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Area-Volume with e

  1. Feb 26, 2007 #1
    1. The problem statement, all variables and given/known data
    Here is the problem.

    Let S be the region in the first quadrant bounded by the graphs of y=e^(-x^2) , y=2x^2, and the y- axis

    a. Find the area of the region S
    b. Find the volume of the solid generated when the region S is rotated about the x axis
    c. The region s is the base of a soid for which each cross section perpendicular to the x-axis is a semi-circle with diameter in the xy plane. Find the volume of this solid.

    2. Relevant equations


    3. The attempt at a solution

    I don't really have any idea how to solve this problem. I understand the process using normal functions however the e is putting me at a loss. I am trying to do it by hand so I can understand it, but am completely lost.

    Heres what I've done. Since e is base of ln.

    lny= -x^2 ln e = lny=-x^2 Therefore x= sqrt(-lny)

    With the other equation in terms of x as I am asuming I am solving for it about the y axis for part a. x = sqrt(y/2)

    From here I get confused, I don't know how to graph x=sqrt(-lny)

    I can't figure out how to find the bounderies of the interval so I can't find the area or the volume. Any help would be appreciated.:uhh: :redface:
  2. jcsd
  3. Feb 26, 2007 #2
    find the upper and lower limits, set x equal to zero and find that e raised to zero will be one and 2x^2=0. Then find the point of intersection of the two graphs and use zero and that point as your interval.

    I admit this looks tough due to the fact that you are working with e^(-x^2) which has no elementary solution.
  4. Feb 27, 2007 #3


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    Science Advisor

    Where in the world did you get this problem! To find the limits of integration, as Plastic Photon said, you need to solve
    [tex]e^{-x^2}= 2x^2[/tex]
    which cannot be done with elementary methods. You might be able to solve it in terms of the Lambert W function. In any case, I see no reasonable way of integrating
    [tex]\int (e^{-x^2}- 2x^2)dx[/tex]
  5. Feb 27, 2007 #4
    maybe YOU don't


    Maple does :cool:

    let's see if my newbie latex skills can do it justice

    > a:=exp(-x^2)- 2*x^2;

    a := [tex]e^{-x^2} - 2x^2[/tex]

    > int(a,x);
    int/indef1: first-stage indefinite integration
    int/indef1: first-stage indefinite integration
    int/indef2: second-stage indefinite integration
    int/exp: case of integrand containing exp
    int/indef1: first-stage indefinite integration
    int/indef2: second-stage indefinite integration
    int/exp: case of integrand containing exp

    [tex]\displaystyle{\frac{1}{2}}\sqrt{\pi}\ erf(x) -\displaystyle{\frac{2x^3}{3}}[/tex]

    ahhh maple...:!!)
    Last edited: Feb 27, 2007
  6. Feb 27, 2007 #5

    I got this problem from my teacher, who got it from a AP calculus practice test. I'm supposed to use my calculator, but I hate doing that when I don't know how to do it by hand.
  7. Feb 27, 2007 #6
    I am so lost in your post...lol :)
  8. Feb 28, 2007 #7
    the erf function, or the error function, is simply defined to be the integral of e^(-x^2) with some constant factors. So, it really wasn't "integrated" but just rewritten in a different way.
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