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anonymous

  • 5 years ago

Use the disk or the shell method to find the volume of the solid generated by revolving the region bounded by the graph of the equations about y = 6. y = 6/x^2 y = 0 x = 1 x = 3

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  1. anonymous
    • 5 years ago
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    I used the washer method, and for some reason, I am continuing to do it wrong

  2. watchmath
    • 5 years ago
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    \(\int_1^3 \pi(6^2-(6-\frac{6}{x^2})^2)\,dx\)

  3. anonymous
    • 5 years ago
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    yes, that's what I started out with, using (top- bottom)

  4. watchmath
    • 5 years ago
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    ok, then how do you know that you continuing it wrong?

  5. anonymous
    • 5 years ago
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    \[\pi \int\limits_{1}^{3} 36 - ((36/x^4) - (72/x^2) +36) dx\]

  6. anonymous
    • 5 years ago
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    when I simplify, I am getting negative numbers, or no right answer at all

  7. anonymous
    • 5 years ago
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    \[\pi \int\limits_{1}^{3}((36/x4)−(72/x2))dx\]

  8. anonymous
    • 5 years ago
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    \[\pi \int\limits_{1}^{3}(-36/x^4) + (72/x^2) dx\]

  9. watchmath
    • 5 years ago
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    \(\pi\int_1^3 72x^{-2}-36x^{-4}\,dx\) \(\pi(-72x^{-1}+12x^{-3}\mid_1^3)\) Did you get that?

  10. anonymous
    • 5 years ago
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    no, I think maybe I did not integrate properly

  11. anonymous
    • 5 years ago
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    Everything you put up so far was right ilovemath

  12. anonymous
    • 5 years ago
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    I think the answer is 328/9 pi, so I will check it really quick

  13. anonymous
    • 5 years ago
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    Thanks for helping! The answer is correct!

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spraguer (Moderator)
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