# Thread: Double Integrals

1. ## Double Integrals

Evaluate $\displaystyle \int\int_Re^{\frac{x-y}{x+y}}$ where $\displaystyle R=\{(x,y)\in\mathbb{R}^2:0\leq x,0\leq y,x+y=1\}$ using $\displaystyle x-y=u,x+y=v$.

I got $\displaystyle \frac{1}{2}\int_0^2\int_u^1e^{\frac{u}{v}}dvdu$ but maple says im wrong.

2. Originally Posted by Abu-Khalil
Evaluate $\displaystyle \int\int_Re^{\frac{x-y}{x+y}}$ where $\displaystyle R=\{(x,y)\in\mathbb{R}^2:0\leq x,0\leq y,x+y=1\}$ using $\displaystyle x-y=u,x+y=v$.

I got $\displaystyle \frac{1}{2}\int_0^2\int_u^1e^{\frac{u}{v}}dvdu$ but maple says im wrong.
Your limits of integration are incorrect see the diagram

So $\displaystyle u=-v..v;v=0..1$

I hope this helps

3. Originally Posted by TheEmptySet
Your limits of integration are incorrect see the diagram

So $\displaystyle u=-v..v;v=0..1$

I hope this helps
Not really >.<

I did $\displaystyle x=\frac{u+v}{2},y=\frac{v-u}{2}$ and then

$\displaystyle y=0\iff u=v$
$\displaystyle y=1-x\iff =\frac{v-u}{2}=1-\frac{u+v}{2}\iff v=1$
$\displaystyle x=0\iff u=0$
$\displaystyle x=1\iff u=2$

could you do it in the same way?

4. Originally Posted by Abu-Khalil
Not really >.<

I did $\displaystyle x=\frac{u+v}{2},y=\frac{v-u}{2}$ and then

$\displaystyle y=0\iff u=v$
$\displaystyle y=1-x\iff =\frac{v-u}{2}=1-\frac{u+v}{2}\iff v=1$
$\displaystyle x=0\iff u=0$
$\displaystyle x=1\iff u=2$

could you do it in the same way?
yes really

On the bottom of the triangle$\displaystyle y=0$ and $\displaystyle x=0..1$

using this we get in the u-v plance

$\displaystyle u=x-y \iff u=x-0 \iff u=x$ and
$\displaystyle v=x+y \iff v=x+0 \iff v=x$ so as x goes from 0 to 1

u and v go to 1. so we get the equation $\displaystyle u=v$

Now on the hypotenuse $\displaystyle y=-x+1$ so we get

$\displaystyle u=x-y \iff u=x-(-x+1) \iff u=2x-1$ and
$\displaystyle v=x+y \iff v=x+(-x+1) \iff v=1$ so as x goes from 1 to 0

So we get the horizontal line segment $\displaystyle v=2$ as u goes from -1 to 1

Finally on the last part we get x=0 y =0..1

$\displaystyle u=x-y \iff u=0-y \iff u=-y$ and
$\displaystyle v=x+y \iff v=y \iff v=y$ so as y goes from 1 to 0

we get the line segment$\displaystyle u=-v$

So we get the triangle in my above post.

5. always a sketch will help to clarify things.

for the $\displaystyle (x,y)$ region you have a triangle whose coordinates are $\displaystyle (0,0),\,(0,1),\,(1,0),$ thus under the transformation it's $\displaystyle (0,0),\,(-1,1),\,(1,1)$ for the $\displaystyle (u,v)$ region.

from there things are a bit clear to take the new region where the double integral is taken.

of course, don't forget the Jacobian.

6. Thanks you both