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YUNoWorkOS

  • one year ago

Find the volume of the solid of revolution generated by revolving the equation \(y=\sqrt(x)\) about the line x=4 (The region is bounded by the x axis and x=4)

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  1. inkyvoyd
    • one year ago
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    I understand if I rewrite the equation as y=\(\sqrt{x-4}\) then integrate with appropriate bounds it will work, but there is some sort of alternate interpretation I am not understanding.

  2. inkyvoyd
    • one year ago
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    *rewrite the equation as y=\(\sqrt{x+4}\)

  3. amistre64
    • one year ago
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    |dw:1365190324765:dw| \[\int 2\pi ~r~h\]

  4. amistre64
    • one year ago
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    r = c-x, h = sqrt(x), and dx from 0 to c

  5. YUNoWorkOS
    • one year ago
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    omg thank you - that applies for shell method as well correct?

  6. amistre64
    • one year ago
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    that IS the shell method :)

  7. YUNoWorkOS
    • one year ago
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    oh shoot - is there any way to do thi with the disk method?

  8. YUNoWorkOS
    • one year ago
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    this problem is listed in the disk method section of my textbook, but i'm not sure how it's done...

  9. amistre64
    • one year ago
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    \[\int~2\pi r\] \[r=g(y)\text{ ; which is the inverse of f(x)}\]and dy from 0 to f(c)

  10. amistre64
    • one year ago
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    inverse of y = sqrt(x); x = y^2

  11. amistre64
    • one year ago
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    inky suggests that we move the graph to something simpler to play with; shift if left by "c" units given us y = sqrt(x+c) r = x = sqrt(y) - c

  12. amistre64
    • one year ago
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    r = x = y^2 - c :)

  13. amistre64
    • one year ago
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    the only issue that may present itself with that setup tho is that you get a negative summation

  14. amistre64
    • one year ago
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    why did i do 2pi r? thats circumference :/ .... area is \[\int pi r^2\]

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