are u trying to calculate ?
Since A is constant this is obviously zero but if you want to verify you can use the divergence theorem:
Hi all.
EDIT: I solved the problem. I am very sorry to have wasted your time. I studied up on that I needed two sides for intigration to work, so the integration is acually over a half circle, which it does not say in my book.
I've had problems with this question for way to long now and im beating myself up because of it
Anyways here it is.
The question
Calculate A = (1,2,3) over the surface x^2+y^2+z^2=1
My attempts:
I know the equation is the one of a sphere. Thus I assume the parametization of
x = r sina cosb
y = r sina sinb
z = cosa
thus I get r(a,b) = (sina cosb, sina sinb, cosa). I set r = 1 as I am only intrested in the surface so integration over r would "fill" the sphere and that would be volume rather than surface. (This is my assumption)
dr/da x dr/db = (sin^2a*cosb,sin^2a*sinb,sina*cosa)
[integral] A * dr/da x dr/db dS.
I've attempted this integral using 0<a<pi and 0<b<2pi. So doing dS = da db, but that did not work. Tried dS = r^2sina da db, which also did not work.
Im going nuts, please help me Where am I doing it wrong?