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anonymous
 3 years ago
Vector calculs question , please help
anonymous
 3 years ago
Vector calculs question , please help

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I actually reworked it using http://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates and got this \[\frac{1}{r*\text{Sin}[\theta ]}(\text{Cos}[\theta ]*\text{Sin}[\theta ]) \hat{r}+\left(\frac{\text{Sin}[\theta ]}{r}\right)\hat{\phi }\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oops, typo. Sorry. The radial component is \[ \frac{1}{r\sin(\theta)} \frac{\partial}{\partial \theta} \sin^2(\theta) = \frac{2\cos(\theta)}{r} \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what about phi component?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Notice 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 \[ \hat{n} = \hat{r} \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh, wow, didn't see that; that does make it a lot easier

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0shouldn't there be sin(theta ) from spherical form of ds?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0http://upload.wikimedia.org/math/8/0/f/80f422b0e2893ef5cc21ad13c71bbc10.png

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh yeah, sorry. Yet another typo.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no problem, your help is greatly appreciated

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The line integral is trivially simple, don't worry :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok, I will try working it out

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0http://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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No, the line integral should be \[ \int_0^{2\pi} r \sin(\theta) \cdot A_\phi d\phi \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0where does this come from?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[ d\vec{l} = \hat{r}dr + \hat{\theta} rd\theta + \hat{\phi} r\sin(\theta) d\phi\] \[\vec{A} = \hat{r} A_r + \hat{\theta} A_\theta + \hat{\phi} A_\phi \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0we don't use r vector becuase it is set to constant?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0just to make sure we are on same page , phi is angle from x axis on xy plane right

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well, A_phi= sin (theta)^2 so integrating it would result in sin(theta)^2*(2*pi)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0since it is in xy plane ,evaluate at pi/2 sin(pi/2)^2 *(2*Pi)=2pi

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0which is what I got for surface integral above

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thanks you , you are not only very smart also very patient
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