General question about surface integrals

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SUMMARY

The discussion centers on evaluating the surface integral \(\int(v \cdot dS)\) where \(v = (3y, 2x^2, z^3)\) and \(S\) is the surface of the cylinder defined by \(x^2 + y^2 = 1\) for \(0 < z < 1\). Participants clarify that \(dS\) is determined by the unit normal vector to the surface, which is horizontal in this case, and that cylindrical coordinates can simplify the evaluation. The confusion regarding the inclusion of \(dr\) in cylindrical coordinates is addressed, emphasizing that for surface integrals, only two differential elements are necessary.

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kevinf
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hi in my engineering mathematics class, we are going over surface integrals again. i have some general question about this subject. sorry for not using the template.

say that i have a problem that goes like this.

"evaluate \int(v*dS) (where the * means dot) where v= (3y,2x^{2},z^{3}) and S is the surface of the cylinder x^{2} + y^{2}=1, 0<z<1. "

to find dS, is it always the unit vector of z=6-2x-2y (gradient of z divide by magnitude of z) multiply by dx and dy? i am not sure if i am right on the dx and dy though.

also some of the problems' solutions use cylindrical coordinates, which i understand, but some of them only use r d\varphi dz. if i remember correclty from my calculus classes, i thought the cylindrical coordinates also included dr?
 
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hi kevinf! :smile:
kevinf said:
also some of the problems' solutions use cylindrical coordinates, which i understand, but some of them only use r d\varphi dz. if i remember correclty from my calculus classes, i thought the cylindrical coordinates also included dr?

Yes, but the integral is a surface integral, so there's only two ds …

r is constant over the surface. :wink:
"evaluate \int(v*dS) (where the * means dot) where v= (3y,2x^{2},z^{3}) and S is the surface of the cylinder x^{2} + y^{3}, 0<z<1. "

sorry, i don't understand what the cylinder is :confused:
 
sorry it was a typo. i corrected it in the original post.
 
ahh! :biggrin:

in that case, i don't understand …
kevinf said:
to find dS, is it always the unit vector of z=6-2x-2y (gradient of z divide by magnitude of z) multiply by dx and dy? i am not sure if i am right on the dx and dy though.

… dS is in the normal direction, which will always be horizontal (no z) :smile:

(and wouldn't it be easier to use cylindrical coordinates?
 

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