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anonymous

  • one year ago

MEDAL***Please help use spherical coordinates to find the volume cut out from the sphere x^2+y^2+z^2=1 by the planes z=1/2 and z=0

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  1. anonymous
    • one year ago
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    @ganeshie8

  2. ganeshie8
    • one year ago
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    Try \(\rho : ~0\to 1 \) \(\theta : ~0\to 2\pi \) \(\phi : ~\frac{\pi}{3}\to \frac{\pi}{2}\)

  3. anonymous
    • one year ago
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    as my limits for triple integration?

  4. ganeshie8
    • one year ago
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    Yes

  5. anonymous
    • one year ago
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    p is dx θ is dy ϕ is dz?

  6. anonymous
    • one year ago
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    whats the functions im integrating?

  7. ganeshie8
    • one year ago
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    Hmm @Loser66 I didn't get your question..

  8. ganeshie8
    • one year ago
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    @jyar what do you know about spherical coordinates ?

  9. anonymous
    • one year ago
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    there related to the cartesian coordinates and its in 3d points (r, θ , ϕ )

  10. ganeshie8
    • one year ago
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    what do \(\rho, \theta\) and \(\phi\) represent ?

  11. anonymous
    • one year ago
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    angles projected from the xyz plane

  12. Loser66
    • one year ago
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    .

  13. anonymous
    • one year ago
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    p^2=x^2+y^2+z^2

  14. ganeshie8
    • one year ago
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    |dw:1436648757792:dw|

  15. ganeshie8
    • one year ago
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    For any point \((\rho, \theta, \phi)\) in space, \(\rho\) is the "distance" from origin \(\theta\) is the angle in xy plane with the positive x axis \(\phi\) is the angle with positive z axis

  16. anonymous
    • one year ago
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    oh okay i understand that

  17. ganeshie8
    • one year ago
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    For our specific problem, it is easy to see that \(\rho\) varies from 0 to 1 because the radius of sphere is 1

  18. ganeshie8
    • one year ago
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    Also \(\theta\) varies from 0 to 2pi is also trivial

  19. ganeshie8
    • one year ago
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    you need to do some work to figure out the bounds for \(\phi\)

  20. anonymous
    • one year ago
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    is θ always 0 to 2pi?

  21. ganeshie8
    • one year ago
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    It depends, from the diagram you can see that \(\theta\) is the angle in xy plane. \(\theta ~: ~0\to 2\pi\) means this angle is swept one full revolution

  22. anonymous
    • one year ago
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    2 full revolution would be 4pi?

  23. ganeshie8
    • one year ago
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    You never want to do 2 full revolutions as that might duplicate the volume

  24. ganeshie8
    • one year ago
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    btw you're correct about 4pi being two full revolutions

  25. anonymous
    • one year ago
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    oh okay so when finding the limits of ϕ what would i do

  26. ganeshie8
    • one year ago
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    the part of sphere between z=1/2 and z=0 looks like below? |dw:1436649362434:dw|

  27. anonymous
    • one year ago
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    yeah i see that

  28. ganeshie8
    • one year ago
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    z=0 represents the xy plane, whats the angle \(\phi\) for xy plane ?

  29. ganeshie8
    • one year ago
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    Remember, \(\phi\) is the angle from z axis

  30. anonymous
    • one year ago
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    pi/3 to 0?

  31. ganeshie8
    • one year ago
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    z axis is perpendicular to xy plane, yes ?

  32. anonymous
    • one year ago
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    yes

  33. ganeshie8
    • one year ago
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    so whats the angle between xy plane and positive z axis ?

  34. anonymous
    • one year ago
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    its looks 60 degrees

  35. anonymous
    • one year ago
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    its not that ?

  36. ganeshie8
    • one year ago
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    Easy, xy plane makes 90 degrees with the positive z axis.

  37. ganeshie8
    • one year ago
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    |dw:1436649826870:dw|

  38. ganeshie8
    • one year ago
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    that means pi/2 is the ending angle lets find the starting angle

  39. anonymous
    • one year ago
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    its not under pi.2 its a little further away...so 17pi/12?

  40. ganeshie8
    • one year ago
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    |dw:1436649896094:dw|

  41. ganeshie8
    • one year ago
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    see if you can find \(\theta\) in that right triangle

  42. ganeshie8
    • one year ago
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    Honestly I have no idea how you got 17pi/12

  43. anonymous
    • one year ago
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    60 i got

  44. ganeshie8
    • one year ago
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    Yes starting angle is 60 degrees, which is same as pi/3

  45. anonymous
    • one year ago
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    yes

  46. ganeshie8
    • one year ago
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    so do the bounds make sense ? \(\rho : ~0\to 1 \) \(\theta : ~0\to 2\pi \) \(\phi : ~\frac{\pi}{3}\to \frac{\pi}{2}\)

  47. ganeshie8
    • one year ago
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    The volume is given by \[\large V = \int\limits_{0}^1 ~\int\limits_{0}^{2\pi}~ \int\limits_{\pi/3}^{\pi/2}~1~\color{blue}{\rho^2\sin\phi~d\phi~d\theta~d\rho}\]

  48. anonymous
    • one year ago
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    ok! the first integrations foes with p^2dp, second sinϕ dϕ, third dθ?

