Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding volumes of solids involving exponentials

  1. Sep 29, 2004 #1


    User Avatar
    Gold Member

    I'm attempting to find the volume of the solid obtained by rotating the region under the curve:

    [tex]e^{-x^2}[/tex] Bounded by y = 0, x = 0, and x = 1.

    I've done quite a few of these problems before, however, none of them have involved an exponential. If I recall correctly [tex]e^{x^2}[/tex] cannot be integrated, at least symbolically. So, therein lies the problem.

    So far this is what I've done:

    [tex]\int_0^1 \pi\left( e^{-x^2}\right)^2 dx[/tex]
    [tex]\pi\int_0^1 e^{-2x^2} dx[/tex]

    I've integrated that integral with Mathematica, and it returns a function that uses the error function, which I doubt is anything that I'm expected to come up with in this class.

    Therefore, the only way I believe this problem can be completed is by finding the area under [tex]e^{2x^2}[/tex] on the interval [tex]\left[0 ,1\right][/tex] using a Riemann Sum.

    Any thoughts?

    Edit: Fixed a typo involving a constant in the integral.
    Last edited: Sep 29, 2004
  2. jcsd
  3. Sep 29, 2004 #2


    User Avatar
    Gold Member

    I must be falling asleep. I decided that I was going about finding the volume the wrong way. For one, the question asks that you rotate the function around the y-axis, therefore, using the method of cylindrical shells would be easier than solving for x and integrating over dy.

    I'm still going to run into the same problem, though - integrating [tex]e^{-x^2}[/tex].

    It's pretty amazing how just talking about a problem can clear a few things up.
  4. Sep 29, 2004 #3
    In answer to 1. You cannot find a primitive, but you can evaluate definite integrals using multiple integrals. What u need to do is change variables so u get ur integral in terms of [tex]e^{-x^2}[/tex], then calculate that integral with the fact that
    [tex]\int_0^1 e^{-x^2}dx=\left(\int_0^1 \int_0^1e^{-x^2-y^2}dxdy\right)^{1/2}[/tex] and then use polar coordinates. The domain can be a little tricky though.

    In answer to two. Its easier that way cus the limits of integration are now from 0 to infinitum, wich makes the integra much easier. am i right? (im falling asleep too :P)
    Last edited: Sep 29, 2004
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook