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August 1st, 2017, 12:27 AM   #11
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Alternatively, substituting u = x + y, v = x - y leads to $\displaystyle \int_{-1}^1\!\int_{-1}^1 (u^2 + v^2)/4\,dudv$, which is 2/3.
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August 1st, 2017, 12:22 PM   #12
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Originally Posted by romsek View Post
that's an integration of the square with corners at

$(-1,-1), (-1,1), (1,1), (1,-1)$

His integration area is a square with corners

$(-1,0), (0,1), (1,0), (0,-1)$
My apologies for my blunder.
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double integral, wrong

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