Why does using the Jacobian to change variables of integration work.

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poorform
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Suppose you wanted to calculate some double integral over some region \displaystyle D we would have \displaystyle \iint_D f(x,y)dA=\iint_D f(x,y)dxdy then suppose we wish to change into polar coordinates to calculate the integral we then get \displaystyle \iint_D f(x,y)dxdy= \iint_{D'} f(rcos\varphi,rsin \varphi)rdrd \varphi where we get the \displaystyle rdrd\varphi by multiplying \displaystyle \det {\bf J} by \displaystyle drd\varphi where \displaystyle \mathbf J(r, \varphi) = \begin{bmatrix}  \dfrac{\partial x}{\partial r} & \dfrac{\partial x}{\partial\varphi}\\[1em]  \dfrac{\partial y}{\partial r} & \dfrac{\partial y}{\partial\varphi} \end{bmatrix}= \begin{bmatrix}  \cos\varphi & - r\sin \varphi \\  \sin\varphi &   r\cos \varphi \end{bmatrix} and so \displaystyle dxdy becomes \displaystyle rdrd \varphi after changing variables.

We can do a similar process for changing into spherical coordinates or polar coordinates in three dimensions.

I understand how to do this but I really don't know why the Jacobian is used for such a process. Is there a video or at least some intuitive explanation as to why we do these steps when switching from one coordinate system to another to calculate the area. I mean I wouldn't expect it to be a straight sway considering the region is likely to look a lot different under the transformation but I would like to understand (or at least have an appreciation of) why we do what we do. For example the extra \displaystyle r that pops up when changing from cartesian to polar is not something that is obvious to me at all.
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atsruser
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(Original post by poorform)

I understand how to do this but I really don't know why the Jacobian is used for such a process. Is there a video or at least some intuitive explanation as to why we do these steps when switching from one coordinate system to another to calculate the area.
Consider the 1-d case first where we find the area under y=x^2 from x=1 to x=4.

We draw a graph and label the x-axis and y-axis and draw the curve. We then compute the integral:

I = \int_{x=1}^{x=4} x^2 \ dx = 21

We give the picture to our friend Sue. She doesn't like our x-axis for some reason, and relabels it so that it is called the u-axis and where we wrote 1 she writes 2, where we wrote 4, she writes 8; in short, she creates a u=2x axis.

She wants to find the area under the graph and she knows that it must be 21 - relabelling a graph can't change the real, measurable area, which we could find by counting squares on grid paper. She realises that she needs the height of the graph on her new axis not to change so she does the integral:

I = \int_{u=2}^{u=8} \frac{u^2}{4} \ du

This makes sense since when u=2, \frac{u^2}{4} = 1 which is the same height when working with the corresponding x value of 1. Also when u=8, \frac{u^2}{4} = 16 which corresponds to x=4 \Rightarrow x^2=16.

She does the integral and it comes to 42, which twice as big as it should be. Susan is sad for a while, but then realises that while the y heights are fine, the distances on her u-axis are double those for the original x-axis. She now realises how to fix it up: she needs a width fudge factor of 1/2 to bring the widths in u-land back to the right scale. So she now calculates:

I = \int_{u=2}^{u=8} \frac{u^2}{4} \frac{1}{2} \ du = 21

and everything is fine.

Sue now tries to relabel the x-axis in other more complex ways. For example, she sets u=x^2. She notes that now the u-axis is stretched in a non-linear way compared to the last relabelling - this tells her that her width fudge factor has to vary along the u-axis, and in fact must get bigger along the u-axis since the x-gap from 1-2 goes to the u-gap 1-4 (size 3) but the x-gap 2-3 goes to the u-gap 4-9 (size 5). And more, she can see it must vary continuously along the u-axis so it must be some function of u.

Her integral is now:

I=\int_{u=1}^{u=16} u g(u) \ du

where g(u) is her width fudge factor function. She figures out by looking at infinitesimal gaps on the x-axis that she needs to set

g(u)=\frac{dx}{du}

to make the widths the right size all along the u-axis. (I'll leave you to check this).

So, with u=x^2\Rightarrow x=\sqrt{u} \Rightarrow \frac{dx}{du} = \frac{1}{2\sqrt{u}}, her integral is now:

I=\int_{u=1}^{u=16} u \frac{1}{2\sqrt{u}} \ du = \int_{u=1}^{u=16} \frac{\sqrt{u}}{2} \ du = [\frac{u^{3/2}}{3}]^{16}_1 = 21

Now, it should be clear to you that in the 2 and 3-d cases, we will need a similar area or volume fudge factor to ensure that the values of our integrals come out right when we make a change of variables. In this case, the fudge factors are called Jacobians, and you can look up the details of how they are computed, but in essence, they merely describe the area or volume ratios of an infinitesimal chunk of the region to be integrated, in the original and transformed axis systems.
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alow
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Are you asking why you have to use an area element when changing coordinate systems?

