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One of my students asked me a question that has perplexed me. I would be grateful for help.

let me start at a fairly advanced stage of the standard solution, that is :-

##I^2=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}dydx## ...(

*a)*

Then we apply a conversion to polar coordinates to get the equation

##I^2=\int_{0}^{2\pi}\int_{}^{\infty}re^{-r^2}dr~d{\theta}## ......

__(b)__

1.

__Observation__

We know that the curve ##re^{-r^2}## denotes the Rayleigh Distribution Curve and that it lends itself to very easy integration. The inner integral above, i.e. ##\int_{0}^{\infty}re^{-r^2}dr##

gives us the area under the Rayleigh Distribution Curve from r=0 to r=infinity as a neat 1/2

Now if we execute the outer integral we get a solid of revolution that is kinda fountain shaped with its volume equal to ##I^2=\pi##

Now look at that ! The volume under our bell shaped solid turns out to be exactly equal to the volume of our 'fountain'-shaped solid both equal to ##\pi## !

2.

__Problem Statement__

Now consider only one half of the curve ##e^{-x^2}dx## which lies on the right hand side of the z axis. The area under the curve is ##I/2 = \int_{0}^{\infty}e^{-x^2}dx##

If we take this half curve as a polar curve and integrate or revolve it around the z-axis by ##2\pi## radians, we get a bell shaped solid that is identical in every way to the bell shaped solid we got from the integral

*(a)*above. We get its volume as ##I/2 * 2\pi## or ##I\pi##.

This too should be of the same volume as our fountain-shaped solid which is seen to be ##\pi##

So we get an equation ##I\pi = \pi##

So we get the area of our initial curve ##z=e^{-x^2}## as ##I=1## !

Where have I gone wrong ?

*(sorry. I have edited my post to correct the limits of integration for the rayleigh distribution curve)*