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Ok, I've seen some questions similar to mine but it didn't really get me what I want so I figured I'd ask. I was given the following problem to solve by making the change of variables $ u = x-y, v = x+y$ in the following integral

$ I = \int_0^1dy\int_0^{1-y}e^\frac{x-y}{x+y}dx$

This is an iterated integral over an region of the x-y plane. The Jacobian of the transformation is 1/2 so the integral becomes

$I = \frac{1}{2}\int\int_Ae^\frac{u}{v}dudv $

At this point I had some difficulties. I eventually realized that if I played with the numbers, that u would have upper and lower bounds of 1 and -1 while v would have upper and lower bounds of 1 and 0. So my first attempt was this:

$ I = \frac{1}{2} \int_0^1dv\int_{-1}^1e^\frac{u}{v}du$

however evaluating this integral was practically impossible (if you don't believe me, try finding an antiderivitive for $xsinh(1/x)$ with elementary techniques!), eventually I realized the integral would simplify if changed the limits as follows:

$ I = \frac{1}{2} \int_0^1dv\int_{-v}^ve^\frac{u}{v}du$

this yielded the correct answer [$I=0.5sinh(1)$] however I would like to know how to determine how to find regions of integration on change of variables in ways other than guessing at the answer! Is there an algorithmic procedure? Is there an easy graphical procedure? I know I could graph in the u-v plane but that seems like a lot more work than necessary.

edit: the way I am thinking about this now is that the first set of boundaries are incorrect because they correspond to a region that is a box (rectangle) in the uv plane. Cleary, the region is triangular in the x-y plane, and I'm assuming it is also triangular in the uv plane. How do I determine the dependence between u and v in of this region?

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it is clear that u and v have some dependence, but I'm not sure how to determine exactly that dependence. Is it possible to get u strictly as a function of v from these equtions? –  Timtam Aug 4 '12 at 2:41

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This is an attempt to describe a picture by using a thousand words. The original integral is over the triangle with corners $(0,0)$, $(1,0)$, and $(0,1)$.

Because of the shape of the integrand, we let $u=x-y$ and $v=x+y$. To see what this means for the geometry, note that $u=c$ is the equation of a line $x-y=c$, that is, of a line with slope $1$. Similarly, $v=c$ is the equation of a line $x+y=c$ with slope $-1$.

The $u$-axis (that is, the line $v=0$) is the line with slope $-1$ through the origin. Similarly, the $v$-axis is the line with slope $1$ through the origin.

Now turn our diagram through $45^\circ$, or twist one's neck a bit. Our original triangle is symmetrical about the $v$-axis. The $x$-axis is the line $u=v$, and the $y$-axis is the line $u=-v$. The other boundary of our triangle of integration is just the line $x+y=1$, or more simply $v=1$. In terms of $u$ and $v$, our triangle has coordinates $(0,0)$, $(1,1)$, and $(-1,1)$.

A natural way to integrate over our triangle is to integrate first from $u=-v$ to $u=v$, and then to integrate from $v=0$ to $v=1$. That is exactly what you arrived at.

Remark: Can't argue with success, you got to the right integral. But I would be unable to get at the answer without using the above-described geometry. The point is that we use a geometric transformation, in this case a simple rotation, to make the problem collapse.

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clearly it is a 45 degree rotation... I actually was able to realize this... I'm a little concerned though: How would I solve a problem under less easy to graph transformations? Do I have to make a graph every time I change variables? I'd just like to know if there are some other techniques? –  Timtam Aug 4 '12 at 3:42
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For these things I am quite geometrically oriented. After one is familiar with the basic linear transformations, one can begin to visualize at least some non-linear ones. If you use $u=f(x,y)$, $v=g(x,y)$, a useful first step is to visualize the curves $u=c$, $v=d$ where $c$ and $d$ are constants. –  André Nicolas Aug 4 '12 at 3:48

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