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amonoconnor

  • 7 months ago

Would someone be willing to check my answer for the following problem involving volumes of solids? y = 2/(x^1/2), y = 2, x = 4; about the x-axis. Any and all help is greatly appreciated! :)

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  1. faisalalif1999
    • 7 months ago
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    so basically

  2. faisalalif1999
    • 7 months ago
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    \[y=\frac{ 2 }{ \frac{ x }{ 2 } }\]

  3. faisalalif1999
    • 7 months ago
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    so y =\[\frac{ 4 }{ x }\]

  4. amonoconnor
    • 7 months ago
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    No... one of the equations defining the area to be revolved is y = 2/(x^(1/2)) and another serving as a limit, and boundary value for the integral, is y = 2.

  5. AccessDenied
    • 7 months ago
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    We might start with a sketch of our situation. This one is pretty tough to visualize, so a graphing calculator might help if that is available. Here is Wolfram's plot: http://www.wolframalpha.com/input/?i=plot+y%3D2%2Fx%5E%281%2F2%29%2C+y%3D2%2C+x%3D4 You want to revolve this region about the x-axis. First, did you have any ideas on how you might set this one up? :)

  6. AccessDenied
    • 7 months ago
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    Since the region is not connected to the x-axis, I think the washer method is looking good here. We need to figure out the limits, and then decide what our inner and outer radius are. If you need more help, I can return to this if you reply soon (I just saw you are offline), or just bump it and someone else can help. Good luck!

  7. amonoconnor
    • 7 months ago
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    Sorry, my iPad does show me as inactive in OS even when I'm not some times... Annoying :p Anyways, yes, I used the Washer Method. I determined... (And please tell me if any of this is incorrect) Boundaries: since we are revolving around the x-axis, they are x-values of: 1, 4. R: 2 r: (2/(x^(1/2))

  8. AccessDenied
    • 7 months ago
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    I agree with that information, yes. :)

  9. amonoconnor
    • 7 months ago
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    After taking the anti-derivative, but before plugging in the boundaries to solve, I had: "pi(4x - 4(ln(x))) 1|4" ?

  10. AccessDenied
    • 7 months ago
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    Looks good!

  11. amonoconnor
    • 7 months ago
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    Sweeeet! :)

  12. amonoconnor
    • 7 months ago
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    And... For a final answer, I am getting "4pi(4 - ln(4))" ??

  13. amonoconnor
    • 7 months ago
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    *units cubed

  14. AccessDenied
    • 7 months ago
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    Looks like something was lost. What was the steps after you plugged in your limits of integration?

  15. amonoconnor
    • 7 months ago
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    What I have on my paper: = pi(4x - 4(ln(x))) 1|4 = [pi(4(4) - 4(ln(4)))] - [pi(4(1) - 4(ln(1)))] = [pi(16 - 4(ln4))] - [pi(4 - 0)] = 16pi - 4pi(ln4) - 4pi = 4pi(4 - ln(4)) units^3

  16. amonoconnor
    • 7 months ago
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    Oh!!! I think I se a mistake... Is it that I didn't subtract 4pi from 16pi, so to should be: 4pi(3 - ln(4))

  17. AccessDenied
    • 7 months ago
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    Yep, there you go! I was thinking either that or you made the other x=1 limit all 0 (just because ln 1 = 0). But that fixes it! :D

  18. amonoconnor
    • 7 months ago
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    Alright, sweet, so the final answer is indeed 4pi(3 - ln(4)) ? :D

  19. AccessDenied
    • 7 months ago
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    That's what I got as well.

  20. amonoconnor
    • 7 months ago
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    Awesome! Thank you very much for your patience and willingness to work with me! It helped, and I fully understand what went wrong, I'm not just confused. Thank you!

  21. AccessDenied
    • 7 months ago
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    Always glad to help! :D

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