# Area-Volume with e

## 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.

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:

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
Homework Helper
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
$$\int (e^{-x^2}- 2x^2)dx$$

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...:!!)

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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
$$e^{-x^2}= 2x^2$$
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.

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 :)

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.