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modphysnoobBest ResponseYou've already chosen the best response.1
I actually reworked it using http://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates and got this \[\frac{1}{r*\text{Sin}[\theta ]}(\text{Cos}[\theta ]*\text{Sin}[\theta ]) \hat{r}+\left(\frac{\text{Sin}[\theta ]}{r}\right)\hat{\phi }\]
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.1
Oops, typo. Sorry. The radial component is \[ \frac{1}{r\sin(\theta)} \frac{\partial}{\partial \theta} \sin^2(\theta) = \frac{2\cos(\theta)}{r} \]
 one year ago

modphysnoobBest ResponseYou've already chosen the best response.1
what about phi component?
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.1
Notice that it doesn't actually matter, because you'll be taking the dot product with the normal vector to the surface of the hemisphere, which is just \[ \hat{n} = \hat{r} \]
 one year ago

modphysnoobBest ResponseYou've already chosen the best response.1
oh, wow, didn't see that; that does make it a lot easier
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.1
Which hemisphere would you like to integrate over? If you do the one above the z = 0 plane, you'll be integrating \[ \int_0^{\pi/2} \int_0^{2\pi} 2\cos(\theta) d\phi d\theta\]
 one year ago

modphysnoobBest ResponseYou've already chosen the best response.1
shouldn't there be sin(theta ) from spherical form of ds?
 one year ago

modphysnoobBest ResponseYou've already chosen the best response.1
http://upload.wikimedia.org/math/8/0/f/80f422b0e2893ef5cc21ad13c71bbc10.png
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.1
oh yeah, sorry. Yet another typo.
 one year ago

modphysnoobBest ResponseYou've already chosen the best response.1
no problem, your help is greatly appreciated
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.1
Anyway, then you will be integrating the line integral around the circular boundary in the z=0 plane. The two will be equal.
 one year ago

modphysnoobBest ResponseYou've already chosen the best response.1
haha, it better be
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.1
The line integral is trivially simple, don't worry :)
 one year ago

modphysnoobBest ResponseYou've already chosen the best response.1
ok, I will try working it out
 one year ago

modphysnoobBest ResponseYou've already chosen the best response.1
http://upload.wikimedia.org/math/0/a/c/0ac2a22b40a49a17025d8ad27f49cf6f.png we only use middle term,their phi is our theta since we actually have don't have any theta vector it is 0
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.1
No, the line integral should be \[ \int_0^{2\pi} r \sin(\theta) \cdot A_\phi d\phi \]
 one year ago

modphysnoobBest ResponseYou've already chosen the best response.1
where does this come from?
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.1
\[ d\vec{l} = \hat{r}dr + \hat{\theta} rd\theta + \hat{\phi} r\sin(\theta) d\phi\] \[\vec{A} = \hat{r} A_r + \hat{\theta} A_\theta + \hat{\phi} A_\phi \]
 one year ago

modphysnoobBest ResponseYou've already chosen the best response.1
we don't use r vector becuase it is set to constant?
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.1
What do you mean? The path we're taking is a circle in the Z=0 plane, so the direction is purely phi.
 one year ago

modphysnoobBest ResponseYou've already chosen the best response.1
just to make sure we are on same page , phi is angle from x axis on xy plane right
 one year ago

modphysnoobBest ResponseYou've already chosen the best response.1
well, A_phi= sin (theta)^2 so integrating it would result in sin(theta)^2*(2*pi)
 one year ago

modphysnoobBest ResponseYou've already chosen the best response.1
since it is in xy plane ,evaluate at pi/2 sin(pi/2)^2 *(2*Pi)=2pi
 one year ago

modphysnoobBest ResponseYou've already chosen the best response.1
which is what I got for surface integral above
 one year ago

modphysnoobBest ResponseYou've already chosen the best response.1
thanks you , you are not only very smart also very patient
 one year ago
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