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anonymous

  • 5 years ago

PLEASE SOLVE :) let R be the region bounded by the graph of y=x^2 -1 and the graph of x=y^2 .... find the volume of the solid generated when r is rotated about the vertical line x=2, and then write an integral expression that can be used to find the volume of the solid generated when R is rotated about the line y=-1

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  1. anonymous
    • 5 years ago
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    I can get you started, but I may have to leave soon. If I help you set up the volume element, can you integrate from there?

  2. anonymous
    • 5 years ago
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    Well, I'll start anyway...

  3. anonymous
    • 5 years ago
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    I'm going to send through an image of the plots you need to look at.

  4. anonymous
    • 5 years ago
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    ok thanks!

  5. anonymous
    • 5 years ago
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  6. anonymous
    • 5 years ago
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    this is hard without a calculator. I got the integral would be \[\pi*\int\limits_{?}^{?}(2-y^2)^2-(2-\sqrt{y+1})^2dy\] and then i would need a calulator to get the limits where the two function are equal...is it supposed to have such nasty algebra or did I mess up soehwhere?

  7. anonymous
    • 5 years ago
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    Assuming your equations are correct, the shaded region is what you're looking at. I would plan on using cylindrical shells, and doing the integral in two parts: one where\[y=\sqrt{x}\]and one where\[y=-\sqrt{x}\]and then add the volumes up.

  8. amistre64
    • 5 years ago
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    plotting it was easy :) how we find the solutions? or is there another way?

  9. anonymous
    • 5 years ago
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    Actually, put cyl. shells on hold. When I first looked at your problem, I took the y-axis as a natural bound, but that's not the case since it's not in your question. When considering the region shaded, discs may be better. I want to work it out.

  10. anonymous
    • 5 years ago
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    I'm deriving it - it's easy to do on paper, but hard to explain. online

  11. amistre64
    • 5 years ago
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    deriving sounds interesting.... would you see where the slopes are perp to each other?

  12. amistre64
    • 5 years ago
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    been trying to solve it like system of equation....but no luck; then again, i could be missing something glaringly obvious :)

  13. anonymous
    • 5 years ago
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    the integral came out \[\pi*[y^5/5-4y^3/3-y^2/2+8/3*y^(3/2)-y]...\]

  14. amistre64
    • 5 years ago
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    jpick: you just subtract the equations from each other?

  15. anonymous
    • 5 years ago
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    but to get bounds for y i have to solve the values where y^4-y-1=0

  16. amistre64
    • 5 years ago
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    yeah.... been trying that one for hours :)

  17. amistre64
    • 5 years ago
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    just thinking outloud, but would newtons approximation work for us? or is that just to clumsy?

  18. anonymous
    • 5 years ago
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    Okay, it seems whatever method you use here (cylindrical shells or discs) the algebra is going to be crappy.

  19. amistre64
    • 5 years ago
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    loki... :) i agree

  20. anonymous
    • 5 years ago
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    Given that, I'm going to persist with cylindrical shells.

  21. anonymous
    • 5 years ago
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    maybe, some numerical approximation (or algebra solver program) will be needed

  22. amistre64
    • 5 years ago
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    finding the bounds is like pulling teeth, but without the lollipop at the end :)

  23. amistre64
    • 5 years ago
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    is it possible to set your own bounds and still get an answer?

  24. anonymous
    • 5 years ago
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    not sure what you mean? the bounds of integration are set by the intersection, or do you mean the approximation getting close enough will be our "own bounds"

  25. amistre64
    • 5 years ago
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    either way... I was wondering if going further than the bounds would have a canceling out effect, but most likely that is just wishful thinking

  26. anonymous
    • 5 years ago
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    Okay, the basic construction for the cylindrical shell is\[\delta V = 2\pi(radius)(height)(elemental.thickness)\]\[=2\pi (x-2)(\sqrt{x}-(x^2-1)) \delta x\]

  27. anonymous
    • 5 years ago
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    where...

  28. anonymous
    • 5 years ago
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    this element is for the section to the right of the line in the following image.

  29. anonymous
    • 5 years ago
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    1 Attachment
  30. anonymous
    • 5 years ago
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    A similar situation will arise with the section to the left, where you have\[\delta V=2\pi (2-x)(\sqrt{x}-(-\sqrt{x})) \delta x=2\pi(2-x)(2\sqrt{x}) \delta x\]

  31. anonymous
    • 5 years ago
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    There's a mistake in the first element: x-2 should read as 2-x, the radius.

  32. anonymous
    • 5 years ago
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    You need to communicate with me. Is this working for you?

  33. anonymous
    • 5 years ago
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    Your limits of integration for the element I just derived above (the one with 2sqrt{x} for height) will be from x=0 to x=t where \[-\sqrt{t}=t^2-1\] (i.e. get a number out for t).

  34. anonymous
    • 5 years ago
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    that will work, the only thing we need are the actual bounds, either in terms of x or y. I do have one question for you lokisan about the earlier differential equation thread. I got a different answer than you, even though we did it the same way given it was an exact equation

  35. anonymous
    • 5 years ago
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    The limits of the initial section I derived will be from x = the t you just found to x=s where \[\sqrt{s}=s^2-1\](i.e. get an approximation for s).

  36. anonymous
    • 5 years ago
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    Which d.e.?

  37. anonymous
    • 5 years ago
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    The one where he wanted substitution?

  38. anonymous
    • 5 years ago
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    bethanymichalski, you there?

  39. anonymous
    • 5 years ago
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    yeah, I got x^2/2+y^/2+xy-x=c

  40. anonymous
    • 5 years ago
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    That's what I got using exact differential equation.

  41. anonymous
    • 5 years ago
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    Oh, i thought it said log(x+y)...my bad nm :(

  42. anonymous
    • 5 years ago
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    oh, he wanted to use a substitution method in the end. I didn't bother to check if the answers were the same, ;p

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