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anonymous

  • 5 years ago

Find the volume of a solid generated by revolving the graphs of the functions y=6-2x-x^2 and y=x+6 about the line y=2 Intersections points are x=-3 and x=0 I think the radius of the solid should be 6-2x-x^2-(x+6)-2 which is -2-3x-x^2 then I have Vol = pi*Integral from -3 to 0 of (-2-3x-x^2)^2dx I integrate and get pi *[4x65/5+3/2x^4+13x^3/3+6x^2+4x ] evalulated between -3 and 0. I end up with 147.9pi which seems way too big.Can you find my mistake? Thank you so much!!!

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  1. amistre64
    • 5 years ago
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    if i see it right; you went ahead and subtracted 2 to get it to the x axis right?

  2. amistre64
    • 5 years ago
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    y = 4 -2x -x^2 y = x+4 and spin that around the x axis (y=0)

  3. amistre64
    • 5 years ago
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    is your interval good? 6-2x-x^2 -x-6 -x^2 -3x = 0 -x(x+3) = 0; x= [-3,0] is good

  4. anonymous
    • 5 years ago
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    Since it is being revolved around y=2, I was subtracting 2 to eliminate that much verticasl distance from the radius.

  5. amistre64
    • 5 years ago
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    then we can integrate: subtract one function from the other... 4-2x-x^2 - x-4 pi [S] [-x^3 -3x]^2 dx ; [-3,0]

  6. amistre64
    • 5 years ago
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    -x^2....typoed it :)

  7. amistre64
    • 5 years ago
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    pi [S] (x^6 +9x^2 +6x^4) dx right?

  8. amistre64
    • 5 years ago
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    -2-3x-x^2 is your function; why?

  9. anonymous
    • 5 years ago
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    That was my radius

  10. amistre64
    • 5 years ago
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    it doesnt match my radius tho :) so whos right?

  11. amistre64
    • 5 years ago
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    yours is -2 larger

  12. amistre64
    • 5 years ago
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    you see it?

  13. amistre64
    • 5 years ago
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    y=6-2x-x^2 (-2) = 4-2x -x^2 y=x+6 (-2) = x+4

  14. amistre64
    • 5 years ago
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    4 -2x -x^2 -[x+4] -3x -x^2 right?

  15. anonymous
    • 5 years ago
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    so I need to subtact 2 from both functions?

  16. amistre64
    • 5 years ago
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    Yes, becasue you need to move both function pi [S] (x^6 +9x^2 +6x^4) dx pi (x^7/7 +3x^3 + 6x^5/5) ; [-3,0]

  17. anonymous
    • 5 years ago
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    How are you getting thisx^6 +9x^2 +6x^4) dx

  18. amistre64
    • 5 years ago
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    you gotta squre your radius

  19. amistre64
    • 5 years ago
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    pi [S] [f(x)]^2 dx

  20. amistre64
    • 5 years ago
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    [f(x)]^2 = [-2x-x^2]^2

  21. anonymous
    • 5 years ago
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    If the radius is (-3x-x^2) and we square it, don't we get 9x^2-6x^3=x^4?

  22. anonymous
    • 5 years ago
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    9x^2-6x^3+x^4

  23. amistre64
    • 5 years ago
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    -3x(-3x) = 9x^2 -x^2(-x^2) = x^4 -3x(-x^2)(2) = +6x^3

  24. amistre64
    • 5 years ago
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    we could go with -1(3x+x^2) as well i htink

  25. amistre64
    • 5 years ago
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    but then we get - everything

  26. anonymous
    • 5 years ago
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    That 's OK b/c the def integral goes form -3 to 0

  27. amistre64
    • 5 years ago
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    wanna go with - everything? i dont think it makes a diff :)

  28. amistre64
    • 5 years ago
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    pi (-3x^3 -x^5/5 -3x^4/2) then right?

  29. anonymous
    • 5 years ago
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    yes

  30. anonymous
    • 5 years ago
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    I get 8.1

  31. anonymous
    • 5 years ago
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    8.1 pi

  32. amistre64
    • 5 years ago
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    pi * (((-3) * ((-3)^3)) - (((-3)^5) / 5) - ((3 * ((-3)^4)) / 2)) = 25.4469005 is what google spits out ....

  33. amistre64
    • 5 years ago
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    which is 8.1 pi :)

  34. anonymous
    • 5 years ago
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    So do you think that's right?

  35. amistre64
    • 5 years ago
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    i think so; you can double check it if you want.... but im pretty sure we worked the kinks out of it :)

  36. anonymous
    • 5 years ago
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    Thanks so much!!!

  37. amistre64
    • 5 years ago
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    try to integrate one function at a time and subtract one from the other in the end to verify; kinda like making a donut and then cutting out the hole

  38. amistre64
    • 5 years ago
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    youre welcome :)

  39. anonymous
    • 5 years ago
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    Nice way to check! Thanks again!

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