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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 $$ S = \{ (x,y,z): z = \sin(y-x), \text{ and } (x,y) \in D \} $$ – Sasha Aug 27 '12 at 20:34
I hope so.. I found the question in a book so I reported it as it is formulated. – Adam Aug 27 '12 at 21:43

Let's parametrize your surface using $(u,v) \in (0,1) \times (0,1)$ as follows: $$ x(u,v) = \pi u, \quad y(u,v) = \pi (u+v), \quad z(u,v) = \sin( \pi v) $$ The induced measure $\mathrm{d}S$ is defined as follows $$ \mathrm{d}S = n_x \mathrm{d}y \wedge \mathrm{d}z + n_y \mathrm{d} z \wedge \mathrm{d} x +n_z \mathrm{d} x \wedge \mathrm{d} y $$ where $n_x = \frac{\partial}{\partial x}\left(z-\sin(y-x)\right) = \cos(y-x) = \cos(\pi v)$, $n_y = -\cos(y-x) = -\cos(\pi v)$, $n_z = 1$. Furthermore: $$ \begin{eqnarray} \mathrm{d} y \wedge \mathrm{d} z &=& \pi^2 \cos(\pi v) \mathrm{d} u \wedge \mathrm{d} v \\ \mathrm{d} z \wedge \mathrm{d} x &=& = - \pi^2 \cos(\pi v) \mathrm{d} u \wedge \mathrm{d} v \\ \mathrm{d} x \wedge \mathrm{d} y &=& \pi^2 \mathrm{d} u \wedge \mathrm{d} v \end{eqnarray} $$ Combining, we arrive at the expression for induced measure: $$ \mathrm{d} S = \pi^2 \left(1 + 2 \cos^2(\pi v) \right)\mathrm{d} u \wedge \mathrm{d} v $$ Using this measure, the integral becomes: $$ \begin{eqnarray} \int \int \sqrt{1 + 2 \cos^2(y-x) } \mathrm{d} S &=& \pi^2 \int_0^1 \int_0^1 \left(1 + 2 \cos^2(\pi v)\right)^{3/2} \mathrm{d} u \mathrm{d} v \\ &=& \pi^2 \int_0^1 \left(1 + 2 \cos^2(\pi v)\right)^{3/2} \mathrm{d} v \\ &\stackrel{t=\cos(\pi v)}{=}& \pi \int_{-1}^1 \frac{\left(1+2 t^2\right)^{3/2}}{\sqrt{1-t^2}} \mathrm{d} t \\ &=& \frac{2\pi}{3} \left( 8 \mathsf{E}\left(-2\right) - 3 \mathsf{K}\left(-2\right)\right) \end{eqnarray} $$ where $\mathsf{K}(m) = \int_0^{\pi/2} \frac{ \mathrm{d} \phi}{\sqrt{1- m \sin^2(\phi)}}$ is the complete elliptic integral of the first kind, and $\mathsf{E}(m) = \int_0^{\pi/2} \sqrt{1- m \sin^2(\phi)}\mathrm{d} \phi$ is the complete elliptic integral of the second kind.

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Thanks for the answer. But I 'm pretty sure there exists an "easier" one. Any other ways? – Adam Aug 27 '12 at 21:42

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