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anonymous

  • 5 years ago

hey! anyone out there good at drawing triple integrals or good in dzdr(dtheta) plane?

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  1. anonymous
    • 5 years ago
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    Umm. I can draw......but i don't know if i can draw triple integrals... sorry

  2. anonymous
    • 5 years ago
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    what's the equation you're dealing with?

  3. anonymous
    • 5 years ago
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    one is the triple integral of (2pi to 0), (square root of 3 to 0), then (3- r squared) with the inside of the integral as rdzdr(dtheta)

  4. anonymous
    • 5 years ago
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    sorry hope that's understandable, i can't type out a lot of those symbols

  5. anonymous
    • 5 years ago
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    is it \[ \int\limits_{0}^{2\pi}\int\limits_{0}^{\sqrt{3}}\int\limits_{3}^{r^2}rdzdrd \theta\]

  6. anonymous
    • 5 years ago
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    if not try using the equation button on the bottowm of the chat window

  7. anonymous
    • 5 years ago
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    yep everything's right except the last one is (3-r^2) to 0, sorry i forgot to insert the 0

  8. anonymous
    • 5 years ago
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    cool, give me one sec.

  9. anonymous
    • 5 years ago
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    so first we integrate with respect to z, so now it's going to be\[\int\limits_{0}^{2\pi}\int\limits_{0}^{\sqrt{3}}rzdrd \Theta\], then we sub in (3-r^2) for z. Zero cancels out so now it's\[ \int\limits_{0}^{2\pi}\int\limits_{0}^{\sqrt{3}}3r-r^3drd \Theta\]. Then we integrate wrt r, and sub in root 3

  10. anonymous
    • 5 years ago
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    \[\int\limits_{0}^{2\pi}(3/2)r^2-(1/4)r^4d \Theta =\int\limits_{0}^{2\pi}(9/2)-(9/4)d \Theta = \int\limits(9/4)d \Theta\], this = 9pi/2. It's not something you can really draw because it is a hypervolume, but that's the answer though.

  11. anonymous
    • 5 years ago
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    awesome thanks so much! glad to know that i actually got the right answer! haha. are you able to help me with one more?

  12. anonymous
    • 5 years ago
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    no prob, sure, what's the question,

  13. anonymous
    • 5 years ago
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    sweet it has to do with converting a triple integral with rectangular coordinates to cylindrical coordinates, i'll try to type it out in the equation thing...

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

  15. anonymous
    • 5 years ago
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    I don't have some of my notes with me, but I think x =rcos(theta) y= rsin(theta) and z=z in cylindrical right?

  16. anonymous
    • 5 years ago
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    yes that is right

  17. anonymous
    • 5 years ago
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    Then I think there is an extra factor or something like that when integrating too right? like an extra r or something like that?

  18. anonymous
    • 5 years ago
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    yes there is an extra r, sorry the box wasn't letting me type anything for awhile

  19. anonymous
    • 5 years ago
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    I'm looking at it, and I don't want to give you bad advice If I had my notes with me, I could help, but the thing that has got me is the integrating operators, I don't think we can just put r^2 sin(thetat)cos(theta)*r d(theta) in there. and I know we need a dr, but without my notes I'm drawing it up. Sorry about that one.

  20. anonymous
    • 5 years ago
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    that's alright, thanks for your help i really appreciate it!

  21. anonymous
    • 5 years ago
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    I think I've got it, sorry, it's looks like you have to sub rcos(thetat) for x, then times the extra r = r^2cos(theta)dzdrd(theta). your bounds for dz change from r^2 to 4, then dr -2, to 2, and d(theta) from 0 to 2pi, to complete the circle the volume is over. Then it's pretty much like before.

  22. anonymous
    • 5 years ago
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    awesome thanks!

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