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Concentrationalizing Group Title

Convert the triple integral into spherical coordinates.

  • 6 months ago
  • 6 months ago

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  1. Concentrationalizing Group Title
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    \[\int\limits_{-2}^{2}\int\limits_{-\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}}\int\limits_{x^{2}+y^{2}}^{4}xdzdydx\]

    • 6 months ago
  2. RolyPoly Group Title
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    I think you are integrating in this region. Not quite sure. |dw:1389713649402:dw|

    • 6 months ago
  3. Concentrationalizing Group Title
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    Thats what I get, yes. But the problem is more about knowing how to do the conversions, not the actual integration. I was able to get it into cylindrical coordinates, just didnt know how to do it properly with spherical.

    • 6 months ago
  4. RolyPoly Group Title
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    |dw:1389713913315:dw|

    • 6 months ago
  5. RolyPoly Group Title
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    Obviously, \(0\le r\le 4\) and \(0\le \theta \le 2\pi\)

    • 6 months ago
  6. Concentrationalizing Group Title
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    Mmm....0 <= r <= 2

    • 6 months ago
  7. RolyPoly Group Title
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    Epic fail.... Yes Yes... Sorry :(

    • 6 months ago
  8. Concentrationalizing Group Title
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    No worries, lol.

    • 6 months ago
  9. RolyPoly Group Title
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    |dw:1389714352578:dw|

    • 6 months ago
  10. RolyPoly Group Title
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    So, we have \[\tan \phi = \frac{2}{4}\]\[\phi = ... = \phi_{\text{max}}\] \[0\le\phi\le\phi_{\text{max}}\]

    • 6 months ago
  11. Concentrationalizing Group Title
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    Oh wow x_x I feel dumb now, lol. Im so new to these things, so not used to the thought process. For some reason I never thought to just make that triangle from the cone. Okay, so that gets me half of the answer. Now Im trying to see if I can derive the other half.

    • 6 months ago
  12. RolyPoly Group Title
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    I am soooo dumb. I consider myself as quite new to this, although not so new... Good luck!

    • 6 months ago
  13. Concentrationalizing Group Title
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    So, what Im trying to find out now is how they have two integrals in the answer. p is between 0 and 4sec(phi) for the first integral and phi is in between 0 and arctan(1/2). But then theres a 2nd integral that is added that has the range of p going from 0 to cot(phi)*csc(phi) and phi going from arctan(1/2) to pi/2. Im trying to see how to derive that part now.

    • 6 months ago
  14. Concentrationalizing Group Title
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    @RolyPoly think youd be able to figure out this last part, too?

    • 6 months ago
  15. RolyPoly Group Title
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    Why does it look so complicated?! :S

    • 6 months ago
  16. RolyPoly Group Title
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    What is the \(\phi_{\text{max}}\) you get?

    • 6 months ago
  17. Concentrationalizing Group Title
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    Im not sure how id get a phi max here. I just know that phi would be arctan(1/2), im not sure how to find another value for it.

    • 6 months ago
  18. RolyPoly Group Title
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    \[\phi_{\text{max}} = \tan^{-1}(\frac{1}{2})\] And the range of \(\phi\) is \(0\le\phi\le\phi_{\text{max}}\)

    • 6 months ago
  19. Concentrationalizing Group Title
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    Yes, that part I have, but then there is a second integral. Here, this is what the answer says: \[\int\limits_{0}^{2 \pi}\int\limits_{0}^{\arctan(1/2)}\int\limits_{0}^{4 \sec \phi} \rho^{3} \sin^{2} \phi \cos \theta d \rho d \phi d \theta + \int\limits_{0}^{2 \pi}\int\limits_{\arctan(1/2)}^{\frac{ \pi }{ 2 }}\int\limits_{0}^{\cot \phi \csc \phi} \rho^{3} \sin^{2} \phi \cos \phi d \rho d \phi d \theta\]

    • 6 months ago
  20. RolyPoly Group Title
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    Must you write in this.... ugly form?

    • 6 months ago
  21. Concentrationalizing Group Title
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    Im sorry. Im just trying to answer this question......

    • 6 months ago
  22. RolyPoly Group Title
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    What is the exact question?

    • 6 months ago
  23. Concentrationalizing Group Title
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    To convert the triple integral into cylindrical and spherical coordinates and then compute the integral. I was just having trouble understanding the answer for the spherical coordinates conversion. Sure I can see the answer, but Im not sure how it was derived. The triangle on the cone earlier helped me understand the first integral, but I dont understand where the second integral comes from now.

