What is the volume of rotation no2?

I have no idea how to do it but Wolfram|Alpha is pretty cluey.

IF I entered the right input in then maybe you get the right output.    295.331

http://www.wolframalpha.com/input/?i=y%3Dpi*integral_0.6%5E2+3.7%5E%281.2x%5E2%29

I'll give myself a tick but if Alan says it is wrong I will take it away again.  Does that sound fair?  

This is essentially asking to integrate k^(x²), where k is a constant.  There is no analytical solution to this in general unless the limits are -∞ to ∞ (or one of these two limits is replaced by 0).  For other limits a numerical solution can be obtained (but this requires knowing what the limits are).

Hi Alan and Olie96,

I just put this through Wolfram|Alpaha,

http://www.wolframalpha.com/input/?i=y%3D3.7%5E%280.6x%5E2%29

Now it is basically just a flatterned parabola shape. So can you not find an indefinite integral for it?

(a and b are constants of course)     

$$\int a^{bx^2} \;dx$$               

I suppose if you want a volume of revolution the relevant indefinite integral would be

$$\\\pi\int (a^{bx^2})^2\;dx\\\\
=\pi\int a^{2bx^2}\;dx$$

Hi Melody.  No, no-one knows how to generate an analytical solution to this indefinite integral as far as I'm aware.  (If it were known, I strongly suspect Mathematica (Wolfram Alpha) would be able to do it.)

It's sort of like trying to integrate a Gaussian (e^{-x^2}) only worse!  Actually, it was the Gaussian I had in mind with my first answer.  The difference here is that the integral from any finite limit to ∞ is likely to be infinite (in the case of the Gaussian it is finite).

Okay, thanks Alan.

opps sorry, totally forgot the limits, so they are 0.6 and 2.

Yes, I get 295.331 as well.

Did you do it the same way Alan.  I assume that you did.   

I used a different piece of software (Mathcad), if that's what you mean, but of course the integral was:

$$\pi \int_{0.6}^23.7^{2\times0.6\times x^2}dx$$ 

Okay, you did it the same.  Great.  :)