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

Jemurray3
 one year ago
Best ResponseYou've already chosen the best response.1Oops, typo. Sorry. The radial component is \[ \frac{1}{r\sin(\theta)} \frac{\partial}{\partial \theta} \sin^2(\theta) = \frac{2\cos(\theta)}{r} \]

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

Jemurray3
 one year ago
Best ResponseYou've already chosen the best response.1Notice 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 \[ \hat{n} = \hat{r} \]

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

Jemurray3
 one year ago
Best ResponseYou've already chosen the best response.1Which hemisphere would you like to integrate over? If you do the one above the z = 0 plane, you'll be integrating \[ \int_0^{\pi/2} \int_0^{2\pi} 2\cos(\theta) d\phi d\theta\]

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

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

Jemurray3
 one year ago
Best ResponseYou've already chosen the best response.1oh yeah, sorry. Yet another typo.

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

Jemurray3
 one year ago
Best ResponseYou've already chosen the best response.1Anyway, then you will be integrating the line integral around the circular boundary in the z=0 plane. The two will be equal.

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

Jemurray3
 one year ago
Best ResponseYou've already chosen the best response.1The line integral is trivially simple, don't worry :)

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

modphysnoob
 one year ago
Best ResponseYou've already chosen the best response.1http://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

Jemurray3
 one year ago
Best ResponseYou've already chosen the best response.1No, the line integral should be \[ \int_0^{2\pi} r \sin(\theta) \cdot A_\phi d\phi \]

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

Jemurray3
 one year ago
Best ResponseYou've already chosen the best response.1\[ d\vec{l} = \hat{r}dr + \hat{\theta} rd\theta + \hat{\phi} r\sin(\theta) d\phi\] \[\vec{A} = \hat{r} A_r + \hat{\theta} A_\theta + \hat{\phi} A_\phi \]

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

Jemurray3
 one year ago
Best ResponseYou've already chosen the best response.1What do you mean? The path we're taking is a circle in the Z=0 plane, so the direction is purely phi.

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

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

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

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

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