    • 6 months ago
  24. RolyPoly Group Title
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    Actually you need not write that ugly thing, I think. Let me double check something first.

    • 6 months ago
  25. Concentrationalizing Group Title
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    alright

    • 6 months ago
  26. RolyPoly Group Title
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    Do you agree that this is the region? |dw:1389717244057:dw|

    • 6 months ago
  27. Concentrationalizing Group Title
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    I figured it was the cone portion, yes.

    • 6 months ago
  28. RolyPoly Group Title
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    (The cone part)

    • 6 months ago
  29. RolyPoly Group Title
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    Ok, then no matter in which coordinate system, the cone is the region we want to integrate, right?

    • 6 months ago
  30. Concentrationalizing Group Title
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    Yes. I think I can integrate it okay, but I just need to know how it got what it did for the spherical coordinate interpretation.

    • 6 months ago
  31. RolyPoly Group Title
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    Do you have problems in understanding why \(\rho\) changes from 0 to 2 and \(\theta\) changes from 0 to \(2\pi\)?

    • 6 months ago
  32. RolyPoly Group Title
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    And \(\phi\) changes from 0 to \(\tan^{-1}(\frac{1}{2})\) as well.

    • 6 months ago
  33. Concentrationalizing Group Title
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    I dont understand the second integral in the answer I posted. How rho goes from 0 to cot(phi)csc(phi) and how phi goes from arctan(1/2) to pi/2

    • 6 months ago
  34. RolyPoly Group Title
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    Must you use that to integrate? It is really ugly and not necessary to do that step..

    • 6 months ago
  35. Concentrationalizing Group Title
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    Its not required to integrate it, you just have to know how to derive it. The problem can be integrated using any coordinate system, but you still have to know how to convert it to all of them.

    • 6 months ago
  36. RolyPoly Group Title
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    What if I can integrating it using spherical coordinates without getting to that step?

    • 6 months ago
  37. RolyPoly Group Title
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    I mean to write that integral in spherical coordinates.

    • 6 months ago
  38. Concentrationalizing Group Title
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    I just need to know where that second integral even comes from. I mean sure, if you know how to do it another way thats fine, but Id rather understand the answer first.

    • 6 months ago
  39. RolyPoly Group Title
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    If you insist, that's fine. Are you sure that you type that correctly? \[\int\limits_{0}^{2 \pi}\int\limits_{0}^{\arctan(1/2)}\int\limits_{0}^{4 \sec \phi} \rho^{3} \sin^{2} \phi \cos \theta d \rho d \phi d \theta + \int\limits_{0}^{2 \pi}\int\limits_{\arctan(1/2)}^{\frac{ \pi }{ 2 }}\int\limits_{0}^{\cot \phi \csc \phi} \rho^{3} \sin^{2} \phi \cos \phi d \rho d \phi d \theta\]?

    • 6 months ago
  40. Concentrationalizing Group Title
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    Yes.

    • 6 months ago
  41. RolyPoly Group Title
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    For changing the coordinates, all you need is to find the new limits and the Jacobian. I just don't understand why you insist on understand that answer instead of finding the limits by drawing a picture, which is way easier. Are you sure that it is \(\rho^3\sin^2\phi\cos\phi\) for the second integral?

    • 6 months ago
  42. Concentrationalizing Group Title
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    The jacobian is in the section after the one Im in. And yeah, it should be cos theta for the second integral, my bad.

    • 6 months ago
  43. RolyPoly Group Title
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    How do you know you need to multiply \(\rho ^2 \sin\phi\) to the integral then?

    • 6 months ago
  44. Concentrationalizing Group Title
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    Its in the basic conversion to spherical coordinates. There is no explanation really as to why its that, but it says to add an extra r to cylindrical and an extra p^2sin(phi) to spherical. Perhaps the jacobian explains why, but im not there yet.

    • 6 months ago
  45. experimentX Group Title
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    let \[ f(x) = \int_{-\sqrt{4-x^2}}^{-\sqrt{4-x^2}} \int_{x^2+y^2}^4 x dy dz\] first of all note that \( f(-x) = -f(x) \) \[ f(-x) = \int_{-\sqrt{4-(-x)^2}}^{-\sqrt{4-(-x)^2}} \int_{(-x)^2+y^2}^4 (-x) dy dz = -\int_{-\sqrt{4-x^2}}^{-\sqrt{4-x^2}} \int_{x^2+y^2}^4 x dy dz = -f(x)\] for odd functions \[ \int_{-a}^a f(x) dx = 0\]

    • 6 months ago
  46. Concentrationalizing Group Title
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    Well, thats the integral evaluated yes, but I guess all I really needed to understand was that spherical answer I posted. How that conversion was derived. In the long run it may not be important, but I at least want to understand it since it was one of my problems.

