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Suppose $C$ is a smoothly bounded convex body (*) in $\mathbb{R}^d$, and $p$ is a point on the boundary of $C$. Let $r>0$ and let $B(r)$ denote the ball of radius $r$ centered at $p$.

Is it true that $\mbox{vol }(B(r) \cap C) / \mbox{vol }(B(r)) \to 1/2$ as $r \to 0$?

This seems obvious, but I can't seem to state a good reason why it's true. Does it follow from some well-known theorem?

I would guess that we don't need convexity, and that something simlar similar holds for smoothly embedded hypersurfaces in Euclidean space, and maybe one can relax "smooth" to class $C^2$?

(*) My understanding is that "smoothly bounded convex body" means a compact, convex set, with nonempty interior, with a unique supporting hyperplane at each point. I am not sure how close this is to a convex image of a smooth embedding of a $d$-dimensional ball, but again, I expect that the statement probably holds in either case.
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Quantifying that near a point on a smooth hypersurface, it looks like a tangent hyperplane

Suppose $C$ is a smoothly bounded convex body (*) in $\mathbb{R}^d$, and $p$ is a point on the boundary of $C$. Let $r>0$ and let $B(r)$ denote the ball of radius $r$ centered at $p$.

Is it true that $\mbox{vol }(B(r) \cap C) / \mbox{vol }(B(r)) \to 1/2$ as $r \to 0$?

This seems obvious, but I can't seem to state a good reason why it's true. Does it follow from some well-known theorem?

I would guess that we don't need convexity, and that something simlar holds for smoothly embedded hypersurfaces in Euclidean space, and maybe one can relax "smooth" to class $C^2$?

(*) My understanding is that "smoothly bounded convex body" means a compact, convex set, with nonempty interior, with a unique supporting hyperplane at each point. I am not sure how close this is to a convex image of a smooth embedding of a $d$-dimensional ball, but again, I expect that the statement probably holds in either case.