# change of variables using a substitution

Let $D$ be the triangle with vertices $(0,0),(1,0)$ and $(0,1)$. Evaluate $$\iint_D \exp\left( \frac{y-x}{y+x} \right) \,dx\,dy$$ by making the substitution $u=y-x$ and $v=y+x$

My attempt Finding the domain, $D$ $$(x,y) \rightarrow(u,v)$$ $$(0,0) \rightarrow(0,0)$$ $$(0,1) \rightarrow(1,1)$$$$(1,0) \rightarrow(-1,1)$$ Thus the domain is $-1 \le u \le1$ and $0 \le v \le1$ and the jacobian is $-2$ and so the integral I get is $$\int_0^1 \int_{-1}^1 -2e^{\frac{u}{v}} \,du\,dv$$ this is awfully too hard to integrate and i can't figure out where i have gone wrong

• – Ron Gordon Jun 14 '13 at 18:42

## 1 Answer

You're right, except for your limits of integration on $u$. Draw the resulting triangle in the $uv$-plane. What are the limits of $u$ in your picture, or $u$ goes from what to what? It only goes from $-1$ to $1$ when $v = 1$. In general, we have $-v \leq u \leq v$, for $0 \leq v \leq 1$ (draw a horizental line from the left edge of the triangle to the right, for any v between $0$ and $1$). Thus, your integral is $$\int_0^1 \int_{-v}^v -2 e^{\frac{u}{v}} \, dudv$$ You should be able to evaluate this now.

• (+1) You are considering a horizontal strip which is bounded by the lines $u=-v$ and $u=v$. – Mhenni Benghorbal Jun 14 '13 at 18:49
• Oh, I see where i was going wrong. I was taking coordinated from the x,y graph (I don't know how i came up with that) but now I understand. From the new points, I draw them with their new axis and take the integral from that. Thanks guys! :) – amanda Jun 14 '13 at 19:02
• @Amanda: You are welcome. – Mhenni Benghorbal Jun 14 '13 at 19:03