Why absolute values of Jacobians in change of variables for multiple integrals but not single integrals?

If g:[a,b]R is a change of 1D coordinates, then the formula is:
g(b)g(a)f(x)dx=baf(g(t))dxdtdt.(1)

If T={x=f(u,v);y=g(u,v)} is a change of 2D coordinates, then the formula is:

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.

Answer

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.

Attribution
Source : Link , Question Author : ShungChing , Answer Author : StrangerLoop

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