Is multiplication of differentials commutative in integrals?

Question

I was recently working on a double integral that I would like to solve through a change of variables to polar coordinates. The integral in question was:

$\displaystyle I=\int_{x=0}^{x=3}\int_{y=-\sqrt{9-x^2}}^{y=\sqrt{9-x^2}}(x^3+xy^2)dydx\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; (1)$

This integral looked like it was made to be solved using a change of variables. After a change of variables, one arrives at:

$\displaystyle I = \int_{\theta=-\pi/2}^{\theta=\pi/2}\int_{r=0}^{r=3}r^4\cos\theta drd\theta\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; (2)$

This integral does yield the correct answer. However, I find it very illogical that we range over $\theta$ in the outer integral. I asked my profsseor how we could change this to make the outer integral range over $r$, to which she responded that $drd\theta = d\theta dr$, wherefore the order can be inverted by changing the order of multiplication of the differentials, but that we may need to change the integration bounds.

What I do not understand is how $drd\theta =d\theta dr$, if that may entail a change of integration bounds. That suggests to me precisely that they aren't equal. If they were equal in the traditional sense, no change of bounds would be necessary. In fact, I always thought that clusters like $dydx$ were sloppy notation for $dy \wedge dx$. But that does not seem to make sense in our integral, because simply flipping the order of multiplication and adding a negative sign does not yield the correct answer.

Answer 1

In $\int_{\theta=-\pi/2}^{\theta=\pi/2}\int_{r=0}^{r=3}r^4\cos\theta drd\theta$, there is no subterm $drd\theta$, because of the implicit parenthesis: $$\int_{\theta=-\pi/2}^{\theta=\pi/2}\left(\int_{r=0}^{r=3}r^4\cos\theta dr\right)d\theta.$$ This nested double integral is equal, by Fubini's theorem, to $$\int_{r=0}^{r=3}\left(\int_{\theta=-\pi/2}^{\theta=\pi/2}r^4\cos\theta d\theta\right)dr,$$ which is, similarly, denoted by $$\int_{r=0}^{r=3}\int_{\theta=-\pi/2}^{\theta=\pi/2}r^4\cos\theta d\theta dr$$only by abuse of notation.

Answer 2

It's important to maintain the distinction between the integral over a two-dimensional region and the iterated (double) integral.

In the expression

$$ \int_\Omega f(x,y) \ \text dx \text dy $$

we have $\text dx \text dy := |\text dx \wedge \text dy|$ as a singular unit. Note that this is not multiplication, it's just a notation. At this point, the order doesn't really mean anything; we choose such an integral (of an absolute differential form) specifically to avoid a dependence on orientation. So, when we swap the order, nothing happens.

$$ \int_\Omega f(x,y) \ \text dy \text dx = \int_\Omega f(x,y) \ \text dx \text dy $$

When we want to actually do any kind of calculation, though, we usually resolve this into an iterated integral. The resulting expression depends on the order in which that integration takes place, because it changes how we express the geometry of $\Omega$ as limits of integration. Notably, it doesn't change the integrand or the region at all, which is consistent with what we see above.

The resulting expressions

$$ \int_a^b \left( \int_{c(y)}^{d(y)} f(x,y) \ \text dx \right) \text dy = \int_p^q \left( \int_{r(x)}^{s(x)} f(x,y) \ \text dy \right) \text dx $$

necessarily split the differential across the two iterations of the integral (that's the whole point of this method), and so we can no longer talk about what it means to reverse their order without reversing the order of the iterations.

To address the concern with change of variables, we can consider that happening before the resolution into iterated integrals. So, we first note that

$$\int_\Omega f(x,y) \text dx \text dy = \int_\Omega rf_p(r,\theta) \text dr \text d\theta$$

and then we do the iteration method on the latter integral.