anonymous
  • anonymous
Vector calculs question , please help
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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goformit100
  • goformit100
@saifoo.khan HELP
anonymous
  • anonymous
@phi

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anonymous
  • anonymous
@jamesJ
anonymous
  • anonymous
I actually reworked it using http://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates and got this \[\frac{1}{r*\text{Sin}[\theta ]}(\text{Cos}[\theta ]*\text{Sin}[\theta ]) \hat{r}+\left(\frac{\text{Sin}[\theta ]}{r}\right)\hat{\phi }\]
anonymous
  • anonymous
Oops, typo. Sorry. The radial component is \[ \frac{1}{r\sin(\theta)} \frac{\partial}{\partial \theta} \sin^2(\theta) = \frac{2\cos(\theta)}{r} \]
anonymous
  • anonymous
what about phi component?
anonymous
  • anonymous
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 \[ \hat{n} = \hat{r} \]
anonymous
  • anonymous
oh, wow, didn't see that; that does make it a lot easier
anonymous
  • anonymous
Which hemisphere would you like to integrate over? If you do the one above the z = 0 plane, you'll be integrating \[ \int_0^{\pi/2} \int_0^{2\pi} 2\cos(\theta) d\phi d\theta\]
anonymous
  • anonymous
shouldn't there be sin(theta ) from spherical form of ds?
anonymous
  • anonymous
http://upload.wikimedia.org/math/8/0/f/80f422b0e2893ef5cc21ad13c71bbc10.png
anonymous
  • anonymous
oh yeah, sorry. Yet another typo.
anonymous
  • anonymous
no problem, your help is greatly appreciated
anonymous
  • anonymous
Anyway, then you will be integrating the line integral around the circular boundary in the z=0 plane. The two will be equal.
anonymous
  • anonymous
haha, it better be
anonymous
  • anonymous
The line integral is trivially simple, don't worry :)
anonymous
  • anonymous
ok, I will try working it out
anonymous
  • anonymous
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
anonymous
  • anonymous
No, the line integral should be \[ \int_0^{2\pi} r \sin(\theta) \cdot A_\phi d\phi \]
anonymous
  • anonymous
where does this come from?
anonymous
  • anonymous
\[ d\vec{l} = \hat{r}dr + \hat{\theta} rd\theta + \hat{\phi} r\sin(\theta) d\phi\] \[\vec{A} = \hat{r} A_r + \hat{\theta} A_\theta + \hat{\phi} A_\phi \]
anonymous
  • anonymous
we don't use r vector becuase it is set to constant?
anonymous
  • anonymous
What do you mean? The path we're taking is a circle in the Z=0 plane, so the direction is purely phi.
anonymous
  • anonymous
just to make sure we are on same page , phi is angle from x axis on x-y plane right
anonymous
  • anonymous
yep
anonymous
  • anonymous
well, A_phi= sin (theta)^2 so integrating it would result in sin(theta)^2*(2*pi)
anonymous
  • anonymous
since it is in x-y plane ,evaluate at pi/2 sin(pi/2)^2 *(2*Pi)=2pi
anonymous
  • anonymous
which is what I got for surface integral above
anonymous
  • anonymous
Yep, good job.
anonymous
  • anonymous
thanks you , you are not only very smart also very patient
anonymous
  • anonymous
Thank you!

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