Surface integral without using Gauss' theorem

maupassant
Messages
10
Reaction score
0

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!
 
Physics news on Phys.org
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 ;-) !
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top