Area-Volume of e: Find Solutions & Calculate Volumes

[Yowhatsupt]

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.

 

[Plastic Photon]

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.

 

[HallsofIvy]

Where in the world did you get this problem! To find the limits of integration, as Plastic Photon said, you need to solve
e^{-x^2}= 2x^2
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
\int (e^{-x^2}- 2x^2)dx

 

[mybsaccownt]

In any case, I see no reasonable way of integrating
\int (e^{-x^2}- 2x^2)dx

maybe YOU don't

...but...

Maple does

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

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


a := e^{-x^2} - 2x^2

> 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




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

ahhh maple...:!)

 

[Yowhatsupt]

Where in the world did you get this problem! To find the limits of integration, as Plastic Photon said, you need to solve
e^{-x^2}= 2x^2
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
\int (e^{-x^2}- 2x^2)dx

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.

 

[Yowhatsupt]

maybe YOU don't

...but...

Maple does

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

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


a := e^{-x^2} - 2x^2

> 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




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




ahhh maple...:!)

I am so lost in your post...lol :)

 

[tim_lou]

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.