If so, the Wikipedia article is pretty good.
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poorform
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(Original post by atsruser)
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Funny, informative and illuminating. Many thanks to you. Shame it won't let me rep.
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atsruser
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(Original post by poorform)
Funny, informative and illuminating. Many thanks to you. Shame it won't let me rep.
One other point that I forgot: you already know about this 1-d fudge factor approach since it's what you do when you use a substitution to do an integral. It is never explained in a geometric way though, which is a great pity as that approach makes clear why it works, and naturally leads to the idea of the Jacobian in higher dimensions.

So e.g. when you are given:

I = \int \cos x e^{\sin x} \ dx

and you put u = \sin x \Rightarrow du = \cos x dx so that I = \int e^u du then you can see that the \cos x is the width fudge factor that arises when you start with the integral \int e^u \ du and make the substitution u = \sin x. Here of course, we are doing things backwards from how I described it in my earlier post (since that lets us write the original integral-with-fudge-factor in a nicer form without fudge factor)
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poorform
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(Original post by atsruser)
One other point that I forgot: you already know about this 1-d fudge factor approach since it's what you do when you use a substitution to do an integral. It is never explained in a geometric way though, which is a great pity as that approach makes clear why it works, and naturally leads to the idea of the Jacobian in higher dimensions.

So e.g. when you are given:

I = \int \cos x e^{\sin x} \ dx

and you put u = \sin x \Rightarrow du = \cos x dx so that I = \int e^u du then you can see that the \cos x is the width fudge factor that arises when you start with the integral \int e^u \ du and make the substitution u = \sin x. Here of course, we are doing things backwards from how I described it in my earlier post (since that lets us write the original integral-with-fudge-factor in a nicer form without fudge factor)
Yeah that makes sense to me Thanks again.
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atsruser
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(Original post by poorform)
Yeah that makes sense to me Thanks again.
There are a couple of other points I may as well make while I'm thinking about it:

1. When we do the integrals we are computing the limit of the sum of rectangles (Riemann integration and all that). The (infinitesimal) areas of the rectangles are given by (height x width) which are:

f(x) dx in the x-axis system
f(u) du in the u-axis system

But for the substitution u=2x, the infinitesimal du is twice as big as the infinitesimal dx (whereas the heights f(x), f(u) are the same) - it's clear that to get the right area, we need to scale down like so:

du \rightarrow \frac{dx}{du} du (=dx)

i.e. we scale by the ratio of the infinitesimal width in x-land to that in u-land: dx : du. Note that this is precisely the same notation introduced by Leibniz for the derivative that we use all the time:

\int f(x) \ dx = \int f(u) \frac{dx}{du} \ du

With this line of reasoning, we have to be comfortable with considering the du's both as infinitesimal lengths and as part of the indivisible derivative "dx by du" notation.

In brief: \frac{dx}{du} is the ratio of infinitesimal lengths, and it's obvious that the dx must go on top to get the area right.

2. In 2-d, we have to do the same thing but with a Jacobian, which gives us the ratio of infinitesimal areas, and it's now clear what the notation must look like, and what goes on top: if we transform (x,y) \rightarrow (u,v) then we write:

\iint f(x,y) \ dx dy = \iint f(u,v) \frac{\partial(x,y)}{\partial(u,  v)} \ du dv

where \partial(x,y) is the infinitesimal area that we are summing over in xy-land, and the Jacobian is the ratio of that to the infinitesimal area that we are summing over in uv-land, namely \partial(u,v)

To get an actual formula for the Jacobian requires some more thought, of course, but that's not too hard: it is based on a vector expression for the area of a parallelogram (which is what the rectangular dx dy transforms to in u-land, in general) and is thus a cross product form which we evaluate via a determinant.
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poorform
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(Original post by atsruser)
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Ahhh that makes sense I guess. Since dxdy is the area of an infinitesimally small parallelogram in the xy plane so evaluating the cross product will give the area of such a parallelogram. And cross products are determined using matrix determinants hence why we use them to calculate the Jacobian. Hmmm interesting stuff indeed. I think I understand it well enough to be getting on with, I will nail it fully in some time hopefully.

You have been an insane help. Cheers.
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