General question about surface integrals

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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