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Nanoman

  • one year ago

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?

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  1. Nanoman
    • one year ago
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    The main problem I have is that I can't solve for x.

  2. Nanoman
    • one year ago
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    Are you there?

  3. Nanoman
    • one year ago
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    @jim_thompson5910

  4. Callisto
    • one year ago
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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 =...???

  5. Nanoman
    • one year ago
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    0?

  6. Callisto
    • one year ago
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    and??

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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