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monroe17

  • 3 years ago

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

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  1. math>philosophy
    • 3 years ago
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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?

  2. ByteMe
    • 3 years ago
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    Which method u prefer? shell or washer?

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

  4. ByteMe
    • 3 years ago
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    The setup of the integral depends on the method you use....

  5. monroe17
    • 3 years ago
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    ^ I didn't see your comment.. the washer method :)

  6. math>philosophy
    • 3 years ago
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    Why you choose washersssssss D:

  7. monroe17
    • 3 years ago
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    isn't easier? or not? usually for me it is easier..

  8. ByteMe
    • 3 years ago
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    i agree that washer is appropriate here.... hang on....

  9. math>philosophy
    • 3 years ago
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    You guys have fun with that >_>

  10. ByteMe
    • 3 years ago
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    |dw:1361159214153:dw|

  11. ByteMe
    • 3 years ago
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    Can you give me the expression for R and r?

  12. monroe17
    • 3 years ago
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    errr.. ;/ crap.

  13. ByteMe
    • 3 years ago
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    ???

  14. ByteMe
    • 3 years ago
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    no... i don't think it's that.. :)

  15. monroe17
    • 3 years ago
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    haha give me a sec

  16. ByteMe
    • 3 years ago
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    consider this...|dw:1361159501396:dw|

  17. monroe17
    • 3 years ago
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    would it be 1/x^4-4

  18. monroe17
    • 3 years ago
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    squared

  19. ByteMe
    • 3 years ago
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    for R, shouldn't it be 1/x^4 + 4 ???

  20. monroe17
    • 3 years ago
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    - - + .. err yeah my bad

  21. ByteMe
    • 3 years ago
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    ok... you also agree that r = 4 (inner radius) ???

  22. monroe17
    • 3 years ago
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    yupp it is

  23. ByteMe
    • 3 years ago
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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 ....

  24. ByteMe
    • 3 years ago
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    sorry... i gotta leave... :(

  25. ByteMe
    • 3 years ago
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    i'm pretty sure what we have is correct....

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