# How to change variables in a surface integral without parametrizing

This is a doubt that I carry since my PDE classes.

## Some background (skippable):

In the multivariable calculus course at my university we made all sorts of standard calculations involving surface and volume integrals in $$R^3$$, jacobians and the generalizations of the fundamental theorem of calculus. In order to make those calculations we had to parametrize domains and calculate differentials.

A couple of years later I took a PDE course. We worked with Evans' Partial differential equations book. This was my first experience with calculus in $$\mathbb R^n$$ and manipulations like $$\text{average}\int_{B(x,r)}f(y)\,dy= \text{average}\int_{B(0,1)}f(x+rz)\,dz.$$ This was an ordinary change of variables. $$y=x+rz,\,\,dy=r^n\,dz$$ and the mystery was solved. Like in that case, I was able to justify most of these formal manipulations after disentangling definitions.

That aside, I found these quick formal calculations to be very powerful.

## However,

I realized that I wasn't able to justify this: $$\text{average} \int_{\partial B(x,r)}f(y)dS(y)= \text{average}\int_{\partial B(0,1)}f(x+rz)\,dS(z).$$ I have some vague idea of what's happening: the same substitution as before, but this time the jacobian is $$r^{n-1}$$ because the transformation is actually happening between regions which "lack one dimension". Also, I see some kind of pattern: a piece of arc-length in the plane is $$r\,d\theta$$, a piece of sphere-area is $$r^2 \sin\theta \, d\phi \,d\theta$$, "and so on". Maybe some measure-theoretic argument can help me: I know, roughly speaking, that for any measure $$\mu$$, $$\int_\Omega f\circ \phi \,d\mu=\int_{\phi(\Omega)} f \, d(\mu\circ\phi^{-1}).$$ I'd say $$\phi(z)=(z-x)/r$$ and $$\phi^{-1}(y)=ry+x$$, but I actually don't know how $$dS(y)$$ looks like "as a measure" (It's not a product measure or a restriction of one, but it somehow relates to Lebesgue's in $$\mathbb R^n$$...). Why would I conclude that $$dS(y)\circ \phi^{-1}=r^{n-1}dS(z)$$? I have an intuition, but either I lack the mathematical concepts and definitions to express it or I'm just too confused. Is there some theory that I could learn in order to understand? Maybe something about the measure $$dS$$. Is it expressible in terms of the Lebesgue measure in some way? Or set-theoretically, maybe, without having to resort to $$n-1$$ parameters and complicated relations?

Maybe all of this would not have been a problem if I had ever mastered n-dimensional spherical coordinates. But even so, more generally, is there a way of changing variables when I'm integrating over a subregion of "dimension$$" without necessarily parametrizing?

Sorry for the vagueness, but I don't really know what to ask for exactly.

Note: I saw some of the answers to this post, but none of them were deep enough in the direction I intend.

Note II: If there are no general methods or theories, maybe restricting to linear transformations, to Lebesgue measure exclusively, or to subregions defined by simple expressions like $$g(x)=C$$ or $$g(|x|)=C$$ could get me somewhere.

Edit: I have not yet studied differential geometry, which has been mentioned in a comment. I added it to the tags.

• There is a unifying framework for the question that's bothering you, differential-geometry. In this case, we think of $dS$ as an "$(n-1)$-differential form" in $\Bbb R^n$, and integrate it along the "$(n-1)$-chain" which is your sphere. What you require is called a "pull-back" of $dS$ via the differentiable mapping $y\mapsto x+ry$. Hopefully, this might point you in the right direction. – Jonathan Y. Feb 12 '14 at 9:20
• @JonathanY. Do you know any good books on the subject? Rigorous, with good prose, motivations, nice examples, maybe even illustrations (not a requirement, though). Also, is this some specific subtopic? The Wikipedia article mentions "Riemannian manifolds", "tensors", the Theorema Egregium, and "symplectic geometry". I wouldn't know exactly where to put my attention. – dafinguzman Feb 13 '14 at 8:03
• While I haven't formally studied differential geometry in its broadest setting, I have used parts of it and in this aspect the books Introduction to Smooth Manifolds and Riemannian Manifolds by John Lee were very...smooth. =) – Mark Fantini Feb 13 '14 at 8:14
• @dafinguzman myself, I like Spivak's "calculus on manifolds" or "comprehensive introduction to differential geometry". You're looking for integration on chains/manifolds and the (proper instance of the) change of variable theorem. – Jonathan Y. Feb 13 '14 at 12:20

No parametrization is needed, just some careful uncovering of definitions. Maybe the following can help:

Let $S^n(r)=\{x\in\mathbb R^{n+1}|\|x\|=r\}\subset\mathbb R^{n+1}$ (the sphere of radius $r>0$ centered at the origin).

There is a natural volume form $\mu_{r}$ on $S^n(r)$, i.e. a nowhere-vanishing $n$-form, induced by the volume form $\mu=dx_1\wedge\ldots \wedge dx_{n+1}$ on $\mathbb R^{n+1}$ and the "outward" unit normal $N:S^n(r)\to\mathbb R^{n+1}$, $N(x)=x/r$, and is defined as follows:

$$\mu_r(x)(v_1,\ldots,v_n):=\mu(x)(N(x), v_1, \ldots, v_n)$$ for all $v_1, \ldots, v_n\in T_x S^n(r).$

(More informally, we write $\mu_r=\mu/dr$. Note that $dr(N)=1$).

Next let $\Phi:S^n(1)\to S^n(r)$ be defined by $\Phi(x)=rx$. Then $$\Phi^*\mu_r=r^n\mu_1.$$ This you can prove without any parametrization, just from the definitions above and generalities like the chain rule, pull back, etc.

