I recently had a very interesting idea. Would it be possible to calculate the surface area of a parametric surface by looking at the equation for the parametric line that wraps around that surface? If there is a way to define that surrounding line by using sines and cosines as the x and y parameters, then it is possible to wrap that line around the parametric surface as tightly as you want (that is, for each point in the sine and cosine functions, the distance between consecutive layers of the functions can be made as tight as possible). This is most easily seen with an example. Consider the following parametric function: x = Cos[2 [tex]\pi[/tex] t] (1 - t^2/a^2) y = Sin[2 [tex]\pi[/tex] t](1 - t^2/a^2) z = t^2/a^2 This parametric line wraps around a cone of height 1 and radius 1. As a gets larger, the line wraps more and more times around the cone. Consider the distance between two points on this function for a values of sine and cosine, so maybe Cos[t] = 1 and sin[t] = 0 ). As a increases, that distance between "rungs" of the line decreases. In this case, the distance between rungs would be the square root of (x(t) -x(t-1))^2+(y(t) - y(t-1)) + (z(t) - z(t-1)). Now, if we multiply that function by the equation that needs to be integrated in order to find arc length (which is easily found by taking derivatives), we have an equation that can help us find the surface area of that cone. All we need to do is take the integral with respect to t from 0 to infinity and then take the limit as a goes to infinite. Unfortunately, that integral is quite difficult, which takes me to my second point: I know (or at least I think I know) that the limit of an integral is equal to the integral of limit. For example, the integral of (limit(a->infinite)(1+x^2/a)^a) with respect to x is just the integral of e^x^2. However, if the expression (1+x^2/a)^a by itself is integrated with respect to x and then the limit is taken, we get the same result. If we attempt to take the limit as a goes to infinity of the parametric equation described above, we find that the limit evaluates to 0. The limit only produces a non zero result when t is taken to infinity as well, which is what is done once the integral is evaluated. My question is this: is there some way to evaluate this limit so that it does not collapse to zero before taking its integral from zero to infinity?