78

If $g:[a,b]\to\mathbf R$ is a change of 1D coordinates, then the formula is: $$ \int_{g(a)}^{g(b)}\,f(x)\,dx = \int_a^b\,f(g(t))\frac{dx}{dt}\,dt. \qquad\text{(1)}$$

If $T=\{x=f(u,v); y=g(u,v)\}$ is a change of 2D coordinates, then the formula is: $$\iint_{T(R)}\,\phi(x,y)\,dx\,dy = \iint_R\,\phi(f(u,v),g(u,v)) \left| \frac{\partial(x,y)}{\partial(u,v)} \right|\,du\,dv \qquad\text{(2a)}$$

It seems that the formula for 2D (and higher dimensions) are extensions of the 1D version. If so, then the 1D version 'should' also require absolute value. Can anyone explain why, except for 1D, do all higher versions require the absolute value of Jacobians?

Does it have something to do with the way we specify the lower and upper limits of the 1D integral? If so, can anyone elaborate it further?

===== Below were added after this question was flagged as a duplicated ======

This question was flagged as duplicated. But it seems to me that this question has a very clear objective: absolute value for higher dimension, but why not 1D? This question is not about any particular example, it is more about the way the theorem is stated in 1D and higher dimension. Specifically, it is more about the challenge of a practical and clerk-like notation of specifying an orientation of a 2D region in double integral.

As elaborated by StrangerLoop and Hurkyl, 1D region such as an interval can be flipped naively just by noting the magnitudes of the endpoints. We can indicate the 'orientation' in a practical and clerk-like manner (as the limits). For double integrals we need to know which side of the region is the 'right' side. As a follow up, I have the following comment:

There is no universal practical and clerk-like way to specify a 2D region. How to specify an oriented region $T(R)$ in a generic manner so that one always mechanically compute a stand-alone double integral correctly, including the sign.

I assume that if we can do that, then it is possible to generalize the 1D version to 2D without the 'artificial' need of absolute value. Because each double integral, as a stand-alone integral, on either side of (2b) will be correctly signed.

$$\iint_{T(R)}\,\phi(x,y)\,dx\,dy = \iint_R\,\phi(f(u,v),g(u,v)) \frac{\partial(x,y)}{\partial(u,v)} \,du\,dv \qquad\text{(2b)} $$

Thanks for the two answers. They are helpful in helping me understand the "why" part. Any further comment on my observation is appreciated.

ShungChing
  • 891
  • 1
  • 7
  • 5
  • My question is really on why the different formulation of the generalization. The answer, as summarized below, is very different form the other question. Seems the 'artificial' absolute value has more to do with the practicality of numerical computation (where 2D regions are just regions without orientation). The 1D version is stayed in a full manner because we are used to the more informative notation of 1D integral. But the 'generalization', as stayed, is not a true generalization because we don't have a practical way to specify the set and the orientation of a 2D region. – ShungChing Jul 05 '14 at 04:25

2 Answers2

73

You are right. 1-d is special and it has to do with switching the limits.

For example, suppose we make the substitution $u=-x$ in $\int_0^1 dx$. Then we get $\int_0^{-1}-1 \: du = \int_{-1}^0 1 \: du = \int_{[-1,0]} 1 \: du$.

Do you see what is happening? If we only keep track of the REGION which $u$ lies in (which is what we do in 2d), then we DO need the absolute value on the Jacobian.

To recap: in 1-d, you have two options: use the absolute value and put the new limits in order from lesser to greater; OR, don't use the absolute value, but put the limits in the same order as the original integral. The first method is exactly what we do in higher dimensions; the second method only works in 1-d.

StrangerLoop
  • 1,731
  • 14
  • 14
  • What is the difference between ∫[−1,0] (subscript) and ∫0−1? (the latter is supposed to be subscript and subscript) – DKNguyen Nov 27 '19 at 18:57
  • 1
    @DKNguyen: $\int_{[-1,0]}$ means we integrate over the interval $[0,-1]$, without specifying an orientation of the interval (we usually specify the orientation so that $-1$ is on the "left" and $0$ is on the "right.") On the other hand, $\int_{-1}^0$ means we integrate over the interval $[-1,0]$ but the orientation of the interval is such that $-1$ is on the left and $0$ on the right. – Alex Ortiz Jul 24 '20 at 17:51
  • @AlexOrtiz Thanks, but why would you not specify orientation? Wouldn't it only make sense to do that if you knew beforehand that it didn't matter? And if you did know beforehand, is there a reason to make it explicit beforehand that it doesn't matter? – DKNguyen Jul 24 '20 at 18:27
  • @DKNguyen: Expressions like $\int_{[a,b]}f(x),dx$ make sense without using the orientability of $[a,b]$. I think it can be useful for the theory to keep the two notions distinct. I am thinking in particular about measure theory and Lebesgue integration, where you define expressions like $\int_E f(x),dx$ in the most general settings you can, other than in $\mathbb R$, and so the theory makes sense when we use Lebesgue integrals on $\mathbb R$. Then there is a theorem about when you can compute a Lebesgue integral using a Riemann integral, at which point you revert to using $\int_a^b$ notation – Alex Ortiz Jul 24 '20 at 20:36
29

If the Jacobian is negative, then the orientation of the region of integration gets flipped.

However, one usually does not learn about the orientation of a region except for one-dimensional integrals or when one starts learning about differential geometry -- to allow introductory texts to talk about change of variables without having to introduce orientations, they make use of the fact that

$$ \iint_R f \, dA = \iint_{-R} -f \, dA $$

so you get the same result if you flip the orientation of the region back to the positive orientation and you flip the sign of the Jacobian: thus, the use of absolute values.

  • Nice explanation! However, why do we not use this fact in 1D? In particular why do we retain the orientation of an interval even when the scaling factor is negative, which should reverse the orientation? – Allawonder Jul 07 '23 at 09:25