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Nanoman Group Title

What is the volume of a solid obtained by rotating the region between the graph of y = (1/2)sin^2 (x^2) and the x-axis for 0 ≤x≤sqrtpi about the y-axis?

  • one year ago
  • one year ago

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  1. Nanoman Group Title
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    The main problem I have is that I can't solve for x.

    • one year ago
  2. Nanoman Group Title
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    Are you there?

    • one year ago
  3. Nanoman Group Title
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    @jim_thompson5910

    • one year ago
  4. Callisto Group Title
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    Put y =0 0 = (1/2)sin^2 (x^2) sin^2 (x^2) = 0 sin (x^2) = 0 x^2 =...???

    • one year ago
  5. Nanoman Group Title
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    0?

    • one year ago
  6. Callisto Group Title
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    and??

    • one year ago
  7. Nanoman Group Title
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    0 is all I can think of.

    • one year ago
  8. Callisto Group Title
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    |dw:1358903484725:dw|

    • one year ago
  9. Nanoman Group Title
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    Oh, right! Sine hits zero on every multiple of pi

    • one year ago
  10. Nanoman Group Title
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    So x = sqrt(pi), 0?

    • one year ago
  11. Callisto Group Title
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    For that domain, yes.

    • one year ago
  12. Nanoman Group Title
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    Right. The domain was a clue.

    • one year ago
  13. Nanoman Group Title
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    I need to use the washer method right?

    • one year ago
  14. Callisto Group Title
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    I think we need to use shell method?!

    • one year ago
  15. Nanoman Group Title
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    oh, right. just a sec....

    • one year ago
  16. Nanoman Group Title
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    x = sqrtpi for the radis, y = (1/2)sin^2 (x^2) for the height?

    • one year ago
  17. Callisto Group Title
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    x = radius... y = height

    • one year ago
  18. Nanoman Group Title
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    \[2\pi \int\limits_{0}^{\sqrt{\pi}} (\sqrt{\pi}/2) \sin^2 (x^2)\] ?

    • one year ago
  19. Callisto Group Title
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    \[V = 2 \pi \int_a^brh(width)\]r = x, h = y, width = dx

    • one year ago
  20. Callisto Group Title
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    y is a function of x, you know.

    • one year ago
  21. Nanoman Group Title
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    Doesn't x = sqrtpi and y = (1/2)sin^2 (x^2)?

    • one year ago
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