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I recently started learning change of variables for (double/triple) integration. One concept that I've struggled to understand is why the linear transformation from one coordinate system to another changes the shape of our rectangle to a parallelogram. I've read through many articles and my textbook, but none give a direct explanation of why this phenomenon occurs.

The following webpage displays an example of this change in geometry:

https://math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/vcalc/change/change.html

Notice how the transformation causes the rectangle to transform into a parallelogram.

I've spent a lot of time reading various articles, but I still do not understand what causes this change.

I would greatly appreciate it if people could please take the time to explain why this phenomenon occurs and what the benefits of it are to us.

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closed as off-topic by TheGeekGreek, JMP, Juniven, hardmath, Daniel W. Farlow Apr 4 '17 at 14:51

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  • $\begingroup$ How is my question off-topic? I would appreciate advice on how I can improve my question. $\endgroup$ – The Pointer Apr 4 '17 at 7:14
  • $\begingroup$ It is unclear what you mean by "why" a rectangle is transformed (by a linear transformation) into a parallelogram. You seem to say that you've seen examples where this happens, but this doesn't exclude that you might want a proof that the image of a rectangle is always a parallelogram (if not again a rectangle, or something degenerate like a line segment or a point). Asking "why this phenomenon occurs" could leave Readers wondering if you think a linear transformation might create a triangle (or a circle?) instead of a parallelogram, and "what the benefits" are is a bit subjective. $\endgroup$ – hardmath Apr 4 '17 at 14:14
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Linear transformations map lines to lines, this is something you may already know or otherwise can show easily.

Slightly more general than your statement: linear transformations also map parallelograms to parallelograms. Note that a parallelogram is completely determined by three (non-collinear) points (vertices), call them $\vec a$, $\vec b$ and $\vec c$. The parallelogram is then given by the set of all points $\vec x$: $$\vec x = \vec a + \lambda \bigl( \vec b - \vec a \bigr)+ \mu \bigl( \vec c - \vec a \bigr) \quad \quad \left( 0 \le \lambda, \mu \le 1 \right)$$ A linear transformation $L$ maps these points to: $$\begin{align} L(\vec x) & = L\left(\vec a + \lambda \bigl( \vec b - \vec a \bigr)+ \mu \bigl( \vec c - \vec a \bigr)\right) \quad \quad\quad \left( 0 \le \lambda, \mu \le 1 \right)\\ & = L(\vec a) + \lambda \bigl( L(\vec b) - L(\vec a) \bigr)+ \mu \bigl( L(\vec c) - L(\vec a) \bigr) \end{align}$$ And the set of all points $L(\vec x)$ is now a parallelogram determined by the points $L(\vec a)$, $L(\vec b)$ and $L(\vec c)$. Note that these vertices are not collinear since $\vec a$, $\vec b$ and $\vec c$ weren't.


I would greatly appreciate it if people could please take the time to explain why this phenomenon occurs and what the benefits of it are to us.

In the context of integration, you can try to use this to map an ugly parallogram to a nice(r) parallogram, at least as a region of integration. If you can map an arbitrary parallogram to a rectangle in the coordinate space, the integral limits will become simple constants.

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  • $\begingroup$ Thanks for the response. Since you were able to shed further light on this concept, I found an excellent webpage that elaborates on the linear algebra surrounding this idea: mathinsight.org/…. Thank you for the assistance. $\endgroup$ – The Pointer Apr 4 '17 at 10:20

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