# Steiner Formula for Surfaces in $\mathbb{R}^3$ In the slides we derived a Steiner formula for polyhedral surfaces in $\mathbb{R}^3$, by considering the Minkowski sum with a ball and working out expressions for the areas and curvatures associated with vertices, edges, and faces. But we can also get a Steiner formula for smooth surfaces, using the expressions already derived in class. In particular, recall that for a closed surface $f: M \to \mathbb{R}^3$ with Gauss map $N$, we can obtain the basic curvatures by just wedging together $df$ and $dN$ in all possible ways:
$\begin{array}{rcl} df \wedge df &=& 2NdA \\ df \wedge dN &=& 2HNdA \\ dN \wedge dN &=& 2KNdA \\ \end{array}$
Here $H$ and $K$ denote the mean and Gauss curvature (resp.), and dA is the area form induced by $f$. For sufficiently small $t$, taking a Minkowski sum with a ball is the same as pushing the surface in the normal direction a distance $t$. In other words, the surface
$f_t := f + tN$
will describe the “outer” boundary of the Minkowski sum; this surface has the same Gauss map $N$ as the original one. To get its area element, we can take the wedge product
$\begin{array}{rcl} df_t \wedge df_t &=& (df + t dN) \wedge (df + t dN) \\ &=& df \wedge df + 2t df \wedge dN + t^2 dN \wedge dN, \end{array}$
where we have used the fact that $\alpha \wedge \beta = \beta \wedge \alpha$ when $\alpha,\beta$ are both $\mathbb{R}^3$-valued 1-forms. The list of identities above then yields
$df_t \wedge df_t = (2 + 4tH + 2t^2 K)NdA,$
or equivalently,
$\fbox{$dA_t = (1+2tH+t^2K)dA.$}$
In other words, just as in the polyhedral case, the rate at which the area is growing is a polynomial in the ball radius $t$; the coefficients of this polynomial are given by the basic curvatures of the surface (also known as quermassintegrals!).