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anonymous

  • one year ago

Find volume of solid generated

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  1. anonymous
    • one year ago
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  2. welshfella
    • one year ago
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    |dw:1436654820728:dw|

  3. anonymous
    • one year ago
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    I don't understand the infinity part on the left side if it isn't enclosed from there

  4. anonymous
    • one year ago
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    oops sorry i was looking at the wrong part XD

  5. welshfella
    • one year ago
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    |dw:1436654961888:dw|

  6. anonymous
    • one year ago
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    I'm not sure what that is, but that is not even close. Thank you though

  7. anonymous
    • one year ago
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    so it is ln15-e^(ln15)

  8. anonymous
    • one year ago
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    so you have the area of the box ln15*1 - pi integral(e^(x/2)^2)dx

  9. anonymous
    • one year ago
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    oops forgot pi

  10. anonymous
    • one year ago
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    the integral is just pi*integra(e^x)=pi*e^x

  11. anonymous
    • one year ago
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    Thank you as well, but no

  12. anonymous
    • one year ago
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    |dw:1436655396490:dw|

  13. anonymous
    • one year ago
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    Your drawing is incorrect as well. I do appreciate the help, but its not helping

  14. anonymous
    • one year ago
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    whoops...

  15. anonymous
    • one year ago
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    forgot the triangle bit thing with the curvy roof

  16. anonymous
    • one year ago
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    No, it is bounded bounded by 1 you have it bounded by y=0

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  17. anonymous
    • one year ago
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  18. welshfella
    • one year ago
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    V = pi(e^(ln15) - e^0) - ln15 pi = 14pi - ln15pi = 11.29 pi

  19. anonymous
    • one year ago
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    \[V=\ln15-\int\limits_{0}^{\ln15}y^2dx\]

  20. anonymous
    • one year ago
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    YES How did you get that I couldn't make out what yo had up top

  21. anonymous
    • one year ago
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    mathway.com for graphing

  22. anonymous
    • one year ago
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    Thta was to welshfella

  23. welshfella
    • one year ago
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    the diagram you mean?

  24. anonymous
    • one year ago
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    no the answer its pi(14-ln15)

  25. welshfella
    • one year ago
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    yes

  26. anonymous
    • one year ago
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    how did you derive that i forgot to add that its supposed to be the shell method, but I'm not too concerned with that

  27. welshfella
    • one year ago
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    |dw:1436655815791:dw|

  28. welshfella
    • one year ago
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    the shaded area is rotated about x axis so we need the volume of the area down to the x axis minus the area of the cylinder below

  29. anonymous
    • one year ago
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    I'm not doing very well... i kept thinking it was a square and forgot it had to be revolved into a cylinder...

  30. welshfella
    • one year ago
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    |dw:1436655967976:dw|

  31. welshfella
    • one year ago
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    that gives volume of revolution of the shaded area below |dw:1436656237239:dw|

  32. welshfella
    • one year ago
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    - then you need to subtract the volume of the cylinder radius 1 and length ln15

  33. anonymous
    • one year ago
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    ok, thank you

  34. welshfella
    • one year ago
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    yw as for the 'shell method' i dont know what that is ( i cant remember that in school - mind that was a long time ago)

  35. anonymous
    • one year ago
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    its 2pi integral (shell height)(shell radius)

  36. welshfella
    • one year ago
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    ah i see - the method i used was integrating infinitely thin discs

  37. anonymous
    • one year ago
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    yeah the shell is integrating "coffee cans"

  38. anonymous
    • one year ago
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    cylinders

  39. welshfella
    • one year ago
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    ok lol

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