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anonymous

  • 4 years ago

The region bounded above by the line y=6, below by the curve y=sqrt(x) and to the left by the y-axis, rotated around x=36.

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  1. amistre64
    • 4 years ago
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    we should get the same results if im thinking of it correctly

  2. amistre64
    • 4 years ago
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    but just in case; think of this as a solid cylinder that we are removing the under part of the sqrt(x)

  3. amistre64
    • 4 years ago
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    |dw:1327534530288:dw|

  4. amistre64
    • 4 years ago
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    so, cylindar itself; r=36, h=6; pi 36^2*6 is total area of this cylindar, now we gut out the area under the sqrt(x)

  5. anonymous
    • 4 years ago
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    so it's like an upside down bowl

  6. amistre64
    • 4 years ago
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    yes, and lets disc out the underside, i think it might integrate easier that way

  7. anonymous
    • 4 years ago
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    ok after I integrated I got, 18x^2 - 2/3 * x^(3/2)

  8. amistre64
    • 4 years ago
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    |dw:1327534763514:dw|

  9. amistre64
    • 4 years ago
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    r = 36-y^2; when y=0, r=36; when y=6, r=0 that works in my head lol

  10. amistre64
    • 4 years ago
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    V = int (pi [f(y)]^2) from 0 to 6 V = int (pi(36-y^2)^2) from 0 to 6, im just gonna paste this into wolfram to verify :)

  11. anonymous
    • 4 years ago
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    so I couldn't leave the y=sqrt(x), but would have to convert it to y^2 instead?

  12. amistre64
    • 4 years ago
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    yes, for disc; maybe shell is easier?

  13. anonymous
    • 4 years ago
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    Gotcha! So your r should always equal 0 if a limit is plugged in?

  14. amistre64
    • 4 years ago
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    not always, the radius changes as the integration moves across the interval; in this case, with the disc, the radius is determined by 36 over - y^2 back

  15. amistre64
    • 4 years ago
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    but lets do shell and see if my thoughts improve lol

  16. anonymous
    • 4 years ago
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    haha alright

  17. amistre64
    • 4 years ago
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    the shell moves it radius from 0 to 36 as it does, the height is determined by sqrt(x); but i tend to like to move everything to the yaxis to get a clearer picture; so; sqrt(x+36) moves everything to the left by 36 and gets our rotation around the y axis

  18. amistre64
    • 4 years ago
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    |dw:1327535297927:dw|

  19. anonymous
    • 4 years ago
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    Ahh. and that isn't what we want!

  20. amistre64
    • 4 years ago
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    this is another view of what we want; its better to me

  21. amistre64
    • 4 years ago
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    \[V=2pi\int_{0}^{-36}x(\sqrt{x+36})dx\] i believe thats right;

  22. amistre64
    • 4 years ago
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    sorry, but I gotta get; hope i havent messed it up too bad. good luck tho :)

  23. anonymous
    • 4 years ago
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    thank you for your help!

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