It then follows, by the change of variables formula, that for any continuous function $f:S^n(r)\to\mathbb R$, $$\int_{S^n(r)}f\mu_r=\int_{S^n(1)} \Phi^*(f\mu_r)=r^n\int_{S^n(1)} (f\circ\Phi) \mu_1.$$

Makes sense?

• Could you maybe elaborate some definitions? I don't know what $n$-forms, $T_x$ or "pull back" are. As you present it, it seems that $\mu$ is a product measure, somehow. Also, I don't know what $dr(N)$ is supposed to mean. I can guess it means (by the notation) that "surface infinitesimal elements" get amplified by $1$ near the points which are on the sphere. – dafinguzman Feb 13 '14 at 8:20
• @dafinguzman: sorry, cannot elaborate, it's too long. I am using standard language taught in most differential geometry courses. You need to take such a course or pick a textbook (popular ones are Spivak, Lee, Guillemin+Pollack. My personal favorite is Arnold's mathematical methods of classical mechanics, intuitive and rigorous at the same time; but it's not easy as a first textbook). Now I am sure it is possible to translate my answer to a more elementary language (although less rigorous). Perhaps another user of the site will try to do it (or even myself if I find the time and energy). – Gil Bor Feb 13 '14 at 16:08
• OK, after a long time I read through some chapters of Spivak's Calculus with Manifolds and now I get what you are trying to say. Trying to adapt to the definitions in Spivak, though, I find that $(\Phi ^ * \mu_r)(x)((v_1)_x,..., (v_n)_x) = r^n \mu (rx) (x, (v_1)_{rx}, ..., (v_n)_{rx})$. Then I'm tempted to say that equals $r^n \mu (x) (x, (v_1)_{x}, ..., (v_n)_{x})$ by "invariance of $\mu$ under translations", which is known for Lebesgue measure in $R^n$. Is that correct? Thanks for your answer: once the definition of "$dS$" is made explicit everything becomes clearer. – dafinguzman Jun 15 '15 at 22:48

I know this is an old question, but I thought this explanation might be helpful to some.

By definition (in $\mathbb R^3$):

$$\int_{\partial B(\pmb x,r)}f(\pmb y)dS(\pmb y)= \int_U f(\pmb y(s,t))\left\|\frac{\partial\pmb y}{\partial s}\times\frac{\partial\pmb y}{\partial t}\right\|dsdt$$

Now, observe that $f(\pmb y)=f(\pmb x+r(\frac{\pmb y-\pmb x}{r}))$, and that if $\pmb y(s,t)$ is a parametrization of $\partial B(\pmb x,r)$ for $(s,t)\in U$, then $\frac{\pmb y(s,t)-\pmb x}{r}$ is a parametrization of $\partial B(\pmb 0,1)$ for $(s,t)\in U$. Finally we observe that

$$\left\|\frac{\partial\pmb y}{\partial s}\times\frac{\partial\pmb y}{\partial t}\right\|= r^2\left\|\frac{\partial}{\partial s} \left (\frac{\pmb y-\pmb x}{r} \right )\times\frac{\partial }{\partial t} \left (\frac{\pmb y-\pmb x}{r} \right )\right\|$$

So if we let $\pmb z(s,t)=\frac{\pmb y(s,t)-\pmb x}{r}$, then we have

$$\int_U f(\pmb y(s,t))\left\|\frac{\partial\pmb y}{\partial s}\times\frac{\partial\pmb y}{\partial t}\right\|dsdt= r^2\int_U f(\pmb x +r\pmb z(s,t))\left\|\frac{\partial\pmb z}{\partial s}\times\frac{\partial\pmb z}{\partial t}\right\|dsdt\\= r^2\int_{\partial B(\pmb 0,1)}f(\pmb x+r\pmb z)dS(\pmb z)$$

• This is great! I see that the cross product gets in the way of immediately generalising this reasoning to $\mathbb R^n$. I'm guessing this is more or less the role of the wedge products in Gil Bor's answer. – dafinguzman Sep 25 '17 at 6:10
• @dafinguzman I think you can generalize if you use the fact that in $\mathbb R^n$ $\int_{\partial B(\pmb x,r)}f(\pmb y)dS(\pmb y)= \int_U f(\pmb y(\pmb z))\left|\det\left(\frac{\partial\pmb y}{\partial z_1}, \dots,\frac{\partial\pmb y}{\partial z_{n-1}},\pmb n\right)\right|d^{n-1}\pmb z$, where $\pmb n$ is the normal vector to the surface, and that $\pmb n$ does not change when the surface is scaled and translated. – user5753974 Sep 26 '17 at 10:08
• @user5753974 Hi, is $\pmb z$ a $(n-1)$-tuple? And does $d^{n-1}\pmb z$ mean $dz_1 dz_2 ... dz_{n-1}$? Thanks! – Sam Wong Oct 11 '18 at 5:48

I think the measure theoretic approach works fine, note that the surface measure is n-1 dimensional Hausdorff measure, in general for the s-dimensional Hausdorff measure we have $$H^{s} (rA)=r^{s} H^{s}(A)$$ and this measure is translation invariant. Now use the measure theoretic change of variable formula.

• Does this mean that the measure $dS$ in the integral over $\partial B$ is the $n-1$ dimensional Hausdorff measure? I hadn't thought of it that way. – dafinguzman Feb 19 at 21:34
• Yes, the surface measure $dS$ which is used on the boundary of n-dimensional objects is exactly the $n-1$ Hausdorff measure on the boundary. – Arya Jamshidi Feb 20 at 5:10