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Triple integration to find volume of regions

  1. Dec 31, 2011 #1

    sharks

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    The problem statement, all variables and given/known data
    http://s2.ipicture.ru/uploads/20111231/kczcXUuF.jpg

    The attempt at a solution
    So, i'm using the transformation to spherical coordinates (ρ,∅,θ)

    Description of region:

    For θ and ∅ fixed, ρ varies from 0 to 4.
    For θ fixed, ∅ varies from 0 to ∏. (i suspect the error is here)
    θ varies from 0 to 2∏.

    But i don't know what i'm doing wrong. I've checked my limits, but i think that's the problem, since i can't get the correct answer.

    When i calculate ∅ by expanding into (ρrsin∅cosθ)^2 + (ρrsin∅sinθ)^2 = 4
    Substituting r=2, i get the value of ∅ = ∏/6, which i also tested as the limits from 0 to ∏/6 and still got the wrong answer.
    I then tested limits ∏/6 to 5∏/6 and still wrong.

    Here is the answer from my notes:
    http://s2.ipicture.ru/uploads/20111231/B1yTY1Ye.jpg
     
    Last edited: Dec 31, 2011
  2. jcsd
  3. Dec 31, 2011 #2
    I don't know about you but I get an error when I try to see the attachments.
     
  4. Dec 31, 2011 #3

    sharks

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    Try to refresh the page. They load perfectly fine for me. I'm using Firefox browser.
    Here is a mirror of the question (click on the image to enlarge):
    th_45697_triple_122_183lo.jpg
    And the answer:
    th_45696_anstriple_122_218lo.jpg
     
  5. Dec 31, 2011 #4
    Find the volume of the regions described by inside both
    o x2+y2+z2 = 16
    o x2+y2 = 4


    Answer: [itex]\frac{32π(8-33/2)}{3}[/itex]
     
  6. Dec 31, 2011 #5

    sharks

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    I've already drawn the graph but i don't know which limits are wrong and why.

    For θ and ∅ fixed, ρ varies from 0 to 4.
    For θ fixed, ∅ varies from 0 to ∏. (i suspect the error is here)
    θ varies from 0 to 2∏.

    OK, so here are my calculations:

    From
    [tex]x^2+y^2+z^2 = 16[/tex]
    I know, by formula, that
    [tex]\rho^2= 16[/tex]
    So,
    [tex]\rho=4[/tex]
    For θ and ∅, i think it's obvious, at least i think so. I don't know where i went wrong.

    Are you suggesting that i find the volume of the sphere first and then subtract the volume of the cylinder from that? I can find the volume of the sphere but not sure how to find the volume of cylinder using spherical coordinates.
     
    Last edited: Dec 31, 2011
  7. Dec 31, 2011 #6
    If you did so, I mean finding the volume of the cylinder alone, you would miss the two little spherical part over the bases of the cylinder. Anyway, I want to try it on my own, I'll tell then. :D
     
  8. Dec 31, 2011 #7

    sharks

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    It looks like it's going to be a long night...
     
  9. Dec 31, 2011 #8
    I tried something but I did some error, it was my last try for today.
    Anyway, my limits where [0,2π] for θ, [π/6,5π/6] for ø and [1/(2sinø), 4] for p. In fact the minimum of the radius changes with ø and in particular I found that function for p. My result is terrible, I get a negative number, but I don't have time now, I'll try again tomorrow maybe. Anyway, that's the right idea. Good Year!
     
  10. Dec 31, 2011 #9

    SammyS

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    Hello sharks !

    The problem can be done more easily in cylindrical coordinates. If you use spherical coordinates, it looks to me as if you will have to break this up into the sum of two integrals.

    For the part of the region bounded by the sphere, 0 ≤ ρ ≤ 4. In this region, 0 ≤ ɸπ/6 , also, 5π/6 ≤ ɸπ.

    For the part of the region bounded by the cylinder, π/6 ≤ ɸ ≤ 5π/6, and ρ does not go to 4.
    In this region [itex]x^2+y^2=\rho^2-\rho^2\cos^2(\phi)=4\,.[/itex] Use that to figure out the bounds for ρ.​
     
  11. Dec 31, 2011 #10
    Interesting though the triple integration might be, and tempting as it might be to change coordinates, I would just calculate where the sphere meets the cylinder and integrate the area over z to find the volume of the cap.

