Question

Is there a simple way to calculate/estimate the volume of Minkowski sum of an n-dimensional unit ball and an n-dimensional ellipsoid? Even a simple ellipsoid like $\frac{x_1^2}{a^2} + x_2^2 + \ldots + x_n^2 = 1$ will do.

Answer 1

I guess you know that the result can be written as a polynomial. $$w_0+w_1{\cdot}r+\dots+w_n{\cdot}r^n.$$ So your question is how to estmate the coefficients $w_i$; this is so called "cross-sectional measures" and they can be defined for any convex body $K$.

• $w_0$ is the volume of $K$,
• $w_n$ is the volume of unit ball.
• $w_1$ is the area of the boundary of $K$, I think there is no good formula for ellipsoid, but estimates are known and you could write it as an integral.
• I would be surprised if some remaining $w_i$ can be expressed by simple formula.

If you want to write $w_i$ as an integral, check for example Burago--Zalgaller, Geometric inequalities. For example $$w_i=\mathrm{MayBeAConst}\cdot\int\limits_{\partial K} \sigma_{i-1}(k_1,k_2,\dots,k_{n-1})\, d\mathrm{area}.$$ where $\sigma_i$ is the $i$-th elementary symmetric polynomial and $k_i$ are principle curvatures. In your case it is easy to find $k_i$...

Answer 2

For more on this, check out "Surface area and other measures of ellipsoids", by one I. Rivin:

[math/0403375] Surface area and other measures of ellipsoids by I Rivin ► Surface area and other measures of ellipsoids - Elsevier by I Rivin - 2007 Advances in Applied Mathematics 39 (2007)

Answer 3

Because your ellipsoid has all axes but one of equal length we don't need the "full theory":

Let $x_1=:x$ and $(x_2, \ldots, x_n)=:y\in{\mathbb R}^{n-1}$. The Minkowski sum $M$ of the unit ball and the given ellipsoid is rotationally symmetric with respect to the $x$-axis and consists of all points $z=(x'+x'', y'+y'') \in {\mathbb R}^n$ with $${x'^2\over a^2}+|y'|^2\leq 1\ ,\qquad x''^2 +|y''|^2\leq 1\ .$$ A hyperplane $x={\rm const.}$ intersects $M$ in an $(n-1)$-dimensional ball whose radius can be found by maximizing $$|y'+y''|\leq |y'|+|y''|\leq \sqrt{1-x'^2/a^2} + \sqrt{1-x''^2}$$ under the constraint $x'+x''=x$.