  49. ganeshie8
    • one year ago
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    It doesn't matter, only the bounds and the differentials need to agree

  50. ganeshie8
    • one year ago
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    Now that we have moved to spherical, \(\large \color{blue}{\rho^2 \sin\phi}\) is your integrand here

  51. anonymous
    • one year ago
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    ok i got that...could we do the first integration

  52. ganeshie8
    • one year ago
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    start by working the inner integral : \[\large V = \int\limits_{0}^1 ~\int\limits_{0}^{2\pi}~ \color{red}{\int\limits_{\pi/3}^{\pi/2}~1~\rho^2\sin\phi~d\phi}~d\theta~d\rho\]

  53. ganeshie8
    • one year ago
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    that red part

  54. anonymous
    • one year ago
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    the 1 intregrated becomes just ϕ?

  55. ganeshie8
    • one year ago
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    integrate below and plug it in \[\large \color{red}{\int\limits_{\pi/3}^{\pi/2}~1~\rho^2\sin\phi~d\phi}\]

  56. ganeshie8
    • one year ago
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    thats same as \[\large \color{red}{\int\limits_{\pi/3}^{\pi/2} \rho^2\sin\phi~d\phi}\]

  57. ganeshie8
    • one year ago
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    you can pull out \(\color{red}{\rho^2}\) because it is constant with respect to \(\phi\) : \[\large \color{red}{\rho^2\int\limits_{\pi/3}^{\pi/2} \sin\phi~d\phi}\]

  58. anonymous
    • one year ago
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    i got (-1/162) without pulling the p^2 out

  59. anonymous
    • one year ago
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    pulling the p^2 i get (1/2)p^2

  60. anonymous
    • one year ago
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    (-1/2)

  61. ganeshie8
    • one year ago
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    \[\large \color{red}{\rho^2\int\limits_{\pi/3}^{\pi/2} \sin\phi~d\phi} = \color{red}{\rho^2 (-\cos\phi)~ \Bigg|_{\pi/3}^{\pi/2} } = \color{red}{\rho^2 [-(0-\frac{1}{2})]} = \color{red}{\frac{1}{2}\rho^2}\]

  62. anonymous
    • one year ago
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    ok yes...for the next integration i \[\int\limits_{0}^{2\pi} (1/2)p^2 dp\]

  63. ganeshie8
    • one year ago
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    plugging that in the volume integral we get \[\large V = \int\limits_{0}^1 ~\int\limits_{0}^{2\pi}~ \color{red}{\int\limits_{\pi/3}^{\pi/2}~1~\rho^2\sin\phi~d\phi}~d\theta~d\rho = \int\limits_{0}^1 ~\int\limits_{0}^{2\pi}~ \color{red}{ \frac{1}{2}\rho^2}~d\theta~d\rho\]

  64. ganeshie8
    • one year ago
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    Nope, next we work : \[\large \int\limits_{0}^{2\pi} (1/2)p^2 d\color{red}{\theta}\]

  65. ganeshie8
    • one year ago
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    keepin mind \(0\to 2\pi\) are bounds for \(\theta\), not \(\rho\)

  66. anonymous
    • one year ago
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    oh okay so those bounds also how the order of work.... for the next intregation i got (4pi/3)

  67. ganeshie8
    • one year ago
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    \[\large \int\limits_{0}^{2\pi} (1/2)p^2 d\color{red}{\theta} = ?\]

  68. anonymous
    • one year ago
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    i pulled out the (1/2) and the inside i got (2pi^3/3) = (8pi/3)= (1/2)(8pi/3)?

  69. ganeshie8
    • one year ago
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    looks wrong, you can pull out entire thing, everythng is constant there

  70. ganeshie8
    • one year ago
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    \[\large \int\limits_{0}^{2\pi} (1/2)p^2 d\color{red}{\theta} = (1/2)p^2\int\limits_{0}^{2\pi} 1 d\color{red}{\theta} = ? \]

  71. anonymous
    • one year ago
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    (pi) p^2?

  72. ganeshie8
    • one year ago
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    Yes, plug that in the volume integral

  73. anonymous
    • one year ago
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    what if i didnt pull everything out as a constant and instead integrated would that be all wrong?

  74. ganeshie8
    • one year ago
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    plugging that in the volume integral we get \[\begin{align}\large V &= \int\limits_{0}^1 ~\int\limits_{0}^{2\pi}~ \color{red}{\int\limits_{\pi/3}^{\pi/2}~1~\rho^2\sin\phi~d\phi}~d\theta~d\rho = \int\limits_{0}^1 ~\int\limits_{0}^{2\pi}~ \color{red}{ \frac{1}{2}\rho^2}~d\theta~d\rho\\~\\ &=\int\limits_{0}^1 ~\pi \rho^2 ~d\rho\\~\\ &= ? \end{align}\]

  75. anonymous
    • one year ago
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    pi/3?

  76. ganeshie8
    • one year ago
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    Yep!

  77. anonymous
    • one year ago
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    regarding my question for the second integration if i didnt pull everything out as a constant and intregrated evrthhing instead would that be wrong?

  78. ganeshie8
    • one year ago
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    can you show me how exactly are you "integrating" everything ?

  79. anonymous
    • one year ago
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    \[\int\limits_{0}^{2\pi}(1/2)p^2 d\]

  80. anonymous
    • one year ago
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    wait i think i got it ! wow thank you soo much! so helpful

  81. ganeshie8
    • one year ago
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    d what ?

  82. anonymous
    • one year ago
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    d theta

  83. ganeshie8
    • one year ago
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    feel free to ask if you have any questions :)

  84. anonymous
    • one year ago
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    will you be online for rest of today?

  85. ganeshie8
    • one year ago
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    il be around for 1 hour or so, feel free to tag me in ur questions

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