1. ## [Sloved]Change of Variables

Evaluate
$\displaystyle \iint_D {(x+2y)exp(y-x) dxdy}$
where D={y<x<2-2y, 0<y< 2/3}

by letting
u=x+2y
v=y-x

2. Originally Posted by chialin4
Evaluate
$\displaystyle \iint_D {(x+2y)exp(y-x) dxdy}$
where D={y<x<2-2y, 0<y< 2/3}

by letting
u=x+2y
v=y-x
The new region of integration is a triangle bound by $\displaystyle -u \le v \le 0,$ $\displaystyle 0 \le u \le 2$. The Jacobian of the transformation is

$\displaystyle \frac{\partial(u,v) }{\partial(x,y) } = \left|\begin{array}{cc} 1 & 2\\ -1 & 1\end{array}\right| = 3$ so the new integral is $\displaystyle \int_0^2 \int_{-u}^0 u e^v \frac{1}{3} dvdu$.

3. Originally Posted by Danny
The new region of integration is a triangle bound by $\displaystyle -u \le v \le 0,$ $\displaystyle 0 \le u \le 2$. The Jacobian of the transformation is

$\displaystyle \frac{\partial(u,v) }{\partial(x,y) } = \left|\begin{array}{cc} 1 & 2\\ -1 & 1\end{array}\right| = 3$ so the new integral is $\displaystyle \int_0^2 \int_{-u}^0 u e^v \frac{1}{3} dvdu$.
thx for ur help!!