Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Nanoman

  • 3 years 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?

  • This Question is Closed
  1. Nanoman
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The main problem I have is that I can't solve for x.

  2. Nanoman
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Are you there?

  3. Nanoman
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @jim_thompson5910

  4. Callisto
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Put y =0 0 = (1/2)sin^2 (x^2) sin^2 (x^2) = 0 sin (x^2) = 0 x^2 =...???

  5. Nanoman
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    0?

  6. Callisto
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    and??

  7. Nanoman
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    0 is all I can think of.

  8. Callisto
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1358903484725:dw|

  9. Nanoman
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Oh, right! Sine hits zero on every multiple of pi

  10. Nanoman
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So x = sqrt(pi), 0?

  11. Callisto
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    For that domain, yes.

  12. Nanoman
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Right. The domain was a clue.

  13. Nanoman
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I need to use the washer method right?

  14. Callisto
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I think we need to use shell method?!

  15. Nanoman
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh, right. just a sec....

  16. Nanoman
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    x = sqrtpi for the radis, y = (1/2)sin^2 (x^2) for the height?

  17. Callisto
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    x = radius... y = height

  18. Nanoman
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[2\pi \int\limits_{0}^{\sqrt{\pi}} (\sqrt{\pi}/2) \sin^2 (x^2)\] ?

  19. Callisto
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[V = 2 \pi \int_a^brh(width)\]r = x, h = y, width = dx

  20. Callisto
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    y is a function of x, you know.

  21. Nanoman
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Doesn't x = sqrtpi and y = (1/2)sin^2 (x^2)?

  22. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy