I'm am slightly confused while trying to keep everything straight between looking at integration as on a manifold vs. the diffeomorphism change of variables.

Consider a smooth domain $B\subset \mathbb{R}^2$ and the domain $B_s\subset\mathbb{R}^3$ such that $B_s=B\times[0,s]$

Question 1: Consider the integral $$I=\int_{\partial B_s} f(x)\, d\sigma_{B_s}$$ where $\sigma_{B_s}$ indicates the integral over the boundary (surface measure). If we change variables, $(x_1,x_2,x_3)\to (x_1,x_2,s\xi)$, then I think we have $$I=s\int_{\partial B\times (0,1)} f(x_1,x_2,s\xi) d\sigma_B d\xi+\int_{B\times\{0,1\}} f(x_1,x_2,s \xi)dx_1 dx_2$$ (where $d\sigma_B$ is the surface measure on $B\subset\mathbb{R}^2$) is this correct? It bothers me that a $s$ does not comes out of the second integral although I know intuitively why (the top part has no "volume" in the third direction). I can't seem to reconcile this with the diffeomorphism version of the change of variables formula, and I would like to understand it better.

Question 2: If we considered $$A=\int_{\partial B_s} \frac{\partial}{\partial_{x_3}} f(x) d \sigma_x,$$ we know from the chain rule that $\displaystyle\frac{\partial}{s\partial \xi}=\frac{\partial}{\partial_{x_3}}$. So when we change variables, does the problem become: $$A=\int_{\partial B \times (0,1)}\frac{1}{s}\frac{\partial}{\partial \xi}f(x_1,x_2,s\xi) s\, d\xi d\sigma_B+\int_{B \times \{0,1\}}\frac{1}{s}\frac{\partial}{\partial \xi}f(x_1,x_2,h\xi) dx_1 dx_2?$$ I am not positive how derivatives inside of the integral are affected by change of variables

It seems both of my confusions would be solved if I understood integration on manifolds better... Thank you.

  • $\begingroup$ Why are you calling $d\xi$ a coordinate? $\endgroup$
    – Muphrid
    Aug 8, 2013 at 3:22
  • $\begingroup$ @Muphrid I corrected it. $\endgroup$
    – toypajme
    Aug 8, 2013 at 5:00
  • $\begingroup$ @toypajme it seems to me the boundary ought to have three parts: $\partial B \times [0,s]$, $B \times \{ 0 \}$(base) and $B \times \{ s \}$(top). On the base and top the coordinates $x_1,x_2$ serve as parameters while $x_3$ is fixed at $0$(base) or $s$(top). So I disagree with your claim. Or, $B \times \{ 1 \}$(top) if you make the change of variables. Generally, I'm a bit confused about your use of $d\sigma$ you have an $x,d,B$, I think I might know, but a bit more detail about your intent would help the answerer. $\endgroup$ Aug 8, 2013 at 5:22
  • $\begingroup$ @JamesS.Cook Those were remnants from bad notation I originally used. Hopefully I've fixed them. For $d\sigma$ I am trying to indicate the integral over the surface portion (and indicate which surface). Which claim do you disagree with? I wrote $B\times\{0,1\}=B\times\{0\} \cup B\times \{1\}$. $\endgroup$
    – toypajme
    Aug 8, 2013 at 5:45

1 Answer 1


The formula you have in Question 1 is correct. As a sanity check, put $f\equiv 1$ in it: the result should describe the surface area.
$$A=s\int_{\partial B\times (0,1)} d\sigma_B\, d\xi+\int_{B\times\{0,1\}} dx_1\, dx_2 = s\int_{\partial B} d\sigma_B+\int_{B\times\{0,1\}} dx_1\, dx_2 $$ Now the fact that $s$ is present only in front of the first integral makes perfect geometric sense: stretching in vertical direction leaves the areas of top and bottom surfaces unchanged, while the side area is multiplied by the stretch factor.

Concerning question 2: to minimize confusion, it helps to use the notation $f_3(x_1,x_2,x_3)$, where subscript indicates differentiation in the third variable (whatever it is called). Since $f_3$ is just an ordinary function, the formula from Question 1 applies to it:
$$\int_{\partial B_s} f_3(x)\, d\sigma_{B_s} = s\int_{\partial B\times (0,1)} f_3(x_1,x_2,s\xi) d\sigma_B d\xi+\int_{B\times\{0,1\}} f_3(x_1,x_2,s \xi)dx_1 dx_2 \tag1$$ The next step is to relate $f_3(x_1,x_2,s\xi)$ and $\frac{\partial}{\partial \xi}f(x_1,x_2,s\xi)$. This is what the chain rule does: $$\frac{\partial}{\partial \xi}f(x_1,x_2,s\xi) = f_3(x_1,x_2,s\xi)\, \frac{\partial(s\xi)}{\partial \xi} =f_3(x_1,x_2,s\xi)\,s $$ Therefore, $f_3(x_1,x_2,s\xi)$ can be replaced by $\frac{1}{s}\frac{\partial}{\partial \xi}f(x_1,x_2,s\xi)$ everywhere in (1): $$\int_{\partial B_s} \frac{\partial }{\partial x_3}f(x)\, d\sigma_{B_s} = \int_{\partial B\times (0,1)} \frac{\partial}{\partial \xi}f(x_1,x_2,s\xi) d\sigma_B d\xi + \frac{1}{s}\int_{B\times\{0,1\}} \frac{\partial}{\partial \xi}f(x_1,x_2,s \xi)dx_1 dx_2 $$

By the way, it's a good idea to write $\frac{\partial }{\partial \xi} f(x_1,x_2,s\xi)$ (as you have done) instead of $\frac{\partial f}{\partial \xi} (x_1,x_2,s\xi)$. I find the latter notation a frequent source of confusion in calculations like these - do we take derivative first and rescale later, or the other way around?


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