Change of variables; why do we take the absolute value?

[Hiero]

In transforming an integral to new coordinates, we multiply the “volume” element by the absolute value of the Jacobian determinant.

But in the one dimensional case (where “change of variables” is just “substitution”) we do not take the absolute value of the derivative, we just take the derivative, be it positive or negative.

Why is the single-variable method different from the multi-variable method (in that it lacks the absolute value)?

 

[mfb]

In the one-dimensional case your integral borders will also change signs accordingly, so you don't need the absolute value. In the multi-dimensional case you can't use the integral borders to keep track of the sign.

 

[Hiero]

In the multi-dimensional case you can't use the integral borders to keep track of the sign.

Can you elaborate? We still change the boundaries according to the transformation (or it’s inverse) right? Doesn’t it depend on the choice of parameters? A simple example, (x, y)=(-u, v) (I mean a reflection across the y axis,)
$$1 = \int_0^1\int_0^1dxdy = \int_0^1\int_0^{-1}|J|dudv=-1$$
Why is that wrong to do? x = 1 corresponds to u = -1

 

[mfb]

I think in this case you shouldn't change the integration borders, or take the absolute value of the overall integral.

 

[Hiero]

Is it just an unspoken part of the rule that we must make the integral over the region so that every inner integral always ‘runs upwards’? I.e. if we have an integral with lower-bound a and upper-bound b which are functions like b = b(x, y, ..., z) then we must make sure that b > a over the whole domain (x, y, ... z) which we might be integrating over?

Then I can understand it working with the absolute value |df/dx| in the 1d case, as well as how the previous example should go from ₀¹∫∫_-1⁰dudv

 

[Infrared]

The actual statement of the change of variables formula is that $\int_S f dA=\int_T f\circ\phi |J| dA$ where $\phi$ is diffeomorphism from $T$ to $S$ (In your example, $\phi(u,v)=(-x,y)$). Note that this formula doesn't give you iterated integrals with bounds; it only tells you the region that you're integrating over. You're right than when you write it as an iterated integral, each integral should go from the lower bound to the upper bound- this is because, for example, $\iint_{[0,1]\times [-1,0]}dA$ unambiguously equals $\int_0^1\int_{-1}^0 du dv$

 

[Hiero]

when you write it as an iterated integral, each integral should go from the lower bound to the upper bound-

Thanks. I was not distinguishing the idea of a “multiple-integral” from an “iterated-integral,” which caused me confusion.

Also, I should have read “Fubini’s theorem” more carefully, because it does explicitly state the upper bounds are larger than the lower bounds (so it’s not an unspoken rule, it’s just spoken in a different theorem).