a solid of revolution
{title2=$I_z=\frac{\rho\pi}{2}\int r(z)^4\,dz$}
A uniform solid with circular cross-sectional radius $r(z)$ can be sliced into thin <discs>. A slice of thickness $dz$ has <mass> $dm=\rho\pi r(z)^2dz$ and axial <moment of inertia> $dI_z=\tfrac12r(z)^2dm$. Thus
$$
M=\rho\pi\int r(z)^2dz,\qquad I_z=\frac{\rho\pi}{2}\int r(z)^4dz.
$$
All slice axes coincide, so no <parallel axis theorem> correction is needed. For a linearly tapered section of length $l$ with end radii $r_0,r_1$, direct <integration> gives $M=\rho\pi l(r_0^2+r_0r_1+r_1^2)/3$ and $I_z=\rho\pi l(r_0^4+r_0^3r_1+r_0^2r_1^2+r_0r_1^3+r_1^4)/10$.
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