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anonymous

  • one year ago

MEDAL!! Please help...! use triple integration to find the volume of the solid bounded by the cylinders x^2+y^2=9 and y^2+z^2=9

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  1. amistre64
    • one year ago
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    well, what part are you having an issue with?

  2. anonymous
    • one year ago
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    the limit points im using if -3 to 3, -squareroot (9-x^2), squareroot (9-x^2), -squareroot (9-y^2), -squareroot (9-y^2)

  3. anonymous
    • one year ago
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    can you guide the first couple steps, im really confused please

  4. amistre64
    • one year ago
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    trying to see if i can see the boundary |dw:1436633020633:dw|

  5. anonymous
    • one year ago
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    is it the inner part of cylinder we are looking for?

  6. amistre64
    • one year ago
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    are your limits valid? other than that it really is just a mechanical process.

  7. anonymous
    • one year ago
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    there valid but im having trouble with the integration

  8. amistre64
    • one year ago
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    well lets see what we can work then dz from -sqrt(9-y^2) to sqrt(9-y^2) dy from -sqrt(9-x^2) to sqrt(9-x^2) dx from -3 to 3 you agree that this is how we want to move?

  9. anonymous
    • one year ago
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    yes

  10. amistre64
    • one year ago
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    then we simply work it from the inside out \[\int \left(\int\left[\int dz\right]~dy\right)~dx\] \[\int \left(\int z_b-z_a~dy\right)~dx\] to start with

  11. amistre64
    • one year ago
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    \[\int \left(2\int \sqrt{9-y^2}~dy\right)~dx\] let y = 3sin(t) , dy = 3cos(t) dt \[\int \left(2\int \sqrt{9-(3sin(t))^2}~3cos(t)~dt\right)~dx\] \[\int \left(2\int \sqrt{9(1-sin^2(t))}~3cos(t)~dt\right)~dx\] \[\int \left(2\int 3cos^2(t)*3cos(t)~dt\right)~dx\] \[\int \left(18\int cos^3(t)~dt\right)~dx\] etc ..

  12. anonymous
    • one year ago
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    where did the sin^2t and cos come from

  13. amistre64
    • one year ago
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    i have a small error ... but its just using a trig substitution we let y = 3sin(t) and substitute all the y parts. sqrt(1-sin^2) = cos ... not cos^2 tho

  14. anonymous
    • one year ago
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    did you do squareroot (9-y^2) ==> squareroot (9-((9-x^2)^(1/2))?

  15. anonymous
    • one year ago
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    where did you plug in the limits for y?

  16. amistre64
    • one year ago
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    no sqrt(9 - y^2) dy if we let y = 3sin(t), the derivative allows us to replace dy by dt parts dy = 3cos(t) dt by replacing the appropriate parts ... sqrt(9 - (3sin(t))^2) * 3cos(t) dt

  17. amistre64
    • one year ago
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    my only concern for this is your limits ... im just not sure that the intersection of 2 cylindars is a sphere ... and that is what your limits are suggesting.

  18. anonymous
    • one year ago
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    right i see that...so my limits for dz are wrong?

  19. amistre64
    • one year ago
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    i believe your limits are bad ... im just not sure at the moment how we would correct them off hand. the draw tool is not that good at 3d imaging for me

  20. anonymous
    • one year ago
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    is there a formula... in mathetica it demonstrated 2 cylinders crossing eachother like a cross and what i was told to look for the volume of the non-intersecting upper part

  21. amistre64
    • one year ago
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    uh, you originally asked for the volume of the intersection ...

  22. amistre64
    • one year ago
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    |dw:1436634359956:dw| so you want the volume of something like this?

  23. anonymous
    • one year ago
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    1 Attachment
  24. anonymous
    • one year ago
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    it looks like this

  25. amistre64
    • one year ago
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    right, and you want the volume of .. what now?

  26. amistre64
    • one year ago
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    if its just 2 cylndars intersecting, i thing the profile of the intersection is a "pringle" chip .. or a saddle back

  27. amistre64
    • one year ago
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    |dw:1436634627932:dw|

  28. anonymous
    • one year ago
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    yes ignore the red cylinder

  29. anonymous
    • one year ago
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    exactly like that

  30. anonymous
    • one year ago
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    from that part, how do you find the limits?

  31. amistre64
    • one year ago
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    now, you want to find the volume of the top part? or of the intersecting part? if its the top part, then we need to have a plane of something that cuts it off from going to infinity

  32. anonymous
    • one year ago
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    interesting part

  33. amistre64
    • one year ago
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    ok, interSECTing part :) yeah z limits are fine, -sqrt(9-y^2) to sqrt(9-y^2) x limits are fine, -3 to 3 its your y limits that need refined if anything

  34. anonymous
    • one year ago
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    good to hear its not completey wrong. y limits how to find them?

