Evaluate the surface integral int_SF*dS for the given vector field F and the oriented surface S. In other words, find the flux of F across S. For closed surfaces, use the positive (outward) orientation. F(x,y,z)=xi+2yj+3zk S is the cube with vertices (+-1,+-1,+-1)

Evaluate the surface integral
${\int }_{S}F\cdot dS$
for the given vector field F and the oriented surface S. In other words, find the flux of F across S. For closed surfaces, use the positive (outward) orientation.
F(x,y,z)=xi+2yj+3zk
S is the cube with vertices $\left(±1,±1,±1\right)$
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unessodopunsep

Remember that is the unit outward normal vector. Luckily that vector will be constant across each of the 6 faces of the cube. So we just need to break this integral into 6 parts:
${\int }_{S}F\cdot dS={\int }_{\mathrm{\top }}F\cdot {\stackrel{^}{n}}_{\mathrm{\top }}dS+{\int }_{\le ft}F\cdot {\stackrel{^}{n}}_{\le ft}dS+{\int }_{f\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}ward}F\cdot {\stackrel{^}{n}}_{f\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}ward}dS+\dots$
Let's start with the top side of the cube. Along that face, the outward normal points up. I.e.$\stackrel{^}{n}=k$.
Also along the top face we have $-1\le x\le 1,-1\le y\le 1,\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}z=1$.Hence we have
${\int }_{\mathrm{\top }}F\cdot {\stackrel{^}{n}}_{\mathrm{\top }}={\int }_{y=-1}^{y=1}{\int }_{x=-1}^{x=1}\left[xi+2yj+3\left(1\right)k\right]\cdot kdxdy$
$={\int }_{y=-1}^{y=1}{\int }_{x=-1}^{x=1}3dxdy$
$=3{\int }_{y=-1}^{y=1}{\int }_{x=-1}^{x=1}dxdy$
$3\left(2\right){\int }_{y=-1}^{y=1}dy$
=3(2)(2)=12 Along the left face (pretend you're sitting somewhat out on the positive.Thus
${\int }_{\le ft}F\cdot {\stackrel{^}{n}}_{\le ft}dS={\int }_{z=-1}^{z=1}{\int }_{x=-1}^{x=1}\left[xi+2\left(-1\right)j+3zk\right]\cdot -jdxdz=8$
At this point hopefully you get the idea so I'm just going to quickly list the rest of the integrals and their values: