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Surface calculation (no integration!)

  1. Apr 1, 2014 #1
    Given r, Δθ and Δz and ρ, Δφ and Δθ, I think that is possible calculate algebraically those regular surfaces without use integration. Is possbile? If yes, how?

    image.png image.png
     
  2. jcsd
  3. Apr 1, 2014 #2

    SteamKing

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    It depends on what you mean by 'calculate'.
     
  4. Apr 1, 2014 #3
    English isn't my natural idiom, sorry. But, by 'calculate', I understand "add", "multiply", "integrate", "compute", "make the calculations"...
     
  5. Apr 1, 2014 #4

    Mark44

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    Jhenrique, you omitted an important word. What you're asking is, I believe, is it possible to calculate the areas of those surfaces, or are there formulas for their areas?

    For the first surface, to the best of my knowledge, there is no general formula. I am assuming that r is not constant. If r is constant, though, what you have is some portion of the surface of a circular cylinder, and that area can be calculated easily.

    For the second, the surface is approximately a rectangle, so the area would be approximately ##\rho^2 Δθ Δ\phi##, I believe. One of the dimensions is ##\rho Δθ## and the other is ##\rho Δ\phi##. The smaller the two angle increments are, the better the approximation is.
     
  6. Apr 1, 2014 #5
    Yeah, in actually, I wish an algebraic formula for calculate the areas of those surfaces, if those formulas are generated by integration or not, don't import, since the result be algebraic.

    And the radius in those 2 surfaces are constant!
     
  7. Apr 1, 2014 #6

    Mark44

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    The area of the shaded region in the first drawing is ##r Δθ Δz##, which is the arc length measured around the portion of the cylinder, multiplied by the height. To get the total area, sum all the area increments.

    As I said before, the area of the shaded region in the second drawing is ##r^2ΔθΔ\phi##. To get the total area, sum the area increments.

    This isn't really calculus - it's only calculus after you take the limits as all the Δ quantities approach 0, and the summation becomes an integration.
     
  8. Apr 1, 2014 #7

    LCKurtz

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    It isn't clear from the figures, but it looks to me like the ##\theta## in the second figure may represent the cylindrical (polar) ##\theta##. In that case he is describing the spherical element of surface area for constant ##\rho## which would be ##\rho^2\sin\phi \Delta \theta\Delta \phi##.
     
  9. Apr 1, 2014 #8

    Mark44

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    I stand corrected.
     
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