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# Evaluate the surface integral ∫∫_S F*ndA by the divergence theorem. Show the details. F=[e^x, e^y, e^z], S the surface of the cube |x|≤1, |y|≤1, |z|≤1

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The divergence of $\pmb{F}$ is

$\text{div}\pmb{F} = e^x + e^y + e^z$

and the region bounded $T$ bounded by the surface $S$ is

$T: \begin{cases} |x| \leq 1 \\ |y| \leq 1 \\ |z| \leq 1 \end{cases}$

which can be written as

$T: \begin{cases} -1 \leq x \leq 1 \\ -1 \leq y \leq 1 \\ -1 \leq z \leq 1 \end{cases}$

Using the divergence Theorem we get:

\begin{aligned} & \iint\limits_S \pmb{F \cdot n} dA = \int\limits_{-1}^1 \int\limits_{-1}^1 \int\limits_{-1}^1 (e^x+e^y+e^z) dxdydz = \\ & = \int\limits_{-1}^1 \int\limits_{-1}^1 \left( e^x + xe^y +xe^z \right) \big\vert_{-1}^1 dy dz = \int\limits_{-1}^1 \int\limits_{-1}^1 \left( e - e^{-1} + 2 e^y + 2e^z \right) dy dz = \\ & = \int\limits_{-1}^1 \left( y (e - e^{-1}) +2 e^y + 2y e^z \right) \big\vert_{-1}^1 dz = \int\limits_{-1}^1 \left( 2 (e - e^{-1}) +2 (e - e^{-1}) +4 e^z \right) dz = \\ & = 4 \int\limits_{-1}^1 \left( e - e^{-1} + e^z \right) dz = 4 \left( z(e- e^{-1} ) + e^z \right) \Big\vert_{-1}^1 = \\ & = 5 ( 2 (e-e^{-1}) +e - e^{-1} ) =12 (e- e^{-1}) = 24 \frac{ e- e^{-1}}{2} = \textcolor{#4257b2}{ 24 \sinh(1)} \end{aligned}

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