(b) Verify the Divergence Theorem for the vector field: F = (6z, y², 6) and the region r? +y? < 1,0 < z<6 oriented with both outward facing normals on the top and bottom, by computing both S fgF -dS and SS S, div(F)dV directly. The region is shown in the picture below: Hint: For S Ss F · dS, break up the cylinder so that S fg F - dS = § frop F · dS + S Suottom F · dS + S Smiddle F-dS. For S Smiddle F -dS, use the parametrization r(u, v) = (cos(u), sin(u), v) where 0< u< 2m and 0 < v < 6. Then compute Tu, T, and N. Then you will get "(v cos(u) + sin (1))dudv. This integral is too hard to solve, so I'm giving you that this integral equals 0! %3D %3D

Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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(b) Verify the Divergence Theorem for the vector field: F = (6z, y², 6) and the region r? +y? < 1,0 < z<6
oriented with both outward facing normals on the top and bottom, by computing both S fgF -dS and
SS S, div(F)dV directly. The region is shown in the picture below:
Hint: For S Ss F · dS, break up the cylinder so that S f3 F - dS = § frop F · dS + S Suottom F · dS +
S Smiddle F-dS. For S Smiddle F -dS, use the parametrization r(u, v) = (cos(u), sin(u), v) where 0< u<
2m and 0 < v < 6. Then compute Tu, T, and N. Then you will get "(v cos(u) + sin (1))dudv.
This integral is too hard to solve, so I'm giving you that this integral equals 0!
%3D
%3D
Transcribed Image Text:(b) Verify the Divergence Theorem for the vector field: F = (6z, y², 6) and the region r? +y? < 1,0 < z<6 oriented with both outward facing normals on the top and bottom, by computing both S fgF -dS and SS S, div(F)dV directly. The region is shown in the picture below: Hint: For S Ss F · dS, break up the cylinder so that S f3 F - dS = § frop F · dS + S Suottom F · dS + S Smiddle F-dS. For S Smiddle F -dS, use the parametrization r(u, v) = (cos(u), sin(u), v) where 0< u< 2m and 0 < v < 6. Then compute Tu, T, and N. Then you will get "(v cos(u) + sin (1))dudv. This integral is too hard to solve, so I'm giving you that this integral equals 0! %3D %3D
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