Surface integral without using Gauss' theorem

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SUMMARY

The discussion centers on calculating the surface integral of the vector field A = 2y(x^2)i - (y^2)j + 4xzk over a specified region in the first octant, bounded by the equation (y^2) + (z^2) = 9 and x = 2. The user initially obtained a result of 108, while the expected result was 180, leading to confusion regarding the limits of integration and the projection method used. The resolution highlighted the necessity of calculating the surface integral over all three faces of the closed surface, not just one, to achieve the correct result.

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Homework Statement



Calculate §§ A.n dS if
A= 2y(x^2)i-(y^2)j + 4xzk
over the region in the first octant bounded by (y^2)+(z^2) = 9 and x = 2

Homework Equations





The Attempt at a Solution



Let n = (yj + zk) / 3

then A.n = [-(y^3) +4xz^3] / 3

Since we 'll project the surface onto the xy-plane:
|n.k| = z/3 and z = SQRT(9-y^2)

Putting all together I obtain
= §§R (4xz^3 - (y^3))/z dx dy



Now making the appropriate changes and setting up the limits of integration:


§y=30 §x=20 4x(9-y^2) - (y^3)/sqrt(9-y^2) dx dy



However I always obtain 108 as a result and not 180 as my book suggested me (and after verification by Gauss' divergence theorem.

Is there a problem with the limits of integration? Wrong projection? I really have no clue ...
Thanks for the help!
 
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You've only calculated the integral over one side/face of the surface...there are three more faces that make up the closed surface bounding the given region...you need to calculate the surface integral over all 3 of those as well.
 
Thanks a lot!
I finally got it (at least I hope so ;-) !
 

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