# Change of Variables Problem. How Does One Know What to Select as $u$ and $v$?

I just learned change of variables for double and triple integrals, but I'm having difficulty applying it. When I encounter a problem, I am not sure which variables to set as $u$ and $v$. Unlike single variable substitution ($u$-substitution), there doesn't seem to be any indication of which variables should be selected as $u$ and $v$. In fact, most of the example problems that I have encountered online actually give you the variables $u$ and $v$ to use, rather than having to find them yourself. However, for my studies, I am required to find them myself; this is something I am having trouble with.

Take the following problem I am given:

Evaluate $\iint_D \dfrac{y^4}{x} \ dxdy$ over the region D contained between the parabolas $x = 1 - y^2$ and $x = 4(1 - y^2)$.

The textbook solution says to make $x = v(1 - u^2)$ and $y = u$, but it does not give any indication of how it chose these $u$ and $v$! How did the solution to the above problem find/select $u$ and $v$? As a student, when encountering a problem as above, how can I deduce $u$ and $v$? Is there a proper methodology that is analogous to single variable substitution ($u$ substitution)?

I would greatly appreciate it if people could please take the time to help me understand this methodology.

## 1 Answer

Caveat lector -- this answer is somewhat heuristic.

There is no "methodology". There is a certain mysterious and charming creativity in finding a change-of-variables that solves a particular problem. It involves playing around with the problem, drawing little pictures, understanding the functions involved, and lots of patience. Moreover, it requires a mental process similar to that which might have led Van-Gogh to choose yellow rather than orange.

True, with time and accumulation of collective experience, people notice that some change of variables are more useful than others. For example, if your function has some spherical symmetry, it makes great sense to use spherical coordinates. Or if your function has some other symmetry, or other properties that concern the geometry of the underyling integration domain, then it makes sense to try and use them.

• Thanks for the response. I was worried that this may be the case. Would you please be so kind as to explain how one would come to the same conclusion for change of variables for the mentioned problem and solution? – The Pointer Mar 27 '17 at 13:46
• Sure. You see, the integration domain -- the "band" between the parabolas $x=1-y^2$ and $x=4(1-y^2)$ in the $x-y$ coordinates is mapped onto a nice rectangular "band", between $v=1$ and $v=4$ in the $u-v$ coordinates. (Check this, by wrtiting $x$ and $1-y^2$ in terms of $u,v$!) This is the motivation behind the choice of these coordinates. – uniquesolution Mar 27 '17 at 14:15