# Intuition behind “surface measure” in integration formula for polar coordinates

In "Real Analysis" by Stein & Shakarchi, the polar coordinates of a point $x \in \mathbb{R}^d - \{0\}$ are the pair $(r,\gamma)$ where $0 < r < \infty$ and $\gamma$ belongs to the unit sphere $S^{d-1} = \{x \in \mathbb{R}^d : |x| =1\}$, determined by $$r = |x|, \quad \gamma = \frac{x}{|x|}.$$ Furthermore, polar integration is defined as $$\int_{\mathbb{R}^d} f(x)\,dx = \int_{S^{d-1}} \int_0^\infty f(r\gamma) r^{d-1}\,drd\sigma(\gamma).$$ This is explained in the text as the result of defining two measure spaces $(X_1,\mathcal{M}_1,\mu_1)$ and $(X_2,\mathcal{M}_2,\mu_2)$, where $\mathcal{M}_1$ is the collection of Lebesgue measurable sets in $X_1 = (0,\infty)$ and $d\mu_1(r) = r^{d-1}\,dr$.

Regarding the second measure space, we take $X_2 = S^{d-1}$ and the authors say that when $E \subseteq S^{d-1}$, to consider the set $\tilde{E} = \{x \in \mathbb{R}^d : x/|x| \in E, 0 < |x| < 1\}$ to be the subset of the unit ball. If $\tilde{E}$ is Lebesgue measurable, we say that $E \in \mathcal{M}_2$ and define $\mu_2(E) = \sigma(E) = d\cdot m(\tilde{E})$, where $m$ is the Lebesgue measure in $\mathbb{R}^d$.

I am seeking intuition behind the "surface measure." How should I interpret the set $\tilde{E}$; that is, what does this set look like? Furthermore, if the measure $\mu_2 = \sigma$ is defined in terms of the Lebesgue measure in $\mathbb{R}^d$, what is the significance of the factor of $d$? Why not define $\sigma(E) = m(\tilde{E})$?

• Was my answer helpful – mathworker21 Mar 17 '18 at 18:40
• Yes; it's more clear now. – Hayden Mar 17 '18 at 20:56
• lol feel free to upvote my answer – mathworker21 Mar 18 '18 at 3:26

Take $E \subseteq S^{d-1}$. Then $\tilde{E}$ is the set of all points in the unit ball lying on the line segments connecting the origin to a given point of $E$ (except for the origin). The motivation for the definition of $\tilde{E}$ is as follows. The main goal is to put a measure on a $d-1$-dimensional subset of $\mathbb{R}^d$ (namely, $E$). We want, for example, the measure of the upper half of $S^{d-1}$ to be $\frac{1}{2}$. Clearly the Lebesgue measure on $\mathbb{R}^d$ is not what we want to directly measure $E$. So the idea is to introduce that $\tilde{E}$, which clearly preserves these "ratios". Indeed, if we take $E$ to be the upper half of $S^{d-1}$, then $\tilde{E}$ is the upper half of the unit ball in $\mathbb{R}^d$.
So we want to define $\sigma(E) := m(\tilde{E})$ where $m$ is the $d$-dimensional Lebesgue measure. Except for one possible issue. What do we want the measure of the whole $S^{d-1}$ to be? It holds that $S_{n-1} = nV_n$ where $S_n,V_n$ are the surface area and volume of an $n$-dimensional unit ball. So, we want $\sigma(S^{d-1}) = d m(\tilde{E})$, which is where the '$d$' comes from.