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monroe17 Group Title

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis. y=1/x^4, y=0, x=2, x=6, about y=-4

  • one year ago
  • one year ago

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  1. math>philosophy Group Title
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    I would use the method of cylindrical shells. At any f(x) value between 2 and 6, what is the radius from that y point to y = -4?

    • one year ago
  2. ByteMe Group Title
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    Which method u prefer? shell or washer?

    • one year ago
  3. monroe17 Group Title
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    I just need help setting it up to integrate then i can take it from there...

    • one year ago
  4. ByteMe Group Title
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    The setup of the integral depends on the method you use....

    • one year ago
  5. monroe17 Group Title
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    ^ I didn't see your comment.. the washer method :)

    • one year ago
  6. math>philosophy Group Title
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    Why you choose washersssssss D:

    • one year ago
  7. monroe17 Group Title
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    isn't easier? or not? usually for me it is easier..

    • one year ago
  8. ByteMe Group Title
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    i agree that washer is appropriate here.... hang on....

    • one year ago
  9. math>philosophy Group Title
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    You guys have fun with that >_>

    • one year ago
  10. ByteMe Group Title
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    |dw:1361159214153:dw|

    • one year ago
  11. ByteMe Group Title
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    Can you give me the expression for R and r?

    • one year ago
  12. monroe17 Group Title
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    errr.. ;/ crap.

    • one year ago
  13. ByteMe Group Title
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    ???

    • one year ago
  14. ByteMe Group Title
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    no... i don't think it's that.. :)

    • one year ago
  15. monroe17 Group Title
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    haha give me a sec

    • one year ago
  16. ByteMe Group Title
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    consider this...|dw:1361159501396:dw|

    • one year ago
  17. monroe17 Group Title
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    would it be 1/x^4-4

    • one year ago
  18. monroe17 Group Title
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    squared

    • one year ago
  19. ByteMe Group Title
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    for R, shouldn't it be 1/x^4 + 4 ???

    • one year ago
  20. monroe17 Group Title
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    - - + .. err yeah my bad

    • one year ago
  21. ByteMe Group Title
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    ok... you also agree that r = 4 (inner radius) ???

    • one year ago
  22. monroe17 Group Title
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    yupp it is

    • one year ago
  23. ByteMe Group Title
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    so... \(\large dV=\pi(R^2-r^2)dx \) so \(\large V=\pi \int_2^6(R^2-r^2)dx \) you replace the expressions for R and r ....

    • one year ago
  24. ByteMe Group Title
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    sorry... i gotta leave... :(

    • one year ago
  25. ByteMe Group Title
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    i'm pretty sure what we have is correct....

    • one year ago
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