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[SOLVED] Area of a Surface of Revolution--Help Please
1. Problem: The curve y=e^(-x), x>0 is revolvd about the xaxis. Does the resultin surface have finite or infinite area? [Remember tht you can sometimes decide whether improper integral converges w/out calculating it exactly]
2. Surface area over [a,b]= 2(pi)*integral[f(x)*squareroot(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.
3. Surface area= 2(pi)*integral[e^(-x)*squareroot(1+e^(-2x))dx].. are the bounds [0,infinity]? What do I do after that to find out if it's finite or not? I'll appreciate any help. Thanks in advance
1. Problem: The curve y=e^(-x), x>0 is revolvd about the xaxis. Does the resultin surface have finite or infinite area? [Remember tht you can sometimes decide whether improper integral converges w/out calculating it exactly]
2. Surface area over [a,b]= 2(pi)*integral[f(x)*squareroot(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.
3. Surface area= 2(pi)*integral[e^(-x)*squareroot(1+e^(-2x))dx].. are the bounds [0,infinity]? What do I do after that to find out if it's finite or not? I'll appreciate any help. Thanks in advance