Evaluate Surface Integral: Solve x^2 + z^2 = 9, x=0, y=0, z=0 and y=8

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The discussion focuses on evaluating the surface integral of the vector field A=(6z, 2x+y, -x) over a region bounded by the cylinder x^2 + z^2 = 9 and the planes x=0, y=0, z=0, and y=8. The application of Gauss' Theorem leads to the conclusion that the divergence of A is 1, simplifying the integral to the volume of the region. The volume is calculated as pi*r^2*h, where r^2=9 and h=8, resulting in 18pi. The quarter factor arises because the region is effectively a quarter of the cylinder due to the constraints imposed by the bounding planes. Understanding these limits clarifies the origin of the quarter factor in the final answer.
Hoofbeat
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Could someone take a look at this please? Thanks

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Q. Evaluate Integral A.n dS for the following case:
A=(6z, 2x+y, -x) and S is the entire surface of the region bounded by the cylinder x^2 + z^2 = 9, x=0, y=0, z=0 and y=8.
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Using Gauss' (or Divergence) Theorem:

Integral A.n dS = Integral Div A dV

Div A = 1, therefore

Integral 1 dV

When I last did this question, I said that dV = pi*r^2*h where r^2=9 and h=8. However, I've also got a factor of 1/4 to give me the final answer of 18pi. Where the hell did the quarter come from?! I'm guessing it has something to do with the limits and the fact that it's bounded by x=0, y=0 and z=0, but I don't understand why! Anyone offer some advice?!
 
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Hoofbeat said:
Where the hell did the quarter come from?! I'm guessing it has something to do with the limits and the fact that it's bounded by x=0, y=0 and z=0, but I don't understand why! Anyone offer some advice?!

You think it right. The bases of the cylinder are parallel to the (xy) plane at y =0 and y=8. The (z,y) plane at x=0 cuts the cylinder into half. The other plane, (x,y) at z=0 does the same, so you have a quarter of a cylinder of length h=8.

ehild
 
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thanks ever so much! :)
 
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