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anonymous

  • 5 years ago

revolve the region in first quadrant bounded by y = x^2 , y = x , about line x = 2, using washer method

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  1. anonymous
    • 5 years ago
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    \[\pi \int\limits_{0}^{4}(2-x)^{2}-(2-x ^{2})^{2}\]

  2. anonymous
    • 5 years ago
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    thats wrong

  3. anonymous
    • 5 years ago
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    noooooooooo

  4. anonymous
    • 5 years ago
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    srry its been a long time.

  5. anonymous
    • 5 years ago
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    the region is from (0,0) to (1,1)

  6. anonymous
    • 5 years ago
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    oh sorry sorry forgot about the bounds. assumed that x22 was also a bound

  7. anonymous
    • 5 years ago
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    also youre using dy , not x

  8. anonymous
    • 5 years ago
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    ok nvm i rambl.ed

  9. anonymous
    • 5 years ago
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    now you will die

  10. anonymous
    • 5 years ago
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    ure supposed to use dx?

  11. anonymous
    • 5 years ago
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    dy

  12. anonymous
    • 5 years ago
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    i was using DY

  13. anonymous
    • 5 years ago
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    since its horizontal integration

  14. anonymous
    • 5 years ago
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    just the wrong values

  15. anonymous
    • 5 years ago
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    no its wrong in other ways!!!

  16. anonymous
    • 5 years ago
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    ok lets do it again: \[\pi \int\limits_{0}^{1} (2-x)^{2}-(2-x ^{2})^{2}\]

  17. anonymous
    • 5 years ago
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    this HASSSSS TO BE RIGHT

  18. anonymous
    • 5 years ago
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    eh eh?

  19. anonymous
    • 5 years ago
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    no

  20. anonymous
    • 5 years ago
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    its pi* integral ( y - 2)^2 - (sqrt y - 2)^2

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