    [tex]\int^{4}_{√12}\pi(16-z^{2})dz[/tex]
     
  12. Dec 31, 2011 #11
    Sorry, posted wrong advice. Lol. This thing needs to be done in cylindrical coordinates I believe. Or as Joffan suggested, in standard Cartesian coordinates.
     
    Last edited: Dec 31, 2011
  13. Dec 31, 2011 #12

    sharks

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    Is that the ripple effect of the New Year? So many wrong advices in this thread. LOL

    But seriously, this question is found at the end of the triple integral using spherical coordinates section, and it must be solved using transformation to spherical coordinates.

    I guess the main question here is - is it possible to solve the volume of a cylinder using cylindrical coordinates? I think yes.

    SammyS, i am following your idea, which is: sum of the top and bottom cones with spherical hats and concave middle cylindrical shape.
    Is that a mistake? It's supposed to be [itex]x^2+y^2=\rho^2\sin^2(\phi)=4\,,[/itex] so [itex]\rho\sin(\phi)=2\,.[/itex]
     
    Last edited: Jan 1, 2012
  14. Jan 1, 2012 #13

    SammyS

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    It's the same thing, just a different way to arrive at it.

    [itex]\rho^2-\rho^2\cos^2(\phi)=\rho^2(1-\cos^2(\phi))[/itex]
    [itex]=\rho^2\sin^2(\phi)[/itex]​
    ... but the form you have it in will definitely work better.
     
  15. Jan 1, 2012 #14
    I'm drunk, aren't you. For the love of god.
     
  16. Jan 1, 2012 #15

    sharks

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    OK, if i understood correctly, i should be able to derive the final answer as the sum of these 3 triple integrals:

    The top cone:
    [tex]\int^{2\pi}_{0}\int^{\frac{\pi}{6}}_{0} \int^{4}_{0}\rho^2sin\phi\,.d\rho d\phi dz[/tex]

    The bottom cone:
    [tex]\int^{2\pi}_{0}\int^{\pi}_{\frac{5\pi}{6}} \int^{4}_{0}\rho^2sin\phi\,.d\rho d\phi dz[/tex]

    The middle section, lined by the sides of the cylinder:
    [tex]\int^{2\pi}_{0}\int^{\frac{5\pi}{6}}_{\frac{\pi}{6}} \int^{\frac{2}{\sin\phi}}_{0}\rho^2sin\phi\,.d\rho d\phi dz[/tex]

    I added the 3 answers from the 3 integrals above and it's wrong.


    LOL. I've been most of the night. It just took a little longer to kick in and disable my typing skills. :biggrin:
     
    Last edited: Jan 1, 2012
  17. Jan 1, 2012 #16

    SammyS

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    Note that the volume of the top cone equals the volume of the bottom cone.

    What did you get for an answer ?
     
  18. Jan 1, 2012 #17
    I did it in the same way before reading this post and I got the same answer, I mean the correct one posted.
    You may deal with it also by calculating the volume of the outer region of the sphere, I mean that out of the cylinder and inside the sphere, and then subtract the result to the volume of the whole sphere. This way the boundary of p is [2/sinø, 4] and that of ø is [π/6, 5π/6]. I haven't checked this way yet, but it must be right.
     
  19. Jan 1, 2012 #18

    sharks

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    Yes, i got the same answer for both cones:
    [tex]2\pi(\frac{64}{3}-\frac{32\sqrt3}{3})[/tex]
    I found it easier to convert it to a numerical answer (in order to add and find the total volume): 17.95812724

    The volume of the middle concave cylinder section:
    [tex]16\sqrt3 = 174.124739[/tex] (maybe this is the wrong part?)

    So, the total volume = (2 x 17.95812724) + 174.124739 = 210.0409934

    However, the correct numerical answer is: 93.95783414

    EDIT: OK, i got it. I made a mistake when integrating the middle cylinder section. The answer is 58.04157697
    The total volume gives me 93.95783414, which is correct!

    Thanks to SammyS and others for your advice.
     
    Last edited: Jan 1, 2012
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