# Measure of convex hulls

I'm not an expert of this kind of questions, but I can't give a satisfactory answer to the following question.

Pick $x_1\dots x_n \in \mathbb{R}^m$. Is there a formula for the measure of the convex hull of $x_1\dots x_n$?

Of course it can be that the convex hull of $x_1\dots x_n$ has measure zero because it is contained in a hyperplane of $\mathbb{R}^m$, but if you ask the question for the appropriate Hausdorff measure, the question always make sense...

• This isn't a complete answer. You can assume that the convex hull $S$ has non empty interior - otherwise restrict to the affine subspace generated by the points and look at the problem there (or declare the volume to be zero). To simplify, I'm going to talk in $\mathbb{R}^3$. Now identify the faces of $S$ (whose vertices must be among the $x_i$'s), and pick an arbitrary point $O$ in space. Then the volume of $S$ is the sum of the volumes of the pyramids with vertex $O$ and base one of the faces, where the volume is counted positively if $O$ is on the same side of the face as $S$ is, and.... Jan 4, 2016 at 11:19
• ...negatively otherwise. Since we know the formula for the volume of a pyramid ($1/3 \times$ (area of base) $\times$ height), this reduces the problem to finding the area of the faces, which are convex polygons. Similarly, if you were working in $\mathbb{R}^n$, this would reduce the dimension to $n-1$, and you'd repeat the process. Jan 4, 2016 at 11:22
• It could be that there's some way to do this independently of locating the faces, but I don't know if that's the case. Jan 4, 2016 at 11:23
• Please put what you said in an answer David. Jan 4, 2016 at 11:26

You can assume that the convex hull $S$ has non-empty interior - otherwise restrict to the affine subspace generated by the points and look at the problem there (or declare the volume to be zero). To simplify, I'm going to talk in $\mathbb{R}^3$. Now identify the faces of $S$ (whose vertices must be among the $x_i$'s), and pick an arbitrary point $O$ in space. Then the volume of $S$ is the sum of the volumes of the pyramids with vertex $O$ and base one of the faces, where the volume is counted positively if $O$ is on the same side of the face as $S$ is, and negatively otherwise. Since we know the formula for the volume of a pyramid, $\frac{1}{3}(\mathrm{base} \times \mathrm{height})$, this reduces the problem to finding the area of the faces, which are convex polygons. Similarly, if you were working in $\mathbb{R}^n$, this would reduce the dimension to $n-1$, and you'd repeat the process.