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I am preparing for an exam and am having trouble calculating the following integral.

$$\iint\limits_B \exp\bigg(\frac{y-x}{y+x}\bigg)\,dx\,dy$$

where $B$ is the the interior of the triangle with vertices at $(0,0), (0,1),$ and $(1,0)$.

I have made the change of variables $u = y - x$ and $v = y + x$ which gives me:

$$\frac{1}{2}\iint\limits_{B*} e^\frac{u}{v}\,du\,dv$$

However, I am not sure how to find the proper surface, $B_*$, over which to integrate.

Any help with this would be greatly appreciated.

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In the original coordinates, the region was given by the set of inequalities $$0\leq x, 0\leq y\text{ and }x+y\leq 1.$$

All you have to do is to transform these inequalities to your new system of variables. Using $x=\tfrac12 (v-u)$ and $y=\tfrac12 (u+y)$, we have $0\leq x= \tfrac12 (v-u)$ from which it follows that $$u\leq v. \tag{1}$$ Equations (1) and (2) imply that $v\geq0$. Moreover, we need $0\leq y = \tfrac12 (u+v)$ which results in $$ u \geq -v. \tag{2}$$ The last inequality reads $$v= x+y \leq 1. \tag{3}$$

The relations (1), (2), and (3) bound the region for the new set of variables. And you should have no troubles in converting these into boundaries for your integrals.

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  • $\begingroup$ I figured this out right as you posted this, but actually your answer helped me better understand that with which I was having such trouble before. The integral becomes: $\frac{1}{2}\int_0^1 \int_{-v}^v e^\frac{u}{v} dudv$ I thank you very much! $\endgroup$ – Zéychin Mar 26 '12 at 7:29

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