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modphysnoob

  • 3 years ago

Vector calculs question , please help

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  1. modphysnoob
    • 3 years ago
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  2. goformit100
    • 3 years ago
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    @saifoo.khan HELP

  3. modphysnoob
    • 3 years ago
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    @phi

  4. modphysnoob
    • 3 years ago
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    @jamesJ

  5. modphysnoob
    • 3 years ago
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    I actually reworked it using http://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates and got this \[\frac{1}{r*\text{Sin}[\theta ]}(\text{Cos}[\theta ]*\text{Sin}[\theta ]) \hat{r}+\left(\frac{\text{Sin}[\theta ]}{r}\right)\hat{\phi }\]

  6. Jemurray3
    • 3 years ago
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    Oops, typo. Sorry. The radial component is \[ \frac{1}{r\sin(\theta)} \frac{\partial}{\partial \theta} \sin^2(\theta) = \frac{2\cos(\theta)}{r} \]

  7. modphysnoob
    • 3 years ago
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    what about phi component?

  8. Jemurray3
    • 3 years ago
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    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 \[ \hat{n} = \hat{r} \]

  9. modphysnoob
    • 3 years ago
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    oh, wow, didn't see that; that does make it a lot easier

  10. Jemurray3
    • 3 years ago
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    Which hemisphere would you like to integrate over? If you do the one above the z = 0 plane, you'll be integrating \[ \int_0^{\pi/2} \int_0^{2\pi} 2\cos(\theta) d\phi d\theta\]

  11. modphysnoob
    • 3 years ago
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    shouldn't there be sin(theta ) from spherical form of ds?

  12. modphysnoob
    • 3 years ago
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    http://upload.wikimedia.org/math/8/0/f/80f422b0e2893ef5cc21ad13c71bbc10.png

  13. Jemurray3
    • 3 years ago
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    oh yeah, sorry. Yet another typo.

  14. modphysnoob
    • 3 years ago
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    no problem, your help is greatly appreciated

  15. Jemurray3
    • 3 years ago
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    Anyway, then you will be integrating the line integral around the circular boundary in the z=0 plane. The two will be equal.

  16. modphysnoob
    • 3 years ago
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    haha, it better be

  17. Jemurray3
    • 3 years ago
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    The line integral is trivially simple, don't worry :)

  18. modphysnoob
    • 3 years ago
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    ok, I will try working it out

  19. modphysnoob
    • 3 years ago
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    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

  20. Jemurray3
    • 3 years ago
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    No, the line integral should be \[ \int_0^{2\pi} r \sin(\theta) \cdot A_\phi d\phi \]

  21. modphysnoob
    • 3 years ago
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    where does this come from?

  22. Jemurray3
    • 3 years ago
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    \[ d\vec{l} = \hat{r}dr + \hat{\theta} rd\theta + \hat{\phi} r\sin(\theta) d\phi\] \[\vec{A} = \hat{r} A_r + \hat{\theta} A_\theta + \hat{\phi} A_\phi \]

  23. modphysnoob
    • 3 years ago
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    we don't use r vector becuase it is set to constant?

  24. Jemurray3
    • 3 years ago
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    What do you mean? The path we're taking is a circle in the Z=0 plane, so the direction is purely phi.

  25. modphysnoob
    • 3 years ago
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    just to make sure we are on same page , phi is angle from x axis on x-y plane right

  26. Jemurray3
    • 3 years ago
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    yep

  27. modphysnoob
    • 3 years ago
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    well, A_phi= sin (theta)^2 so integrating it would result in sin(theta)^2*(2*pi)

  28. modphysnoob
    • 3 years ago
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    since it is in x-y plane ,evaluate at pi/2 sin(pi/2)^2 *(2*Pi)=2pi

  29. modphysnoob
    • 3 years ago
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    which is what I got for surface integral above

  30. Jemurray3
    • 3 years ago
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    Yep, good job.

  31. modphysnoob
    • 3 years ago
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    thanks you , you are not only very smart also very patient

  32. Jemurray3
    • 3 years ago
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    Thank you!

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