This week's problem was correctly answered by Bacterius, Kiwi, lfdahl, and Opalg.
Here's Kiwi's solution, which uses Green's theorem:
[sp]We have Greens Theorem:
[tex]\oint(Ldx+Mdy)=\int\int\left(\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}\right)dA[/tex]
Identifying [tex]L = \frac{\partial L}{\partial y}=-e^{x+y}[/tex] and [tex]M = \frac{\partial M}{\partial y}=0[/tex]
we can write:
[tex]-\oint e^{(x+y)}dx=\int\int e^{(x+y)}dA[/tex]
The boundary is a rotated square with its corners located at (1,0),(0,1),(-1,0) and (0,-1). We integrate around the boundary in each of the four quadrants separately and sum the contributions together.
In the first quadrant
x+y=1 so
contribution to [tex]-\oint e^{(x+y)}dx=-\int^0_1 edx = e[/tex]
In the second quadrant
y=x+1 so x+y=2x+1 and so
contribution to [tex]-\oint e^{(x+y)}dx=-\int^{-1}_0 e^{2x+1}dx = -[\frac 12 e^{2x+1)}]^{-1}_0=-\frac 12 (e^{-1}-e^1)[/tex]
In the third quadrant
x+y=-1 so
contribution to [tex]-\oint e^{(x+y)}dx=-\int^0_{-1} e^{-1}dx = \int^{-1}_{0} e^{-1}dx=-e^{-1}[/tex]
In the fourth quadrant
y=x-1 so x+y=2x-1 and so
contribution to [tex]-\oint e^{(x+y)}dx=-\int^{1}_0 e^{2x-1}dx = -[\frac 12 e^{2x-1)}]^{1}_0=-\frac 12 (e^{1}-e^{-1})[/tex]
The second and fourth quadrant contributions sum to zero leaving the solution:
[tex]\int\int e^{(x+y)}dA=e-\frac 1e[/tex][/sp]
Here's Opalg's solution, which uses the Jacobian transformation:
[sp]Let $u=x-y$, $v = x+y$. Then $R$ is given by $-1\leqslant u \leqslant 1$, $-1\leqslant v \leqslant 1$. The Jacobian determinant of the change of variables is $$\left|\frac{\partial (u,v)}{\partial(x,y)}\right| = \begin{vmatrix} 1&-1 \\ 1&1 \end{vmatrix} = 2.$$ So the integral becomes $$\iint_R e^{x+y}dx\,dy = \int_{-1}^1 \int_{-1}^1 e^v \frac{du\,dv}{\left|\frac{\partial (u,v)}{\partial(x,y)}\right|} = \frac12 \int_{-1}^1 \int_{-1}^1du\,dv = \int_{-1}^1 e^vdv = e-e^{-1} = 2\sinh1.$$[/sp]