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Vector calculs question , please help

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@saifoo.khan HELP

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I actually reworked it using 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?
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.
haha, it better be
The line integral is trivially simple, don't worry :)
ok, I will try working it out 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
Yep, good job.
thanks you , you are not only very smart also very patient
Thank you!

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