# MvCalc67

$\displaystyle \iint_{R}(x+y)e^{x^2-y^2}dA \,$

where $\displaystyle _{R}\,$ is the rectangle enclosed by the lines

$\displaystyle x - y = 0 \,$

$\displaystyle x - y = 2 \,$

$\displaystyle x + y = 0 \,$

$\displaystyle x + y = 3 \,$

For change of variables, Let

$\displaystyle u = x - y \,$ and

$\displaystyle v = x + y \,$

So then by solving for x and y we get...

$\displaystyle x=\frac{1}{2}(u+v)\,$ and

$\displaystyle y=\frac{1}{2}(v-u)\,$

$\displaystyle 0\le u \le 2 \,$

$\displaystyle 0\le v \le 3 \,$

$\displaystyle \int_{0}^{3}\int_{0}^{2}(v)e^{uv}\cdot |J|dudv \,$ Where |J| is the Jacobian determinant

$\displaystyle \frac{\partial (x,y)}{\partial (u,v)}= \left[\begin{matrix} \frac{1}{2} & \frac{1}{2} \\[6pt] -\frac{1}{2} & \frac{1}{2} \end{matrix}\right]= \frac{1}{2}\,$

$\displaystyle \frac{1}{2}\int_{0}^{3}\int_{0}^{2}(v)e^{uv}dudv \,$

$\displaystyle \frac{1}{2}\int_{0}^{3}[e^{uv}]_{0}^{2}dv \,$

$\displaystyle \frac{1}{2}\int_{0}^{3}(e^{2v}-1)dv \,$

$\displaystyle \frac{1}{2}\int_{0}^{3}e^{2v}dv -\frac{1}{2}\int_{0}^{3}dv\,$

$\displaystyle \frac{1}{2}[\frac{1}{2}e^{2v}]_{0}^{3} -\frac{1}{2}[v]_{0}^{3}\,$

$\displaystyle \frac{1}{2}[\frac{1}{2}e^{6}-\frac{1}{2}] -\frac{3}{2}\,$

$\displaystyle \frac{1}{4}e^{6}-\frac{7}{4} \,$

Hence..$\displaystyle \iint_{R}(x+y)e^{x^2-y^2}dA = \frac{1}{4}(e^{6}-7)\,$