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Let $D$ be the parallelogram on the $(x,y)$ plane ($z=0$) that comes from the intersection of the lines: $y=x$, $x = \pi$, $y = x + \pi$ and $x=0$.

Compute the following integral:

$$ \int_S \sqrt{ 1 + 2 \cos^2(y-x) } dS $$

where $S$ is the surface described by the equation $z = \sin(y-x)$ projected on $D$.

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    I guess you mean to say that the integration is performed over the surface $$ S = \{ (x,y,z): z = \sin(y-x), \text{ and } (x,y) \in D \} $$2012-08-27
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    I hope so.. I found the question in a book so I reported it as it is formulated.2012-08-27

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