Second moment of area ($I$)

The second moment of area of a cross-section about an axis is $I=\int y^{2}dA$, where $y$ is the perpendicular distance from that axis. Its dimensions are length to the fourth power. For a circular section of radius $a$, polar integration gives $I={\int }_0^a{r}^{3}dr{\int }_0^{2\pi }{\sin }^2\theta d\theta =\pi a^4/4$. A slender rod with Young's modulus $E$ has filament bending modulus $B=EI$: the axial strain caused by curvature $\kappa$ is $-y\kappa$, and integrating the linear elasticity energy density $E{y}^2{\kappa }^2/2$ over the section gives $B{\kappa }^2/2$ per unit length.

The polar second moment of a planar area about a perpendicular axis is the integral of squared distance from that axis over area. It has units ${\mathrm{m}}^4$ and equals the sum of second moments about two perpendicular in-plane axes through the same origin. For a disk of radius $a$, $J_p={\int }_0^{a}2\pi r^{3}dr=\pi a^4/2$. It is geometric, not a mass moment of inertia; a uniform cylinder has axial mass moment $\rho L{J}_p$.

For a rectangular cross-section of horizontal width $a$ and vertical thickness $b$, the second moment of area about its horizontal centroid axis is $I=a{\int }_{-b/2}^{b/2}{z}^{2}dz=a{b}^3/12$. Thus increasing vertical thickness is especially effective at reducing the curvature caused by a given bending moment.