Change of Variables over an $n$-dimensional ball

Question

I wish to understand what is going on in this change of variables, specifically what the transformation function and Jacobian are. The result is as follows, if we are integrating a function over a an $n$-dimensional ball of radius $r$ centered at the origin, then the following change of variables holds. Let $t \in [0,r]$ and $\omega \in \partial B_n(0,1)$, setting $y=t\omega$ gives, $$ \int_{B_n(0,r)}f(y)dy=\int_0^r \int_{\partial B_n(0,1)} f(t\omega)t^{n-1}d\omega dt. $$ I will explain my understanding and I would very much appreciate it if anyone could tell me where I'm going wrong. Let's specialise to $n=2$ for ease of calculations. To apply the change of variables we need to find an appropriate transformation function, I will call this $G$. Set $y=(y_1,y_2)=G(t,\omega)=(t\omega_1,t\omega_2)$. Now the problem is our $G$ is a function of 3 variables and not 2. However we can use the fact $\|\omega\|=\sqrt{\omega_1^2+\omega_2^2}=1$, thus we must have $\omega_2=\pm \sqrt{1-\omega_1^2}$. Our change of variables is then, $y=(y_1,y_2)=G(t,\omega)=(t\omega_1,t(\pm \sqrt{1-\omega_1^2}))$, which is now a function of two varibles as required, now we just calculate the Jacobian, $$ \left| \dfrac{\partial G(t,\omega_1)}{\partial(t,\omega_1)}\right|=\begin{vmatrix}t & \omega_1 \\\ \dfrac{t\omega_1}{\pm\sqrt{1-\omega_1^2}}& \pm\sqrt{1-\omega_1^2}\end{vmatrix}= t\sqrt{1-\omega_1^2}-\dfrac{t\omega_1^2}{\sqrt{1-\omega_1^2}}=\dfrac{t}{\sqrt{1-\omega_1^2}}. $$ Thus we get this extra factor of $(\sqrt{1-\omega_1^2})^{-1}$. I don't know how to get rid of this, and I also don't like the fact I have a plus-minus hanging around. This argument easily generalizes to $n$-dimensions, and we get the Jacobian to be, $$ \dfrac{t^{n-1}}{\sqrt{1-(\omega_1^2+\cdots+\omega_{n-1}^2)}}. $$ This must be incorrect but I don't see why, any help is greatly appreciated.

Answer 1

The change of variables formula does not apply directly to the spherical coordinates formula. That requires the Coarea formula which is harder to prove.

Why doesn't it apply? Well, when precisely stated, (a version of) the change of variables formula applies to diffeomorphisms $U\cong V$ of open subsets of $\Bbb R^n$, for some $n$. With spherical coordinates, you're seeking to use $\Bbb R^n\setminus\{0\}\cong(0,\infty)\times S^{n-1}$, which is true but not applicable. There is a problem with dimension (RHS lives most naturally in $\Bbb R^{n+1}$) and with openness ($S^{n-1}$ is not open in $\Bbb R^n$).

A third problem is that, while $\mathrm{d}\omega$ can be made to make sense, it's not the Lebesgue measure and the change of variables formula is a statement about integrating with Lebesgue measure, among other things. If you don't know what I mean by that, then I can rephrase it as this; integration against $\mathrm{d}\omega$ is not "ordinary" Riemann integration and the change of variables formula isn't applicable (directly).

So that's what's going wrong!

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N.B. integration against $\mathrm{d}\omega$ around the sphere $S^{n-1}$ can be viewed (rigorously) as the measure-theoretic integral against the Hausdorff measure viewing $S^{n-1}$ as an $(n-1)$-dimensional metric space, up to a normalisation constant.

Answer 2

NOT AN ANSWER - LONG COMMENT ON THE DEFINITION OF THE SURFACE INTEGRAL

As I mentioned in comments, the problem is that $d\omega$ has not been defined. How are you supposed to compute $$\tag{1} \int_{\partial B(0,1)} f(\omega)\, d\omega,$$ with some explicit $f$ (such as, say $f(\omega)=\omega_1^2$)? If you do not know how to compute that, it makes little sense to try to use this symbol in more advanced computations.

This (1) is an example of a surface integral. Hypersurfaces such as $\partial B(0,1)$ do not have a preferred coordinate system. This is in contrast with the Euclidean space $\mathbb R^n$, in which all definitions can be worked out in the standard Cartesian coordinate system and then pulled back to other coordinates if wanted. Therefore (1) does not really make sense until we choose a coordinate system, so that we can expand it in a ordinary multiple integral.

Tangential remark. Sometimes you cannot choose one single coordinate system for the whole surface. In that case you need to chop the integral in several parts. This is the fundamental idea of differential geometry.

For the $(n-1)$-dimensional unit sphere $\partial B(0, 1)$, which is a subset of $\mathbb R^n$, a convenient coordinate system is the polar coordinates; \begin{equation*} \begin{split} x_n&= \cos \theta, \\ x_{n-1}&=\sin \theta \cos \phi_1, \\ x_{n-2}&=\sin \theta \sin \phi_1\cos \phi_2,\\ \vdots & \\ x_2 &= \sin \theta \sin \phi_1\sin \phi_2\ldots \cos \phi_{n-2}, \\ x_1&= \sin \theta \sin \phi_1\sin \phi_2\ldots \sin \phi_{n-2}. \\ \end{split} \end{equation*}

The $n=3$ case is detailed nicely on MathWorld.

This coordinate system satisfies a very nice recursion relation that allows to define the surface element $d\omega_{n-1}$ on the $(n-1)$-dimensional sphere $\partial B(0,1)$ recursively: $$ d\omega_{n-1}=(\sin \theta)^{n-2}d\theta d\omega_{n-2}.$$ (I have commented a tiny bit on this formula in this older answer, at the very end).

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Only now that $d\omega$ has been defined we are in position to compute the formula $$ \int_{\mathbb R^n} f(y)\, dy = \int_0^\infty \int_{\partial B(0,1)} f(t\omega)t^{n-1}dtd\omega.$$

I had thought of doing so, but I really cannot devote more time to this, so I am stopping here. Hope that what I wrote could be of some use nonetheless.