  35. amistre64
    • one year ago
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    still working on that ....

  36. anonymous
    • one year ago
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    its oki ! im still here

  37. amistre64
    • one year ago
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    https://www.youtube.com/watch?v=rASKzEbMXIQ this might be a bit more useful then me trying to remember things long dead past :) at least for a review

  38. anonymous
    • one year ago
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    thank you so much

  39. anonymous
    • one year ago
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    would you happen to remember greens thm?

  40. amistre64
    • one year ago
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    i never really covered greens thrm in my courses. so no.

  41. anonymous
    • one year ago
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    its oki..would you be staying online today? if its oki to ask you other questions

  42. amistre64
    • one year ago
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    im not going to be online much today .. got family stuff to do. but after the youtube video i do see where the issue was :) it was in the limits

  43. anonymous
    • one year ago
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    oh so the set up is the same?

  44. amistre64
    • one year ago
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    the setup is the same, but we need to better define the limits ... and using symmetry helps x^2+y^2=9 y^2+z^2=9 they both have a y^2 in common so lets define z and x in terms of y x = 0 to sqrt(9-y^2) z = 0 to sqrt(9-y^2) y = 0 to 3 and this defines 1/8 of our volume dz dx dy is a good order of integration

  45. amistre64
    • one year ago
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    \[\int_{y=0}^{y=3}\int_{x=0}^{x=\sqrt{9-y^2}}\int_{z=0}^{z=\sqrt{9-y^2}}~dz~dx~dy\]

  46. anonymous
    • one year ago
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    oki can you show me the frst integrration please

  47. amistre64
    • one year ago
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    well, using your fundamental thrm of calculus ... \[\int_{0}^{b}dz=F(b)-F(0)\]

  48. amistre64
    • one year ago
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    what is the integration of 1?

  49. anonymous
    • one year ago
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    x

  50. amistre64
    • one year ago
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    well, z in this case, since we are integrating with respect to z but yeah so z(b) - z(0) = z(b)

  51. anonymous
    • one year ago
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    so with z...then we plug in the limits of x?

  52. amistre64
    • one year ago
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    limits of z ...

  53. amistre64
    • one year ago
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    \[\int_{0}^{\sqrt{9-y^2}}dz=z_b-z_a=\sqrt{9-y^2}-0\]

  54. amistre64
    • one year ago
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    now, what is the middle integral? \[\int_{0}^{\sqrt{9-y^2}}\sqrt{9-y^2}~dx=??\]

  55. anonymous
    • one year ago
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    \[\int\limits_{0}^{3} (1/2)(\sqrt{9-y^2}y+9\sin^{-1} (y/3)\]

  56. amistre64
    • one year ago
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    not quite sqrt(9-y^2) , with respect to x is just some constant ... so, what is the integration of a constant again?

  57. anonymous
    • one year ago
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    x^2/2

  58. amistre64
    • one year ago
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    no, try again.

  59. amistre64
    • one year ago
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    what was the integration of 1?

  60. anonymous
    • one year ago
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    z

  61. amistre64
    • one year ago
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    the respective variable yes ... in this case we are doing dx, so its an x

  62. amistre64
    • one year ago
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    for some constant k \[\int_{0}^{b}dx=b(k)-0(k)\]

  63. amistre64
    • one year ago
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    in this case, k=sqrt(9-y^2), and b=sqrt(9-y^2) leaving us: \[\int_{0}^{3}9-y^2~dy\]

  64. anonymous
    • one year ago
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    oh ok so this is our middle integral

  65. amistre64
    • one year ago
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    yes, we did the first, then the middle, and we are now left with the end

  66. anonymous
    • one year ago
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    is the final answer 19?

  67. amistre64
    • one year ago
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    dunno, i havent gotten to the end of it yet

  68. amistre64
    • one year ago
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    i dont think 19 is it

  69. anonymous
    • one year ago
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    18

  70. amistre64
    • one year ago
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    what is: \[\int_{0}^{3}9-y^2~dy?\] work the steps for me

  71. anonymous
    • one year ago
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    \[9y-y\]

  72. anonymous
    • one year ago
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    9y-y^3/3

  73. amistre64
    • one year ago
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    9y - y^3/3 , given that y=3 yes 27 - 27/3 = 27(2/3) = 18 but recall that we only worked 1/8 of it, so we need to mulitply this by 8 to get the total volume.

  74. anonymous
    • one year ago
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    144...can you remind me again why we work with 1/8

  75. anonymous
    • one year ago
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    the limits for x and z why are they the same?

  76. amistre64
    • one year ago
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    we dont have to, but due to the symmetry of the intersection, it makes life a little simpler for some.

  77. amistre64
    • one year ago
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    those are better explained in the youtube video i posted.

  78. anonymous
    • one year ago
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    oki i understand...so much more clear. thank you

  79. amistre64
    • one year ago
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    good luck :)

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