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PhoenixFire

  • 3 years ago

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.

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  1. e.mccormick
    • 3 years ago
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    The limits would be the z axis change.

  2. e.mccormick
    • 3 years ago
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    Then you rotate around y, with a floating radius of x.

  3. e.mccormick
    • 3 years ago
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    That would be one way to do it.

  4. genius12
    • 3 years ago
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    Is that z = x^2 + y^2 - 1? @PhoenixFire

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

  6. e.mccormick
    • 3 years ago
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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.

  7. PhoenixFire
    • 3 years ago
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    @e.mccormick I'm not quite sure how you would do that.

  8. e.mccormick
    • 3 years ago
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    Or I think you can... hehe. I have been known to be wrong!

  9. PhoenixFire
    • 3 years ago
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    @Hoa Yes! Give us some direction if you know how to do it.

  10. e.mccormick
    • 3 years ago
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    @Hoa Please do, I am checking my statement! And I may be completely off here.

  11. zugzwang
    • 3 years ago
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    lol hoa dont bring real analysis into integrals confusing!

  12. PhoenixFire
    • 3 years ago
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    @Hoa I think polar coordinates might be the way to do this.

  13. zugzwang
    • 3 years ago
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    no, i'm still studying this

  14. zugzwang
    • 3 years ago
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    im just a chess player lol :D but okay let's see...

  15. zugzwang
    • 3 years ago
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    polar coords okay?

  16. zugzwang
    • 3 years ago
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    it looks less menacing that way \[\large z = r^2 - 1\]

  17. PhoenixFire
    • 3 years ago
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    Maybe it is a triple integral. I thought it would be double.

  18. zugzwang
    • 3 years ago
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    guys u still there?

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

  20. zugzwang
    • 3 years ago
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    i dont have one. :(

  21. zugzwang
    • 3 years ago
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    maybe it does have to be in triple integrals...

  22. PhoenixFire
    • 3 years ago
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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.

  23. PhoenixFire
    • 3 years ago
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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.

  24. zugzwang
    • 3 years ago
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    i have an idea

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

  26. zugzwang
    • 3 years ago
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    if r goes from 1 to square root of 2, then z would go from 0 to 1.

  27. zugzwang
    • 3 years ago
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    im sorry i didn't see besides you probably deleted it

  28. zugzwang
    • 3 years ago
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    yes we meet lol hi im Kevin Nguyen and you are...?

  29. zugzwang
    • 3 years ago
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    american, vietnamese dad

  30. zugzwang
    • 3 years ago
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    no i dont speak tieng viet

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

  32. zugzwang
    • 3 years ago
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    big brother/sister lol

  33. zugzwang
    • 3 years ago
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    dont know what that means

  34. zugzwang
    • 3 years ago
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    maybe I will you better not have cussed at me tho lol

  35. zugzwang
    • 3 years ago
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    No. I'm a college freshman.

  36. PhoenixFire
    • 3 years ago
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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.

  37. zugzwang
    • 3 years ago
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    then I gots no clue sorry :)

  38. zugzwang
    • 3 years ago
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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

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

  40. zugzwang
    • 3 years ago
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    that works too

  41. zugzwang
    • 3 years ago
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    but yeah, it's not gonna be straight

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

  43. PhoenixFire
    • 3 years ago
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    Yeah, it's definitely not going to be straight.

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