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Ruled Surfaces

  1. May 29, 2005 #1
    I've been trying to figure out how to calculate the exact area of a ruled surface. I think I've come to a solution that works, but I'm not sure it's totally accurate. The procedure is as follows:

    Consider two curves. A ruled surface is constructed by connecting each point on one curve to a corresponding point on the other curve. Now, taking a point on curve 1 (a), the corresponding point on curve 2 (b), and another point on curve 1 arbitrarily close to the first point (c), we form a triangle. Also, by taking another point on curve 2 close to point b (d), we can form triangle bcd. Adding the areas of abc and bcd, we have one element of area. Integrating these areas, you get the whole area.

    The question is, is this exact, or just an approximation?
  2. jcsd
  3. May 31, 2005 #2
    Depends on how small u are making that base of the triangle. If u are making it very small, then it should be exact otherwise it would be an approximation. (Your method corresponds very close to how computers generate surfaces on computer)

    -- AI
  4. May 31, 2005 #3
    I'm not sure what you mean by taking a point that is arbitrarily close. Here is one way to carry out your program.

    Partition one of the lines with points [itex]\{a_n\}[/itex] and the other line with points [itex]\{b_n\}[/itex], where [itex]b_n[/itex] is the point associated with [itex]a_n[/itex]. Take the sum of the areas of the triangles formed by the points [itex]a_1, b_1, a_2[/itex] and by [itex]b_1, a_2, b_2[/itex], etc. to the end of the lines. Note that, in general, such sums are approximations to the area because, the sides of the triangles are straight lines, but the surface may be curved. Take the limit of such sums as the distance between the points in each partition goes to zero. The limit will be the exact area.
  5. May 31, 2005 #4
    Thanks. Just what I wanted to know.

  6. Jun 30, 2005 #5
    Let's say you have two parametric functions (represented via positional vectors) [itex] \vec r_1 \left( t \right) [/itex] and [itex] \vec r_2 \left( t \right) [/itex], continuous [itex] \forall t \in \left( {a,b} \right) [/itex] , where [itex] t [/itex] is your parameter. Now, you connect [itex] \vec r_1 \left( t \right) [/itex] and [itex] \vec r_2 \left( t \right) [/itex] via line segments corresponding to equivalent parameter values --- i.e., connect point [itex] \vec r_1 \left( t \right) [/itex] to [itex] \vec r_2 \left( t \right) [/itex] (corresponding to equivalent t's) for all t's from t=a to t=b. What I do below is just divide the (a,b) interval into smaller units of just [itex] {\Delta t} [/itex].
    *Then, by Moo of Doom's method, the area would be:
    [tex] \frac{1}
    {2}\mathop {\lim }\limits_{\Delta t \to 0^+} \sum\limits_{n = 0}^{2\left( {\frac{{b - a}}
    {{\Delta t}}} \right) - 1} {\left\{ {\left[ {\vec r_1 \left( {a + A_n \Delta t} \right) - \vec r_2 \left( {a + B_n \Delta t} \right)} \right] \times \left[ {\vec r_1 \left( {a + B_n \Delta t} \right) - \vec r_2 \left( {a + C_n \Delta t} \right)} \right]} \right\}} [/tex]
    [tex] \left\{ \begin{gathered}
    A_n = \frac{{\left( { - 1} \right)^n + 2n - 1}}
    {4} \hfill \\
    B_n = \frac{{ - \left( { - 1} \right)^n + 2n + 1}}
    {4} \hfill \\
    C_n = \frac{{\left( { - 1} \right)^n + 2n + 3}}
    {4} \hfill \\
    \end{gathered} \right\} [/tex]
    However, this method assumes no intersections of the line segments connecting [itex] \vec r_1 \left( t \right) [/itex] and [itex] \vec r_2 \left( t \right) [/itex] vectors.
    My question is: How would this Riemann sum be converted into an integral?
    Last edited: Jul 1, 2005
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