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Concentrationalizing

  • 2 years ago

Convert the triple integral into spherical coordinates.

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  1. Concentrationalizing
    • 2 years ago
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    \[\int\limits_{-2}^{2}\int\limits_{-\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}}\int\limits_{x^{2}+y^{2}}^{4}xdzdydx\]

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

  3. Concentrationalizing
    • 2 years ago
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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.

  4. RolyPoly
    • 2 years ago
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    |dw:1389713913315:dw|

  5. RolyPoly
    • 2 years ago
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    Obviously, \(0\le r\le 4\) and \(0\le \theta \le 2\pi\)

  6. Concentrationalizing
    • 2 years ago
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    Mmm....0 <= r <= 2

  7. RolyPoly
    • 2 years ago
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    Epic fail.... Yes Yes... Sorry :(

  8. Concentrationalizing
    • 2 years ago
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    No worries, lol.

  9. RolyPoly
    • 2 years ago
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    |dw:1389714352578:dw|

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

  11. Concentrationalizing
    • 2 years ago
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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.

  12. RolyPoly
    • 2 years ago
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    I am soooo dumb. I consider myself as quite new to this, although not so new... Good luck!

  13. Concentrationalizing
    • 2 years ago
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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.

  14. Concentrationalizing
    • 2 years ago
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    @RolyPoly think youd be able to figure out this last part, too?

  15. RolyPoly
    • 2 years ago
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    Why does it look so complicated?! :S

  16. RolyPoly
    • 2 years ago
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    What is the \(\phi_{\text{max}}\) you get?

  17. Concentrationalizing
    • 2 years ago
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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.

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

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

  20. RolyPoly
    • 2 years ago
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    Must you write in this.... ugly form?

  21. Concentrationalizing
    • 2 years ago
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    Im sorry. Im just trying to answer this question......

  22. RolyPoly
    • 2 years ago
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    What is the exact question?

  23. Concentrationalizing
    • 2 years ago
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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.

  24. RolyPoly
    • 2 years ago
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    Actually you need not write that ugly thing, I think. Let me double check something first.

  25. Concentrationalizing
    • 2 years ago
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    alright

  26. RolyPoly
    • 2 years ago
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    Do you agree that this is the region? |dw:1389717244057:dw|

  27. Concentrationalizing
    • 2 years ago
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    I figured it was the cone portion, yes.

  28. RolyPoly
    • 2 years ago
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    (The cone part)

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

  30. Concentrationalizing
    • 2 years ago
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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.

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

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

  33. Concentrationalizing
    • 2 years ago
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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

  34. RolyPoly
    • 2 years ago
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    Must you use that to integrate? It is really ugly and not necessary to do that step..

  35. Concentrationalizing
    • 2 years ago
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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.

  36. RolyPoly
    • 2 years ago
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    What if I can integrating it using spherical coordinates without getting to that step?

  37. RolyPoly
    • 2 years ago
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    I mean to write that integral in spherical coordinates.

  38. Concentrationalizing
    • 2 years ago
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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.

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

  40. Concentrationalizing
    • 2 years ago
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    Yes.

  41. RolyPoly
    • 2 years ago
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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?

  42. Concentrationalizing
    • 2 years ago
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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.

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

  44. Concentrationalizing
    • 2 years ago
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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.

  45. experimentX
    • 2 years ago
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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 \[ \int_{-a}^a f(x) dx = 0\]

  46. Concentrationalizing
    • 2 years ago
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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.

  47. RolyPoly
    • 2 years ago
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    Consider the black triangle |dw:1389718930585:dw|

  48. RolyPoly
    • 2 years ago
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    We get \[\cos\phi = \frac{4}{r}\]\[r=\frac{4}{\cos\phi}=4 \sec\phi\]

  49. Concentrationalizing
    • 2 years ago
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    Mhm, thats rho in the first integral, I managed that part, lol.

  50. RolyPoly
    • 2 years ago
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    And \(\phi\) in the first part |dw:1389719149953:dw|

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

  52. Concentrationalizing
    • 2 years ago
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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.

  53. RolyPoly
    • 2 years ago
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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!

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

  55. Spacelimbus
    • 2 years ago
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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.

  56. RolyPoly
    • 2 years ago
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    I also have to go now. Will check this post again as soon as possible.

  57. experimentX
    • 2 years ago
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    |dw:1389729705792:dw|

  58. experimentX
    • 2 years ago
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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.

  59. Concentrationalizing
    • 2 years ago
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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 ?

  60. experimentX
    • 2 years ago
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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|

  61. experimentX
    • 2 years ago
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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.

  62. Mendicant_Bias
    • one year ago
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    What we're graphing is a paraboloid with a circular cross-section, right? The thing that initially confused me, looking at the first few posts, is that a cone is depicted.

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