Evaluate the surface integral ∫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)

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Evaluate the surface integral   ∫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)

unessodopunsep · Accepted Answer

Remember that F⋅ds=F⋅n^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: ∫SF⋅dS=∫⊤F⋅n^⊤dS+∫≤ftF⋅n^≤ftdS+∫forwardF⋅n^forwarddS+… Let&#039;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 −1≤x≤1,−1≤y≤1,andz=1.Hence we have ∫⊤F⋅n^⊤=∫y=−1y=1∫x=−1x=1[xi+2yj+3(1)k]⋅kdxdy =∫y=−1y=1∫x=−1x=13dxdy =3∫y=−1y=1∫x=−1x=1dxdy 3(2)∫y=−1y=1dy =3(2)(2)=12 Along the left face (pretend you&#039;re sitting somewhat out on the positive(axis looking at the cube),the out ward is−jandthe value sofx,y,andz are−1≤x≤1,y=−1,and−1≤z≤1.Thus ∫≤ftF⋅n^≤ftdS=∫z=−1z=1∫x=−1x=1[xi+2(−1)j+3zk]⋅−jdxdz=8 At this point hopefully you get the idea so I&#039;m just going to quickly list the rest of the integrals and their values:

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)

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