1
$\begingroup$

When learning about double integrals, we learn that the only way we can do them is if Fubini's theorem applies (or maybe I should say, the only way I learned to do them in that one class). But then when we went over surface integrals, we could always find a parametrization so that we could convert our surface integral (which looks an awful lot like a double integral) into an iterated integral.

Is Fubini's theorem the reason we can always convert a surface integral into an iterated integral? If so, how?

$\endgroup$
0
$\begingroup$

No it isn't. There's a theorem that says that all "manifolds" (in your case, ignore this word and substitute the word "surfaces") are locally graphs. That is to say: we can always cut them up into pieces that look like

$$y=f(x_1,\ldots, x_n)$$

in your case this means pieces that look like

$$z=f(x,y)$$

which is a parametrization for that piece of the surface in question. Then you can add up the pieces.

Fubini's theorem only refers to the ability to interchange the order of integration on a given piece.

Edit: I should emphasize: do not try to do this in practice. The theorem is out there to ensure that the parametrizations actually exist so we can satisfy the definition, in real life we find ways around this to do integration in a more intelligent way, and I was just answering the question you asked which was: "is Fubini responsible?"

$\endgroup$
2
  • $\begingroup$ This is misleading, as typically we do surface integrals parametrically (e.g., spherical coordinates $(\phi,\theta)$ or cylindrical coordinates $(z,\theta)$) over a region in the parameter space. It is less common to split into explicit graphs. $\endgroup$ – Ted Shifrin Jul 11 '14 at 20:57
  • $\begingroup$ Yes, I tried to emphasize this is only "possible" from an abstract standpoint, I'll edit to try and put more oomph into "don't actually try to do this in practice." $\endgroup$ – Adam Hughes Jul 11 '14 at 21:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.