Two definitions of integral on a sphere: equal? How do I prove that? In class, we defined $\int_{S^{N-1}}f\mathrm{d}\sigma$ in the following way. Locally, $S^{N-1}$ is the graph of a map from $\mathbb{R}^{N-1}$, for example the inverse projections, one of which is $\phi(x)=\sqrt{1-|x|^2}$. Locally, we may say:
$$\int_{S^{N-1}}f\mathrm{d}\sigma:=\int_{B_{N-1}(1)}f(x,\phi(x))\sqrt{1+\nabla\phi(x)^2}dx.$$
If $f$ is not supported on one of the sets where the $S^{N-1}$ is a graph, use a partition of unity to reduce the integral to several integrals with integrands supported on such a set, and the use the above expression. Willem's book defines:
$$\int_{S^{N-1}}f\mathrm{d}\sigma:=N\int_{B_N}f\left(\frac{x}{|x|}\right)dx,$$
where $B_N$, I guess, is the $N$-dimensional unit ball. This definition is used to prove polar coordinates integration (cfr. here), and I would like to add a proof of this fact to my notes, so I would need to show these two are equal and then the proof is from the book. But how are these equal? I was thinking of starting with indicators and then extending to simple functions by linearity, positive functions by monotone convergence, and arbitrary functions again by linearity. But how do I go about the indicator case?
 A: Here are some additional details on the straightforward brute force approach that I suggested in the comments:
Let's run a substitution $x\to (r,t)$, with $0\le r\le 1$ and $t\in B_{N-1}$, and these variables are related by $x=r(t,\phi(t))$. Notice that (except for the origin) this sets up a bijective correspondence between the range indicated and $x\in B_N$. We obtain that
$$
N\int_{B_N} f(x/|x|)\, dx = N \int_0^1 dr\int_{B_{N-1}}dt\, f(t,\phi(t))J(r,t) ,
$$
with
$$
J(r,t) = |\det (\partial x_j/\partial t_k)| = \begin{vmatrix} t_1 & r & 0 && \ldots & 0 \\ t_2 & 0 & r & 0 & \ldots & 0 \\ &&&\ldots &&\\ t_{N-1} & 0& 0 &&\ldots & r\\ \phi &r\phi_1 & r\phi_2 && \ldots & r\phi_{N-1}
\end{vmatrix} .
$$
By induction on $N$ (expand along the first row), we find that $J=r^{N-1}|\phi - t\cdot \nabla\phi|$. Since
$$
\nabla\phi(t) = -\frac{t}{\sqrt{1-|t|^2}},
$$
we can rewrite this as
$$
\phi-t\cdot\nabla\phi = \frac{1}{\sqrt{1-|t|^2}}=\sqrt{\frac{1-|t|^2+|t|^2}{1-|t|^2}} =\sqrt{1+|\nabla\phi|^2} ,
$$
so the first formula is obtained after integration with respect to $r$.
