Area under Normal Distribution Curve: I^2 = INT -inf to + inf exp(-x^2)dx*INT -inf to

[gm_navy]

I am trying to teach this to a nephew and am getting confused myself. I would grateful for help.

I am familiar with the procedure. I consider two curves standing upright on the xyz coordinate system, one stretched along the x axis and one along the y axis. I is the area to be found,
I^2 = INT -inf to + inf exp(-x^2)dx*INT -inf to + inf exp(-y^2)dy = INT -inf to + inf [exp^ (x^2+y^2)*dydx] which gives us the volume of the resultant 3D bell shape.

After this we do this clever hop over to polar coordinates and we get I^2 =INT 0 to 2pi {Int 0 to inf rexp(-r^2)*dr]*dTheta.

OK so here is my doubt. Int 0 to inf rexp(-r^2)*dr] is actually the Rayleigh Distribution Curve, and the area under it computed from 0 to inf simply 1/2.

Now if we integrate it with respect to Theta from 0 to 2pi, we get a fountain shaped solid of revolution of volume pi.

I know the Math says so, but I find it difficult relating the volume of the bell shaped solid of revolution with the volume of that fountain shaped solid we get from the Rayleigh curve.

Also doesnt this show that half a normal curve (from 0 to inf) has the same area under it as the rayleigh distribution curve from 0 to inf ?

I know it cannot be so, but where am I going wrong in my reasoning ?

 

[Dr.Peterson]

I am trying to teach this to a nephew and am getting confused myself. I would grateful for help.

I am familiar with the procedure. I consider two curves standing upright on the xyz coordinate system, one stretched along the x axis and one along the y axis. I is the area to be found,
I^2 = INT -inf to + inf exp(-x^2)dx*INT -inf to + inf exp(-y^2)dy = INT -inf to + inf [exp^ (x^2+y^2)*dydx] which gives us the volume of the resultant 3D bell shape.

After this we do this clever hop over to polar coordinates and we get I^2 =INT 0 to 2pi {Int 0 to inf rexp(-r^2)*dr]*dTheta.

OK so here is my doubt. Int 0 to inf rexp(-r^2)*dr] is actually the Rayleigh Distribution Curve, and the area under it computed from 0 to inf simply 1/2.

Now if we integrate it with respect to Theta from 0 to 2pi, we get a fountain shaped solid of revolution of volume pi.

I know the Math says so, but I find it difficult relating the volume of the bell shaped solid of revolution with the volume of that fountain shaped solid we get from the Rayleigh curve.

Also doesnt this show that half a normal curve (from 0 to inf) has the same area under it as the rayleigh distribution curve from 0 to inf ?

I know it cannot be so, but where am I going wrong in my reasoning ?

You didn't quite state the problem you are trying to solve, but I think it is INT -inf to + inf exp(-x^2)dx. (This is not, strictly speaking, the normal distribution, but an unnormalized form of it, the Gaussian integral. The work you describe is what is shown on this page (under Computation), right?

You seem to be thinking that in doing the integration in polar form, the inner integral represents an area that is part of your problem. It doesn't. It is just an integral that arises in the process of evaluating the volume you are calculating. There is no "fountain shaped solid of revolution" whose volume is being found. Volumes in polar form don't work that way.

 

[HallsofIvy]

So you are actually using cylindrical coordinates with \(\displaystyle z= e^{-r^2}\) for all \(\displaystyle \theta\). The "differential of area" in polar coordinates is \(\displaystyle r drd\theta\) so the volume you want is \(\displaystyle \int_{\theta= 0}^{2\pi}\int_{r= 0}^\infty e^{-r^2} rdrd\theta\). The integral is simply \(\displaystyle 2\pi \int_0^\infty e^{-r^2}rdr\). Because of that "r" in the integral, it is much easier. Let \(\displaystyle u= r^2\) so \(\displaystyle du= 2r dr\), \(\displaystyle r dr= \frac{1}{2}du[\) and this becomes \(\displaystyle \pi\int_0^\infty e^{-u}du= \pi\left[-e^{-u}\right]_0^\infty= \pi\).