?I\'m trying this again because the previous answer that I had received for this

Question

?I'm trying this again because the previous answer that I had received for this question was completely unhelpful. The question that I have listed is for no. 27 however it's more of a general problem that I'm having and that is finding the parameterizations/ cross product. I realize for this question that the cross product for S1 is 2xi-j+2zk and for S2 it's j but for the life of me I can't figure how someone gets to the point. Where does the rx/ry come from etc. I don't have this problem with just this problem but with all problems and have looked everywhere for an explanation but have yet to find anything that explains what is going on so if anyone can explain it in the simplest possilbe way, I'd really really appreciate it!!

Explanation / Answer

Ans.27

By the “Divergence Theorem” : [S] F•n dS = div(F) dV where V is the volume enclosed by the closed surface S.
For this problem div(F) = F_x/x + F_y/x + F_z/x = 0 + 1 1 = 0
[S] F•n dS = div(F) dV = 0 … a useful check on integration results.
…..................................

S is in two parts, the paraboloid (S1) with its axis as the y-axis and the disk (S2) which is the cross-section of the paraboloid at y=1.
If (x,y,z)=r(u,v) is the parameterisation of the surface (which if possible defines S in terms of a rectangular region in the (u,v) plane) then ...

Surface Integral is [S] F•dS where dS = (r/uXr/v) dudv.
We evaluate the in two parts.

S1 : Let x=v cos(u), y=v², z=v sin(u) so S1 is covered by [v=0,1] and [u=0,2]
r/u X r/v = ( vsin(u), 0, vcos(u) ) X ( cos(u), 2v, sin(u) )
So dS = ( 2v²cos(u), v, 2v²sin(u) ) dudv

Since y-component must be for outward pointing normal we change sign

dS = ( 2v²cos(u), v, 2v²sin(u) ) dudv
F•dS = ( 0, v², vsin(u) )•( 2v²cos(u), v, 2v²sin(u) ) = v³2v³sin²(u) = v³{2cos(2u)}

{S1] = [v=0,1], [u=0,2] v³{2cos(2u)} du dv
wrt v : [v=0,1], [u=0,2] ¼v{2cos(2u)} du = ¼ [u=0,2] {2cos(2u)} du
wrt u : ¼ [u=0,2] {2u ½sin(2u)} =

Note - Wrt means with respect to V/U

S2 : Let x=vcos(u), y=1, z=vsin(u) so S1 is covered by [v=0,1] and [u=0,2]
r/u X r/v = ( vsin(u), 0, vcos(u) ) X ( cos(u), 0, sin(u) ) = ( 0, v, 0 )
dS = ( 0, v, 0 ) du dv

Since v>0 this is pointing away from enclosed region.

F•dS = ( 0, 1, vsin(u) )•( 0, v, 0 ) = v
[S2] = [v=0,1], [u=0,2] v dudv =

[S] = + = 0

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