We define polar coordinates in $\mathbb{R}^{n}$\ $\{ 0\}$ by $x=ry$, where $r=|x|>0$ and $y \in \partial B(0,1)$ is a point on the unit sphere. In the coordinates, Lebesgue measure has the representation $dx = r^{n-1}drdS(y)$. Where $dS(y)$ is the surface area measure on the unit sphere.
Does anyone know where the differential $dx = r^{n-1}drdS(y)$ comes from and why $dS(y)$ is used instead of just $dy$?
Thanks for any help!
It is because the Lebesgue measure is a product of two measures, one of which is the surface measure in $S^{n-1}$. The construction is as follows (taken verbatim from Folland's book) :
Define $\varphi : \mathbb{R}^n\setminus\{0\} \to (0,\infty)\times S^{n-1}$ by $$ \varphi(x) = \left (|x|, \frac{x}{|x|} =: x'\right ) $$ Define $m_{\ast}$ on $(0,\infty) \times S^{n-1}$ by $$ m_{\ast}(E) = m(\varphi^{-1}(E)) $$ and $\rho_n$ on $(0,\infty)$ by $$ \rho_n(E) = \int_E r^{n-1}dr $$ Then, there is a unique measure $\sigma_{n-1}$ on $S^{n-1}$ such that $$ m_{\ast} = \rho_n \times \sigma_{n-1} $$ We want this measure to satisfy the following : If $f\geq 0$ or $f\in L^1(\mathbb{R}^n)$, one has $$ \int_{\mathbb{R}^n} f(x)dx = \int_0^{\infty}\int_{S^{n-1}} f(rx')r^{n-1}d\sigma_{n-1}(x')dr \qquad\qquad \text{(1)} $$ (This gives the expression you refer to.)
To that end, the measure $\sigma_{n-1}$ is defined as follows :
For any Borel set $E \subset S^{n-1}$, write $$ E' = \varphi^{-1}((0,1]\times E) = \{rx' : 0 < r \leq 1, x' \in E'\} $$ Now if the formula (1) should hold for $\chi_{E'}$, then it must follow that $$ \sigma_{n-1}(E) = n m(E') $$ So we take this to be the definition of $\sigma_{n-1}$
Now one checks that the $m_{\ast} = \rho_n \times \sigma_{n-1}$ by checking it on all subsets of the form $(a,b]\times E$ and then extending it to all Borel sets. Once that is done (1) holds for all simple measurable functions, and then one applies the usual monotone/dominated convergence tricks to extend it to all $f\geq 0$ or $f \in L^1$