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El_Arrow

  • one year ago

Need help please!

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  1. El_Arrow
    • one year ago
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    find the volume of the solid formed by revolving the region bounded by the graph of f(x) = x^2 and g(x) = 12-x to the right of x=1 about the y-axis

  2. El_Arrow
    • one year ago
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    first thing i did was set f(x) and g(x) equal to one another and i got -4 and 3

  3. El_Arrow
    • one year ago
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    |dw:1440982705148:dw|

  4. freckles
    • one year ago
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    \[x^2=12-x \\ x^2+x-12=0 \\ (x+4)(x-3)=0 \\ x=-4 \text{ or } x=3 \] (-4)^2=16 (3)^2=9 |dw:1440982735696:dw|

  5. freckles
    • one year ago
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    |dw:1440982829162:dw| we are looking at this region (the shaded region) right?

  6. El_Arrow
    • one year ago
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    yes

  7. El_Arrow
    • one year ago
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    so it would be the integral from 1 to 3 @freckles ?

  8. freckles
    • one year ago
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    yea

  9. freckles
    • one year ago
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    so it looks like you were trying to use shells method

  10. El_Arrow
    • one year ago
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    yes thats correct

  11. freckles
    • one year ago
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    |dw:1440983126638:dw|

  12. El_Arrow
    • one year ago
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    so something like this |dw:1440983268893:dw|

  13. freckles
    • one year ago
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    yes. |dw:1440983363623:dw|

  14. El_Arrow
    • one year ago
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    but why do you have the square root over the x why not x^2

  15. freckles
    • one year ago
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    lol

  16. freckles
    • one year ago
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    I forgot what function we were using

  17. freckles
    • one year ago
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    sqrt(x) is suppose to be x^2

  18. El_Arrow
    • one year ago
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    oh okay lol

  19. freckles
    • one year ago
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    i'm sorry i do idiot things sometimes

  20. El_Arrow
    • one year ago
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    its okay

  21. freckles
    • one year ago
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    |dw:1440983615720:dw|

  22. El_Arrow
    • one year ago
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    the answer is 116pi/3 but i got 16pi?

  23. freckles
    • one year ago
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    hmm so the radius should have been just x

  24. El_Arrow
    • one year ago
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    i think so let me see if it works

  25. freckles
    • one year ago
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    |dw:1440983787477:dw|

  26. freckles
    • one year ago
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    |dw:1440983907300:dw| so you were right to begin with you just needed 1 to 3 instead of 0 to 1

  27. El_Arrow
    • one year ago
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    yeah i got right now thanks alot

  28. freckles
    • one year ago
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    it does make sense I think if I had looked at that whole thing from 0 to 3 and subtract out the part that is 0 to 1 we would have seen it better \[\int\limits_0^3 2\pi(x)(12-x-x^2) dx-\int\limits_0^1 2\pi(x)(12-x-x^2)dx \\ =\int\limits_1^3 2 \pi (x) (12-x-x^2) dx\]

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