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anonymous

  • 5 years ago

im stumped calculus circumference arc. how come we cant use integral C(x)dx to find surfacea area. C(x) is the circumference so its 2pi*r , and dx is the thickness. so why isnt surface area integral 2pi * r * dx instead it is integral 2pi y sqrt ( 1 + dy/dx^2)dx

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  1. anonymous
    • 5 years ago
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    You're trying to compute the area of a circle?

  2. anonymous
    • 5 years ago
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    so say you have f(x)>0 in the first quadrant

  3. anonymous
    • 5 years ago
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    im doing the formula for surface area , made by revolving a curve about an axis .

  4. anonymous
    • 5 years ago
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    generated i should say

  5. anonymous
    • 5 years ago
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    Ok

  6. anonymous
    • 5 years ago
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    I don't understand that second integral you wrote.

  7. anonymous
    • 5 years ago
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    Oh wait, yes I do.

  8. anonymous
    • 5 years ago
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    integral 2pi * f(x) sqrt ( 1 + f ' (x) ^2 ) dx

  9. anonymous
    • 5 years ago
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    yeah

  10. anonymous
    • 5 years ago
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    Polpak, when you get finished can you assist me?

  11. anonymous
    • 5 years ago
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    so my question is , why cant we just use the easier formula integral 2pi f(x) dx

  12. anonymous
    • 5 years ago
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    what r u guys parternes?

  13. anonymous
    • 5 years ago
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    Are you rotating around the x axis?

  14. anonymous
    • 5 years ago
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    yes in this case

  15. anonymous
    • 5 years ago
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    say y = x^2 from 0 to 2 , but trying to find surfacea area

  16. anonymous
    • 5 years ago
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    But you still need to account for the way f(x) is changing over your dx

  17. anonymous
    • 5 years ago
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    You would be fine if it were flat washers

  18. anonymous
    • 5 years ago
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    But along the top edge there can be a lot of slopes that will change the area of the final surface.

  19. anonymous
    • 5 years ago
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    thats true but

  20. anonymous
    • 5 years ago
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    when you go to zero, dont they approach 2pi * r , the height

  21. anonymous
    • 5 years ago
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    Yes, but they still have an angle.

  22. anonymous
    • 5 years ago
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    You're ok computing the volume that way

  23. anonymous
    • 5 years ago
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    but not the surface area.

  24. anonymous
    • 5 years ago
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    actually its not volume

  25. anonymous
    • 5 years ago
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    it would be the surface of a cylinder type figure, i guess, not sure

  26. anonymous
    • 5 years ago
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    for example here, we have using my way integral 2pi x^2 dx from 0 to 2

  27. anonymous
    • 5 years ago
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    See you're way makes the assumption that along the top edge for a tiny dx, that the length of the line of the function = dx.

  28. anonymous
    • 5 years ago
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    the books way is integral 2pi x^2 sqrt ( 1 + (2x)^2) dx

  29. anonymous
    • 5 years ago
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    come back

  30. anonymous
    • 5 years ago
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    but it doesn't. The length of the line for a small dx is proportional the the square root of the square of dy^2 + dx^2

  31. anonymous
    • 5 years ago
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    err I said that wrong

  32. anonymous
    • 5 years ago
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    But you see what I mean.

  33. anonymous
    • 5 years ago
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    We are ok with the area under the curve assuming that as we take smaller and smaller values for dx, the area approaches f(x) * dx

  34. anonymous
    • 5 years ago
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    But for the line integral it doesn't come close

  35. anonymous
    • 5 years ago
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    yeah, why is that

  36. anonymous
    • 5 years ago
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    Let me see if I can explain.

  37. anonymous
    • 5 years ago
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    http://www.dabbleboard.com/draw?b=Guest666913&i=0&c=d86fb2255acc9e8490af7e397d12fe9fc183a40a

  38. anonymous
    • 5 years ago
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    You there?

  39. anonymous
    • 5 years ago
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    http://www.twiddla.com/529576

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