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modphysnoob
Vector calculs question , please help
I actually reworked it using http://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates and got this \[\frac{1}{r*\text{Sin}[\theta ]}(\text{Cos}[\theta ]*\text{Sin}[\theta ]) \hat{r}+\left(\frac{\text{Sin}[\theta ]}{r}\right)\hat{\phi }\]
Oops, typo. Sorry. The radial component is \[ \frac{1}{r\sin(\theta)} \frac{\partial}{\partial \theta} \sin^2(\theta) = \frac{2\cos(\theta)}{r} \]
what about phi component?
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 \[ \hat{n} = \hat{r} \]
oh, wow, didn't see that; that does make it a lot easier
Which hemisphere would you like to integrate over? If you do the one above the z = 0 plane, you'll be integrating \[ \int_0^{\pi/2} \int_0^{2\pi} 2\cos(\theta) d\phi d\theta\]
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
Anyway, then you will be integrating the line integral around the circular boundary in the z=0 plane. The two will be equal.
The line integral is trivially simple, don't worry :)
ok, I will try working it out
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
No, the line integral should be \[ \int_0^{2\pi} r \sin(\theta) \cdot A_\phi d\phi \]
where does this come from?
\[ d\vec{l} = \hat{r}dr + \hat{\theta} rd\theta + \hat{\phi} r\sin(\theta) d\phi\] \[\vec{A} = \hat{r} A_r + \hat{\theta} A_\theta + \hat{\phi} A_\phi \]
we don't use r vector becuase it is set to constant?
What do you mean? The path we're taking is a circle in the Z=0 plane, so the direction is purely phi.
just to make sure we are on same page , phi is angle from x axis on x-y plane right
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
thanks you , you are not only very smart also very patient