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PhoenixFire Group Title

Double Integrals Find the volume of the solid \(z=(x^2 + y^2)-1\) between the two plates z=0 and z=1 I'm not sure how to start. I've graphed the solid, but that hasn't helped me figure out what limits to use.

  • one year ago
  • one year ago

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  1. e.mccormick Group Title
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    The limits would be the z axis change.

    • one year ago
  2. e.mccormick Group Title
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    Then you rotate around y, with a floating radius of x.

    • one year ago
  3. e.mccormick Group Title
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    That would be one way to do it.

    • one year ago
  4. genius12 Group Title
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    Is that z = x^2 + y^2 - 1? @PhoenixFire

    • one year ago
  5. PhoenixFire Group Title
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    Yeah, I don't know why they put the ( ) there. it's the same as \[z=x^2+y^2-1\]

    • one year ago
  6. e.mccormick Group Title
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    It would be a shift on the z axis. You can still reduce it to a single axis problem, which was what I was thinking of.

    • one year ago
  7. PhoenixFire Group Title
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    @e.mccormick I'm not quite sure how you would do that.

    • one year ago
  8. e.mccormick Group Title
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    Or I think you can... hehe. I have been known to be wrong!

    • one year ago
  9. PhoenixFire Group Title
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    @Hoa Yes! Give us some direction if you know how to do it.

    • one year ago
  10. e.mccormick Group Title
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    @Hoa Please do, I am checking my statement! And I may be completely off here.

    • one year ago
  11. zugzwang Group Title
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    lol hoa dont bring real analysis into integrals confusing!

    • one year ago
  12. PhoenixFire Group Title
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    @Hoa I think polar coordinates might be the way to do this.

    • one year ago
  13. zugzwang Group Title
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    no, i'm still studying this

    • one year ago
  14. zugzwang Group Title
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    im just a chess player lol :D but okay let's see...

    • one year ago
  15. zugzwang Group Title
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    polar coords okay?

    • one year ago
  16. zugzwang Group Title
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    it looks less menacing that way \[\large z = r^2 - 1\]

    • one year ago
  17. PhoenixFire Group Title
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    Maybe it is a triple integral. I thought it would be double.

    • one year ago
  18. zugzwang Group Title
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    guys u still there?

    • one year ago
  19. PhoenixFire Group Title
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    Yup, just looking through my textbook to see if maybe it is a Triple. But keep going with your solution

    • one year ago
  20. zugzwang Group Title
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    i dont have one. :(

    • one year ago
  21. zugzwang Group Title
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    maybe it does have to be in triple integrals...

    • one year ago
  22. PhoenixFire Group Title
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    No I actually do not have answers. It's part of my assignment. We haven't learned triple integrals yet, so maybe I'm getting ahead of myself.

    • one year ago
  23. PhoenixFire Group Title
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    Alright guys, thanks for trying. I'm going to close this question and do some research into it. I'll come back and ask again in a few days if it's not sorted.

    • one year ago
  24. zugzwang Group Title
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    i have an idea

    • one year ago
  25. zugzwang Group Title
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    how about \[\Large \int\limits_{\theta = 0}^{2\pi}\int\limits_{r=1}^{\sqrt 2}(r^2-1) \ rdrd\theta\]

    • one year ago
  26. zugzwang Group Title
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    if r goes from 1 to square root of 2, then z would go from 0 to 1.

    • one year ago
  27. zugzwang Group Title
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    im sorry i didn't see besides you probably deleted it

    • one year ago
  28. zugzwang Group Title
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    yes we meet lol hi im Kevin Nguyen and you are...?

    • one year ago
  29. zugzwang Group Title
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    american, vietnamese dad

    • one year ago
  30. zugzwang Group Title
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    no i dont speak tieng viet

    • one year ago
  31. zugzwang Group Title
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    you gave one, i gave it again only wait for his reaction are you a high school student?

    • one year ago
  32. zugzwang Group Title
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    big brother/sister lol

    • one year ago
  33. zugzwang Group Title
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    dont know what that means

    • one year ago
  34. zugzwang Group Title
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    maybe I will you better not have cussed at me tho lol

    • one year ago
  35. zugzwang Group Title
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    No. I'm a college freshman.

    • one year ago
  36. PhoenixFire Group Title
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    @zugzwang That will give me the region |dw:1367636348171:dw| Right? Which isn't what we want. |dw:1367636411100:dw| This has to be a triple integral. I'll see if I understand it next week when we do triple integral.

    • one year ago
  37. zugzwang Group Title
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    then I gots no clue sorry :)

    • one year ago
  38. zugzwang Group Title
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    wait, why does it have to be straight? it's not gonna be straight, it's an elliptic paraboloid for heavens sakes

    • one year ago
  39. PhoenixFire Group Title
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    Can I not find the area of this and then rotate it around Z from 0 -> 2pi? |dw:1367636933452:dw|

    • one year ago
  40. zugzwang Group Title
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    that works too

    • one year ago
  41. zugzwang Group Title
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    but yeah, it's not gonna be straight

    • one year ago
  42. zugzwang Group Title
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    the straight (like a cone) graph would be \[\large z = r\]or \[\large z = \sqrt{x^2+y^2}\]

    • one year ago
  43. PhoenixFire Group Title
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    Yeah, it's definitely not going to be straight.

    • one year ago
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