    • 6 months ago
  47. RolyPoly Group Title
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    Consider the black triangle |dw:1389718930585:dw|

    • 6 months ago
  48. RolyPoly Group Title
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    We get \[\cos\phi = \frac{4}{r}\]\[r=\frac{4}{\cos\phi}=4 \sec\phi\]

    • 6 months ago
  49. Concentrationalizing Group Title
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    Mhm, thats rho in the first integral, I managed that part, lol.

    • 6 months ago
  50. RolyPoly Group Title
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    And \(\phi\) in the first part |dw:1389719149953:dw|

    • 6 months ago
  51. Concentrationalizing Group Title
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    Yes, arctan(1/2) because of the sides of the black triangle you drew.

    • 6 months ago
  52. Concentrationalizing Group Title
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    I gotta go :/ If either of you can explain that 2nd integral in the answer I posted, then please do and post it for me. Thanks for your time.

    • 6 months ago
  53. RolyPoly Group Title
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    @Spacelimbus Sir, sorry for tagging you here. Here's the problem: Show that \[\int\limits_{-2}^{2}\int\limits_{-\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}}\int\limits_{x^{2}+y^{2}}^{4}xdzdydx\]\[=\int\limits_{0}^{2 \pi}\int\limits_{0}^{\arctan(1/2)}\int\limits_{0}^{4 \sec \phi} \rho^{3} \sin^{2} \phi \cos \theta d \rho d \phi d \theta + \int\limits_{0}^{2 \pi}\int\limits_{\arctan(1/2)}^{\frac{ \pi }{ 2 }}\int\limits_{0}^{\cot \phi \csc \phi} \rho^{3} \sin^{2} \phi \cos \phi d \rho d \phi d \theta\] We have problem in understanding the second integral on the right. Please give us if possible. Thanks in advance!

    • 6 months ago
  54. RolyPoly Group Title
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    Amendment to the fifth line. The fourth and the fifth line should be \[\int\limits_{-2}^{2}\int\limits_{-\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}}\int\limits_{x^{2}+y^{2}}^{4}xdzdydx\]\[=\int\limits_{0}^{2 \pi}\int\limits_{0}^{\arctan(1/2)}\int\limits_{0}^{4 \sec \phi} \rho^{3} \sin^{2} \phi \cos \theta d \rho d \phi d \theta + \int\limits_{0}^{2 \pi}\int\limits_{\arctan(1/2)}^{\frac{ \pi }{ 2 }}\int\limits_{0}^{\cot \phi \csc \phi} \rho^{3} \sin^{2} \phi \cos \theta d \rho d \phi d \theta\]

    • 6 months ago
  55. Spacelimbus Group Title
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    I will take a look into this in a few, seems to take a while to play with the limits of integration to get it in the desired form.

    • 6 months ago
  56. RolyPoly Group Title
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    I also have to go now. Will check this post again as soon as possible.

    • 6 months ago
  57. experimentX Group Title
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    |dw:1389729705792:dw|

    • 6 months ago
  58. experimentX Group Title
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    |dw:1389729892139:dw||dw:1389729934880:dw| solve \[ x^2 + y^2 + z^2 =r^2 \\ z = x^2 +y^2\] for 'r' and obtain \( r = \cot(\phi) \csc(\phi) \) easy way, since it is symmetric on theta, put theta = 0. then put x=r sin(phi), z=r cos(phi) and solve.

    • 6 months ago
  59. Concentrationalizing Group Title
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    Alrighty. So why is the second integral necessary? Iwouldve thought that just the first integral wouldve covered it all, but for some reason two integrals were needed. Do you know why? @experimentX ?

    • 6 months ago
  60. experimentX Group Title
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    this is due to two different values of 'r' in two different regions. for one region you have 4sec(phi) and for other you have cot(phi)csc(phi) |dw:1389785515049:dw||dw:1389785535697:dw|

    • 6 months ago
  61. experimentX Group Title
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    note that in first region r goes from to to .. that straight line, on the second region ... r goes from 0 to that parabola.

    • 6 months ago
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