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

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- Feb 4th 2010, 05:50 AMchialin4[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 - Feb 4th 2010, 06:52 AMJester
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 $\displaystyle \int_0^2 \int_{-u}^0 u e^v \frac{1}{3} dvdu$. - Feb 4th 2010, 05:27 PMchialin4