General question about surface integrals

[kevinf]

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?

 

[tiny-tim]

hi kevinf!

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.

"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

 

[kevinf]

sorry it was a typo. i corrected it in the original post.

 

[tiny-tim]

ahh!

in that case, i don't understand …

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)

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