#6**+8 **

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?

Melody
Oct 7, 2014

#1**+5 **

This is essentially asking to integrate k^(x^{2}), 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).

Alan
Oct 7, 2014

#2**0 **

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$$

Melody
Oct 7, 2014

#3**+5 **

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).

Alan
Oct 7, 2014

#6**+8 **

Best Answer

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?

Melody
Oct 7, 2014