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@saifoo.khan HELP

what about phi component?

oh, wow, didn't see that; that does make it a lot easier

shouldn't there be sin(theta ) from spherical form of ds?

http://upload.wikimedia.org/math/8/0/f/80f422b0e2893ef5cc21ad13c71bbc10.png

oh yeah, sorry. Yet another typo.

no problem, your help is greatly appreciated

haha, it better be

The line integral is trivially simple, don't worry :)

ok, I will try working it out

No, the line integral should be
\[ \int_0^{2\pi} r \sin(\theta) \cdot A_\phi d\phi \]

where does this come from?

we don't use r vector becuase it is set to constant?

just to make sure we are on same page , phi is angle from x axis on x-y plane right

yep

well,
A_phi= sin (theta)^2
so integrating it would result in
sin(theta)^2*(2*pi)

since it is in x-y plane ,evaluate at pi/2
sin(pi/2)^2 *(2*Pi)=2pi

which is what I got for surface integral above

Yep, good job.

thanks you , you are not only very smart also very patient

Thank you!