Area-Volume with e

  • Thread starter Yowhatsupt
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  • #1
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Homework Statement


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.



Homework Equations



y=e^(-x^2)
y=2x^2


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:
 

Answers and Replies

  • #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.
 
  • #3
HallsofIvy
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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]
 
  • #4
In any case, I see no reasonable way of integrating
[tex]\int (e^{-x^2}- 2x^2)dx[/tex]
maybe YOU don't

...but...

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:
  • #5
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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]

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.
 
  • #6
34
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maybe YOU don't

...but...

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...:!!)
I am so lost in your post...lol :)
 
  • #7
682
1
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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