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Math Help - [SOLVED] [Calculus]--Help with surface area of revolution please

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    [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]

    Solution Attempt:
    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.


    So,
    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
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    Quote Originally Posted by coe236 View Post
    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]

    Solution Attempt:
    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.


    So,
    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
    The integral limits are x = 0 to x = +oo.

    To do the integral, make the substitution u = e^{-x}:

    1. dx = -e^{-x} \, du \Rightarrow dx = - \frac{du}{u}.

    2. x = 0 \Rightarrow u = 1 and x = +\infty \Rightarrow u = 0.

    Therefore:

    S = 2 \pi \int_{1}^{0} u \, \sqrt{1 + u^2} \, \left( - \frac{du}{u} \right) = -2 \pi \int_{1}^{0} \sqrt{1 + u^2} \, du = 2 \pi \int_{0}^{1} \sqrt{1 + u^2} \, du ......
    Last edited by mr fantastic; March 11th 2008 at 07:23 PM.
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    Quote Originally Posted by mr fantastic View Post
    The integral limits are x = 0 to x = +oo.

    To do the integral, make the substitution u = e^{-x}:

    1. dx = -e^{-x} \, du \Rightarrow dx = - \frac{du}{u}.

    2. x = 0 \Rightarrow u = 1 and x = +\infty \Rightarrow u = 0.

    Therefore:

    S = - 2 \pi \int_{1}^{0} u \, \sqrt{1 + u^2} \, \left( - \frac{du}{u} \right) = -2 \pi \int_{1}^{0} \sqrt{1 + u^2} \, du = 2 \pi \int_{0}^{1} \sqrt{1 + u^2} \, du ......
    Thanks very much--really quick--this might be a dumb question, but why is it negative 2(pi) instead of positive?
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    Quote Originally Posted by coe236 View Post
    Thanks very much--really quick--this might be a dumb question, but why is it negative 2(pi) instead of positive?
    Whoops. My mistake with cut and paste. I've edited my reply, correcting that error.
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    Gotcha. Thanks again
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