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anonymous

  • 5 years ago

revolve the region bounded by x axis , y = sqrt ( x+ 4) , and x=0 about y axis using shell method. compare it to the disc method

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  1. dumbcow
    • 5 years ago
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    \[=2 \pi \int\limits_{0}^{4}x \sqrt{x+4} dx\]

  2. anonymous
    • 5 years ago
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    no thats not the region

  3. anonymous
    • 5 years ago
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    the region is bounded by y = sqrt ( x + 4) , the y axis, its in quadrant 2

  4. anonymous
    • 5 years ago
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    so using that integral , its from -4 to 0

  5. dumbcow
    • 5 years ago
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    yes but we are revolving around y-axis so it covers both quadrants symmetrically and this way you get a positive volume

  6. anonymous
    • 5 years ago
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    it isnt symmertric though

  7. dumbcow
    • 5 years ago
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    x represents radius of 3-d figure, as you move away from y-axis the radius goes from 0 to 4

  8. anonymous
    • 5 years ago
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    the area from 0 to 4 is not equal to the area from -4 to 0

  9. anonymous
    • 5 years ago
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    how about this, revolve region bounded by y = sqrt x, x = 4, axis, about the line y = -4

  10. dumbcow
    • 5 years ago
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    oh i see, ok think of you reflected the function over y-axis replace x with (-x) f(x) = sqrt(4-x) i didn't do that in the intergral

  11. anonymous
    • 5 years ago
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    yes, that will do it,

  12. dumbcow
    • 5 years ago
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    sorry

  13. anonymous
    • 5 years ago
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    im trying to avoid these cases. are there other situations i can make up,

  14. anonymous
    • 5 years ago
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    no you did fine :)

  15. anonymous
    • 5 years ago
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    i dont seem to have a problem with disc method, since we have pi*r^2, but the shell method does produce negative

  16. dumbcow
    • 5 years ago
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    i guess always try to reflect onto positive quadrant when possible with volumes you want to integrate over positive values

  17. dumbcow
    • 5 years ago
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    make radius positive in shell method then adjust f(x) accordingly

  18. anonymous
    • 5 years ago
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    but i think the disc method in that case works

  19. anonymous
    • 5 years ago
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    using horizontal slices, we have x = y^2 - 4

  20. anonymous
    • 5 years ago
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    for y = sqrt ( x + 4

  21. dumbcow
    • 5 years ago
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    yeah discmethod \[=\pi \int\limits_{0}^{2}(y^{2}-4)^{2}dy\]

  22. anonymous
    • 5 years ago
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    so we get integral pi (

  23. anonymous
    • 5 years ago
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    pi integral ( 0 - (y^2-4))^2

  24. anonymous
    • 5 years ago
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    from -4 to 0

  25. dumbcow
    • 5 years ago
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    i prefer disc method

  26. anonymous
    • 5 years ago
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    oh my bad

  27. anonymous
    • 5 years ago
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    0 to 2

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