Problem of the Week #121 - July 21st, 2014

[Chris L T521]

Here's this week's problem!

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Problem: Evaluate $\displaystyle \iint\limits_R \exp(x+y)\,dA$ where $R$ is given by the inequality $|x|+|y|\leq 1$.

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[Chris L T521]

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:

\oint(Ldx+Mdy)=\int\int\left(\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}\right)dA

Identifying L = \frac{\partial L}{\partial y}=-e^{x+y} and M = \frac{\partial M}{\partial y}=0

we can write:

-\oint e^{(x+y)}dx=\int\int e^{(x+y)}dA

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 -\oint e^{(x+y)}dx=-\int^0_1 edx = e

In the second quadrant
y=x+1 so x+y=2x+1 and so

contribution to -\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)

In the third quadrant
x+y=-1 so

contribution to -\oint e^{(x+y)}dx=-\int^0_{-1} e^{-1}dx = \int^{-1}_{0} e^{-1}dx=-e^{-1}

In the fourth quadrant
y=x-1 so x+y=2x-1 and so

contribution to -\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})

The second and fourth quadrant contributions sum to zero leaving the solution:

\int\int e^{(x+y)}dA=e-\frac 1e[/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]