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For a solid of revolution formed by rotating $0\le y\le r(x)$ about the horizontal axis, with nonnegative continuous radius $r$ on $[a,b]$, each perpendicular slice is a disk of area $\pi r(x{)}^2$. Adding the slice volumes by a definite integral gives $V=\pi {\int }_a^{b}r(x{)}^{2}dx$. If the rotated region instead has inner and outer radii $0\le r(x)\le R(x)$, subtract the missing inner disks: $V=\pi {\int }_a^b[R(x{)}^2-r(x{)}^2]dx$. This is often called the washer method. A complete revolution is already included in the circular cross-sectional area, so no further factor of $2\pi$ belongs in the volume integral.

Consider the uniform solid of revolution about the $x_3$-axis whose boundary isFor total mass $M$, the center of mass lies a distance $3h/5$ from the pointed origin, and the principal moments about that origin are

Put $z=x_3$. The body is a solid of revolution with disk radiusIts volume and uniform mass density are

By the continuous center of mass formula, the distance of the center of mass from the pointed origin is

For a disk of radius $R$, $\int r^{2}dA=\pi R^4/2$. Therefore the axial moment isFor the transverse $x_1$-axis, $\int x_2^{2}dA=\pi R^4/4$, while every slice also contributes $z^2$ times its mass. HenceBy axial symmetry $I_2=I_1$. These values agree with the moment of inertia of a pointed cubic-profile solid of revolution.