Willmore energy

Remarks

For the sphere of radius $r$, one has $H^2 = 1/r^2$ and area $4\pi r^2$, so $\mathcal{W} = 4\pi$, for every $r$. For the cylinder $x^2 + y^2 = r^2$, which is not compact, $H^2 = 1/(4r^2)$ is constant and the area is infinite, so the integral diverges.

The same formula defines $\mathcal{W}$ for a smooth immersion of a compact surface into $\R^3$; the value does not depend on the orientation, since $H^2$ does not. Since $H^2 - K = (k_1 - k_2)^2/4$ and $\int_S K \, dA = 2\pi\chi(S)$ by the Gauss–Bonnet theorem, $$\mathcal{W}(S) = 2\pi\chi(S) + \frac{1}{4} \int_S (k_1 - k_2)^2 \, dA.$$ Some authors call $\int_S (H^2 - K)\, dA$ the Willmore energy; it differs from $\mathcal{W}$ by the topological constant $2\pi\chi(S)$. For every such $S$, $\mathcal{W}(S) \ge 4\pi$, with equality exactly for round spheres.