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anonymous

  • 5 years ago

revolve the region bounded by y = x^2 , y = 1, about the y axis .

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  1. anonymous
    • 5 years ago
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    before you say anything, its not possible to do. x = sqrt y, and x = -sqrt y , but i have

  2. dumbcow
    • 5 years ago
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    \[V = \pi \int\limits_{0}^{1}(\sqrt{y})^{2} dy\]

  3. dumbcow
    • 5 years ago
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    what do you mean?

  4. anonymous
    • 5 years ago
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    but what about -1 to 0

  5. dumbcow
    • 5 years ago
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    function is not defined for y<0

  6. anonymous
    • 5 years ago
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    but the region is defined

  7. anonymous
    • 5 years ago
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    ok here is another one like it, revolve y = x^2 and y = x+2 bounded, about y axis

  8. anonymous
    • 5 years ago
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    it exists in the math universe, not in the real world ( you can only revolve one revolution at a time about an axis )

  9. anonymous
    • 5 years ago
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    dumbcow, do you see y = x^2 , y = x+2

  10. dumbcow
    • 5 years ago
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    oh yes i see, sorry i was also assuming it was bounded by x=0

  11. anonymous
    • 5 years ago
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    so does it make sense , you get extra volume, correct ?

  12. dumbcow
    • 5 years ago
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    wait no it would still be same volume if rotating around y-axis the whole region is from y=0 and y=1

  13. anonymous
    • 5 years ago
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    but you have extra volume in the second quadrant

  14. dumbcow
    • 5 years ago
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    hmm ok its double the volume then

  15. anonymous
    • 5 years ago
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    not quite double though

  16. anonymous
    • 5 years ago
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    its a real feather of a problem, lets use shell method , see what we get

  17. anonymous
    • 5 years ago
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    easier than using disc method here

  18. anonymous
    • 5 years ago
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    integral 2pi * x ( x+2 - x^2)

  19. dumbcow
    • 5 years ago
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    ok \[=2 \pi \int\limits_{-1}^{1}x(x ^{2})dx\]

  20. dumbcow
    • 5 years ago
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    oh we are doing the other one

  21. anonymous
    • 5 years ago
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    as i suspected, we get negative volume from -1 to 0

  22. anonymous
    • 5 years ago
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    i think in the shell method, its important that the radius remain positive

  23. anonymous
    • 5 years ago
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    here is a similiar question, revolve region y = (x+2)^2 and y = 1 about y axis, using shell method. you get a negative answer

  24. dumbcow
    • 5 years ago
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    yes but volume can't be neg in the practical sense

  25. anonymous
    • 5 years ago
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    right

  26. anonymous
    • 5 years ago
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    so there are limitations to the revolving region problems

  27. dumbcow
    • 5 years ago
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    you just switch the bounds from neg to positive, because when revolving around axis there is symmetry

  28. dumbcow
    • 5 years ago
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    not in every case though

  29. anonymous
    • 5 years ago
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    what about y = x^2 - 4 and bounded by y = 0, revolve it about x axis

  30. anonymous
    • 5 years ago
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    i think that works, we only get problems with the shell method because the radius is assumed to be positive

  31. anonymous
    • 5 years ago
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    and in the disc method we have radius squared, so its more flexible, but why cant we integrate y = x^2 and y=1 about y axis, using horizontal discs ?

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