# Finding the double integral over a specified region.

Let $D$ be the region in the xy-plane that is bounded by the coordinate axes and the line $x+y =1$ , we need to find : $\displaystyle\iint (x-y)e^{ x^{2} + y^{2}}dydx$ over $D$.

I am trying the change of variables technique here :

Let $u = x-y$ and $v = x + y$ and by considering the region given the integral came out to be :

$\displaystyle\int_0^1 \int_0^v ue^{ \frac{ u^{2} + v^{2}}{2}}dudv$ ,

First integrating w.r.t $u$ gives :

$\displaystyle\int_0^1 e^{v^{2}} - e^{\frac{v^{2}}{2}}dv$ ,

That is the point where I am stuck ..

Could anyone help me with this ?

Keeping $u$ constant , i.e changing the order of integration isn't helping..

The integral must be $0$. You can get $$\iint_D (x-y)e^{ x^{2} + y^{2}}dydx = \iint_D (y-x)e^{ x^{2} + y^{2}}dydx$$ because $D$ is symmetric along the line $x=y$
(Use change of variables switching $x$ and $y$)
p.s. You did a mistake; you must have $$\int_0^1 \int_{-v}^v ue^{ \frac{ u^{2} + v^{2}}{2}}dudv$$ so you can see easily that the integral is $0$.
• Finally got it.. I neglected the fact that the function $$ue^{ \frac{ u^{2} + v^{2}}{2}}$$ is not symmetric about $u$.. Thanks ! Nov 8 '15 at 18:21