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)

Reeves 2020-11-03 Answered
Evaluate the surface integral
SFdS
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)
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unessodopunsep
Answered 2020-11-04 Author has 105 answers

Remember that Fds=Fn^dS where n^ 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:
SFdS=Fn^dS+ftFn^ftdS+forwardFn^forwarddS+
Let's start with the top side of the cube. Along that face, the outward normal points up. I.e.n^=k.
Also along the top face we have 1x1,1y1,andz=1.Hence we have
Fn^=y=1y=1x=1x=1[xi+2yj+3(1)k]kdxdy
=y=1y=1x=1x=13dxdy
=3y=1y=1x=1x=1dxdy
3(2)y=1y=1dy
=3(2)(2)=12 Along the left face (pretend you're sitting somewhat out on the positive(axis looking at the cube),the out ward isjandthe value sofx,y,andz are1x1,y=1,and1z1.Thus
ftFn^ftdS=z=1z=1x=1x=1[xi+2(1)j+3zk]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:

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