Why does using the Jacobian to change variables of integration work.
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Suppose you wanted to calculate some double integral over some region
we would have
then suppose we wish to change into polar coordinates to calculate the integral we then get
where we get the
by multiplying
by
where
and so
becomes
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
that pops up when changing from cartesian to polar is not something that is obvious to me at all.






![\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} \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}](https://www.thestudentroom.co.uk/latexrender/pictures/86/864f7fb4b26964698ced1d4c1ed4b5c9.png)


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

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#2
(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.
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.



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

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

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:

This makes sense since when



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:

and everything is fine.
Sue now tries to relabel the x-axis in other more complex ways. For example, she sets

Her integral is now:

where


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

![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 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](https://www.thestudentroom.co.uk/latexrender/pictures/2a/2a5c545c5dd30e68c62f76bf23acaf51.png)
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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#3
Are you asking why you have to use an area element when changing coordinate systems?
If so, the Wikipedia article is pretty good.
If so, the Wikipedia article is pretty good.
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(Original post by atsruser)
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#5
(Original post by poorform)
Funny, informative and illuminating. Many thanks to you. Shame it won't let me rep.
Funny, informative and illuminating. Many thanks to you. Shame it won't let me rep.
So e.g. when you are given:

and you put





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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:
and you put
so that
then you can see that the
is the width fudge factor that arises when you start with the integral
and make the substitution
. 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)
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:

and you put






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#7
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:


But for the substitution





i.e. we scale by the ratio of the infinitesimal width in x-land to that in u-land:


With this line of reasoning, we have to be comfortable with considering the

In brief:


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


where


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

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(Original post by atsruser)
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You have been an insane help. Cheers.
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