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anonymous

  • 5 years ago

find the volume of the solid obtained by rotating about the y-axis the region bounded by the curve y=2x^2-x^3 and the x-axis. How do I do this?

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  1. amistre64
    • 5 years ago
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    find the the x and y intercepts

  2. anonymous
    • 5 years ago
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    ok.

  3. amistre64
    • 5 years ago
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    y = x^2(2-3x) right?

  4. amistre64
    • 5 years ago
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    y=0 at 0 and 3/2 right?

  5. anonymous
    • 5 years ago
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    no y=x^2(2-x)

  6. amistre64
    • 5 years ago
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    thats right lol.... i transposed that 3 dint i :)

  7. amistre64
    • 5 years ago
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    so we have bounds of x=0 and x=2 then right?

  8. anonymous
    • 5 years ago
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    yes.

  9. amistre64
    • 5 years ago
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    we need to get these in terms of y to calculate the y axis spin; lets do an inverse

  10. anonymous
    • 5 years ago
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    ok sounds good.

  11. amistre64
    • 5 years ago
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    x = 2y^2 -y^3 ; solve for y :)

  12. amistre64
    • 5 years ago
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    wolfram might help with this :)

  13. anonymous
    • 5 years ago
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    good idea. I will get back to you a little later on this I have a baseball game.

  14. amistre64
    • 5 years ago
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    are you sure its spun around the y axis? and not the x axis?

  15. amistre64
    • 5 years ago
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    let me read up on the shell method; its prolly easier with this setup..

  16. amistre64
    • 5 years ago
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    2pi [S] x[f(x)] dx ; from [0,2]

  17. amistre64
    • 5 years ago
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    x(2x^2-x^3) = 2x^3 -x^4 soo... we integrate that to get; (5x^4 -2x^5)/10. At 0 it is zero; so we just worry about the 2

  18. amistre64
    • 5 years ago
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    i get 8/5 as the volume of the torus....if i did it right lol

  19. amistre64
    • 5 years ago
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    so i am going to try this on firefox..... and it seems to be working good so far :)

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