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MathSofiya Group Title

just making sure I understand it.... \[S=\int_{a}^{b} 2\pi y \sqrt{1+{(\frac{dy}{dx})}^2} dx\] \[S=\int_{c}^{d} 2\pi y \sqrt{1+{(\frac{dx}{dy})}^2} dy\] \[S=\int_{a}^{b} 2\pi x \sqrt{1+{(\frac{dy}{dx})}^2} dx\] One is for revolution about the x axis...the other for revolution about the y- axis...and then we have one more equation....why?

  • 2 years ago
  • 2 years ago

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  1. Libniz Group Title
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    what the devil is this?

    • 2 years ago
  2. MathSofiya Group Title
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    Area of a Surface of Revolution

    • 2 years ago
  3. Libniz Group Title
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    surface area ?

    • 2 years ago
  4. MathSofiya Group Title
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    yes

    • 2 years ago
  5. Libniz Group Title
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    let me make some drawing

    • 2 years ago
  6. MathSofiya Group Title
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    @TuringTest

    • 2 years ago
  7. Libniz Group Title
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    |dw:1341858624540:dw|

    • 2 years ago
  8. Libniz Group Title
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    we are taking circumference of each plate , they have height(radius) of 'y' 2 Pi r= 2 Pi y

    • 2 years ago
  9. MathSofiya Group Title
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    what if we're rotating about the y axis...would it still be 2 pi y

    • 2 years ago
  10. Libniz Group Title
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    no

    • 2 years ago
  11. Libniz Group Title
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    it would be much more complicated

    • 2 years ago
  12. MathSofiya Group Title
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    2 pi x

    • 2 years ago
  13. Libniz Group Title
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    not that simple

    • 2 years ago
  14. Libniz Group Title
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    first ,you gotta define function in term of x

    • 2 years ago
  15. MathSofiya Group Title
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    ok

    • 2 years ago
  16. MathSofiya Group Title
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    as in x= .....y....

    • 2 years ago
  17. MathSofiya Group Title
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    or x=g(y)

    • 2 years ago
  18. Libniz Group Title
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    it is too complicated unlike finding volume , we just use x axis

    • 2 years ago
  19. MathSofiya Group Title
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    ok

    • 2 years ago
  20. MathSofiya Group Title
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    |dw:1341859119434:dw|

    • 2 years ago
  21. MathSofiya Group Title
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    \[S=\int 2\pi x ds\]

    • 2 years ago
  22. MathSofiya Group Title
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    but the book still has (dy/dx)^2

    • 2 years ago
  23. MathSofiya Group Title
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    @helder_edwin please help

    • 2 years ago
  24. Libniz Group Title
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    it is 3 dimensional

    • 2 years ago
  25. helder_edwin Group Title
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    give a second to check my books it's been a long time

    • 2 years ago
  26. MathSofiya Group Title
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    ok

    • 2 years ago
  27. MathSofiya Group Title
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    Everyone abandoned me :'(

    • 2 years ago
  28. helder_edwin Group Title
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    no

    • 2 years ago
  29. TuringTest Group Title
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    I didn't get your ping... weird I'll have to ask administration about that... ok, what do we ave here, let me read...

    • 2 years ago
  30. TuringTest Group Title
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    so you want to know why we have 4 formulas, right? my answer is that there are really only two, but each one can be seen from two different perspectives...

    • 2 years ago
  31. MathSofiya Group Title
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    ok

    • 2 years ago
  32. TuringTest Group Title
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    first consider the arc length formula:\[ds=\sqrt{1+[f'(x)]^2}dx\]now this is the formula for arc length taken from the perspective of y being a funcion of x but arc length is the same regardless of whether you look at the function as f(x) or g(y) since the arc itself will still have the same length. so we can also write\[ds=\sqrt{1+[g'(y)]^2}dy\] and as long as we are talking about the same curve they should be equal, since the arc can only have one length. make sense so far?

    • 2 years ago
  33. MathSofiya Group Title
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    Yes

    • 2 years ago
  34. TuringTest Group Title
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    now for a revolution, the formula is\[A=\int2\pi yds\]or\[A=\int2\pi xds\]depending on which axis we are going around but as I just explained above, ds (the arc length differential) can always be written two ways depending on whether we consider y a function of x or vice-versa, so each of these formulas is potentially two depending on how we look at our ds

    • 2 years ago
  35. helder_edwin Group Title
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    can you read spanish?

    • 2 years ago
  36. TuringTest Group Title
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    yo si

    • 2 years ago
  37. MathSofiya Group Title
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    sorry I can't

    • 2 years ago
  38. SmoothMath Group Title
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    lol. That guy.

    • 2 years ago
  39. TuringTest Group Title
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    here is a nice full explanation if you care to dig deeper http://tutorial.math.lamar.edu/Classes/CalcII/SurfaceArea.aspx

    • 2 years ago
  40. MathSofiya Group Title
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    I will , thanks @TuringTest .

    • 2 years ago
  41. TuringTest Group Title
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    welcome :)

    • 2 years ago
  42. SmoothMath Group Title
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    Sofiya, try to break down each integral like this: First, look at the end part, the variable you're integrating with respect to. Then, look at the limits, so you say to yourself, "Okay, travelling along x from a to b." or something like that. Then, look at the function inside and try to break that down, and what that means at each particular x. For these particular integrals, the insides have 2 basic parts. The first part has the form 2pi*something, where that something is the radius. The second part is the arclength formula. So it's calculating the arclength, and then it's multiplying that in a circle.

    • 2 years ago
  43. helder_edwin Group Title
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    sorry i wanted to send you something but it's in spanish. but i goes much in the same way as what @TuringTest did

    • 2 years ago
  44. MathSofiya Group Title
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    oh ok....thanks everyone Thanks @SmoothMath !!!

    • 2 years ago
  45. SmoothMath Group Title
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    My pleasure =D

    • 2 years ago
  46. TuringTest Group Title
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    @helder_edwin me lo mandas por favor? quiero aprender mas la terminologia en espanol

    • 2 years ago
  47. helder_edwin Group Title
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    claro!

    • 2 years ago
  48. helder_edwin Group Title
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    @TuringTest recibiste el pdf?

    • 2 years ago
  49. TuringTest Group Title
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    no, donde lo pusiste? podrias mandarme un ("link"?) o url ? ya puesto que estoy tu "fan" es posible mandar mensajes privadas

    • 2 years ago
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