[Intergal]Change of Variables

**chialin4** · February 4th 2010, 05:50 AM

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

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

**Danny** · February 4th 2010, 06:52 AM

Originally Posted by chialin4

Evaluate
$\iint_D {(x+2y)exp(y-x) dxdy}$
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 $-u \le v \le 0,$ $0 \le u \le 2$ . The Jacobian of the transformation is

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

**chialin4** · February 4th 2010, 05:27 PM

Originally Posted by Danny

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

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

thx for ur help!!

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