Finding volumes of solids involving exponentials

[graphic7]

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

$$e^{-x^2}$$ 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 $$e^{x^2}$$ cannot be integrated, at least symbolically. So, therein lies the problem.

So far this is what I've done:

$$\int_0^1 \pi\left( e^{-x^2}\right)^2 dx$$
$$\pi\int_0^1 e^{-2x^2} dx$$

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 $$e^{2x^2}$$ on the interval $$\left[0 ,1\right]$$ using a Riemann Sum.

Any thoughts?

Edit: Fixed a typo involving a constant in the integral.

 

[graphic7]

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

It's pretty amazing how just talking about a problem can clear a few things up.

 

[ReyChiquito]

If I recall correctly $$e^{-x^2}$$ cannot be integrated, at least symbolically.

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 $$e^{-x^2}$$, then calculate that integral with the fact that
$$\int_0^1 e^{-x^2}dx=\left(\int_0^1 \int_0^1e^{-x^2-y^2}dxdy\right)^{1/2}$$ 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, which makes the integra much easier. am i right? (im falling asleep too :P)