[SOLVED] [Calculus]--Help with surface area of revolution please
Problem: The curve y=e^(-x), x>0 is revolvd about the xaxis. Does the resulting surface have finite or infinite area? [Remember tht you can sometimes decide whether improper integral converges w/out calculating it exactly]
Surface area over [a,b]= 2(pi)*integral[f(x)*sqrt(1+f'(x)^2)dx]
Comparison test for improper integrals assuming f(x)>g(x)>0 for x>a: if integral[f(x)dx] on [a,infinity] converges, then integral[g(x)dx] on [a,infinity] also converges.
Surface area= 2(pi)*integral[e^(-x)*sqrt(1+e^(-2x))dx].. are the bounds [0,infinity]? What do I do after that to find out if it's finite or not? Do I use the comparison theorem? I'll appreciate any help. Thanks in advance