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amistre64

  • 5 years ago

Find the volume of the solid created by the bounded region of (x=y-y^2) and (x=y^2 -3) when it is spun around the line (x=-4).

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  1. anonymous
    • 5 years ago
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    Ah this question still o.O

  2. amistre64
    • 5 years ago
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    Yeah :) but I found out what I was doing wrong.... other than trying to drink and derive :)

  3. anonymous
    • 5 years ago
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    So we at least know you can add 4 to all terms to change the bounds and have it revolved around x = 0

  4. anonymous
    • 5 years ago
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    And it's a torus, right? So you need to integrate the area of the torus around the line in a full circle.

  5. anonymous
    • 5 years ago
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    A cross section of the torus.

  6. amistre64
    • 5 years ago
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    yep...

  7. anonymous
    • 5 years ago
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    Can't you just integrate it?

  8. amistre64
    • 5 years ago
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    its pi (S) [4+F(y)]^2 - [4+G(y)]^2 | -1, 3/2 I kept trying: pi (S) [(4+F(y)) - (4+G(y))]^2 | -1, 3/2

  9. amistre64
    • 5 years ago
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    math: yes, but couldnt figure out how to last night :)

  10. anonymous
    • 5 years ago
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    \[x = y - y^2 + 4\] \[x = y^2 + 1\]

  11. anonymous
    • 5 years ago
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    Okay so the bounds are y = -3, y = 1/2

  12. anonymous
    • 5 years ago
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    No...

  13. anonymous
    • 5 years ago
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    What?

  14. amistre64
    • 5 years ago
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    Krb: goog good. the bounds of integration are -1 and 3/2

  15. anonymous
    • 5 years ago
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    Yeah those are the bounds ^

  16. anonymous
    • 5 years ago
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    i need to figure out how to use the equations to find the solids of revolution..if this is x=-4, then the line we get is a vertical one right? so its been revolved upwards?

  17. amistre64
    • 5 years ago
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    you got the equation right; now square them individually before you subtract them..

  18. anonymous
    • 5 years ago
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    So \[\int\limits_{(-1)}^{(3/2)}(y - y^2 + 4) - (y^2 + 1)\]

  19. anonymous
    • 5 years ago
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    That's what i would think

  20. anonymous
    • 5 years ago
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    Krb you're always on this site. I like how when I look at paul's notes, you were the first one on open study.

  21. amistre64
    • 5 years ago
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    think; the rotation is around the "y axis" so to speak

  22. anonymous
    • 5 years ago
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    yeah..thats what i mean, so its like integrating the difference of squares of the equations right? and these equations signify -----?

  23. amistre64
    • 5 years ago
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    Krb: getting closer. we need to integrate [F(y)]^2 - [G(x)]^2 in the bounds

  24. anonymous
    • 5 years ago
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    Yes, normally when integrating with respect to x you are finding the are underneath the curve that drops to the x axis. So in this case you are finding the area "to the side of the curve" that drops to the y axis

  25. amistre64
    • 5 years ago
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    Think; exactly :)

  26. anonymous
    • 5 years ago
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    I'm confused as to why it would be squared

  27. amistre64
    • 5 years ago
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    typo...[G(y)]^2

  28. anonymous
    • 5 years ago
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    i guess it should be squared...shouldn't it?

  29. amistre64
    • 5 years ago
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    Krb: if we do it seperately for each on, there is no real "area" so to speak until we combine them. Does that make sense?

  30. amistre64
    • 5 years ago
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    pi* (S) F(y)^2 dy - (S) G(y)^2 dy The equations ARE our Radiusessess

  31. anonymous
    • 5 years ago
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    OH

  32. anonymous
    • 5 years ago
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    if y1 & y2 are the curves then , \[\int\limits_{}^{}[y1^{2}-y2^{2}]\] gives the volume of the solid? where y1=upper curve & y2 the lower

  33. anonymous
    • 5 years ago
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    multiplied by a pi

  34. anonymous
    • 5 years ago
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    SO we are looking at it from a top down view, and then you need pi * r^2 for both "circles"

  35. amistre64
    • 5 years ago
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    exactly

  36. anonymous
    • 5 years ago
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    on efor inner and the other for outer 'circle' i guess

  37. amistre64
    • 5 years ago
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    think: thats the gyst of it yeah

  38. anonymous
    • 5 years ago
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    thnx!!

  39. anonymous
    • 5 years ago
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    \[[\pi \int\limits_{-1}^{(3/2)}(y - y^2 + 4)^2] - [\pi \int\limits_{-1}^{(3/2)}(y^2 + 1)^2]\]

  40. amistre64
    • 5 years ago
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    Krb: your a genius :)

  41. anonymous
    • 5 years ago
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    No, you are :)

  42. anonymous
    • 5 years ago
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    u both are :)

  43. amistre64
    • 5 years ago
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    we should start a mensa club right here lol

  44. anonymous
    • 5 years ago
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    the open study math mensa

  45. anonymous
    • 5 years ago
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    :) :) so krb ur'e pursuing your masters too?

  46. amistre64
    • 5 years ago
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    C1 - C2 is what we were missing last night. 2pi(R) - 2pi(r) does not equal 2pi(R-r)

  47. amistre64
    • 5 years ago
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    it does, but it doesnt when you integrate lol

  48. anonymous
    • 5 years ago
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    wer those constants that u'll missed out yesterday?

  49. anonymous
    • 5 years ago
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    Ends up being 85.9029

  50. anonymous
    • 5 years ago
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    think: I'm honestly not sure

  51. anonymous
    • 5 years ago
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    krb: that's alright

  52. anonymous
    • 5 years ago
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    i dunno why it gets posted twice always, when i click only once!

  53. anonymous
    • 5 years ago
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    I think I probably would like to get a Masters

  54. anonymous
    • 5 years ago
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    good

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