# Area under the Gaussian Curve

I have been teaching undergrad students informally, and one of the math problems that I have always enjoyed introducing them to is how to compute the area under a gaussian curve, or to keep it simple, the area under the curve $z=e^{-x^2}$

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)

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$I^2=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}dydx$ ...............(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)
All these have dimensions of area and not volume!

Last edited by a moderator:
phyzguy
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 calculation of the volume is simply not correct. Imagine replacing your half curve with the curve z=1 when x>1 and x<2, and z=0 elsewhere. The integral under this curve is 1. Now revolve it around the z-axis as before. By your logic, this cylinder would have a volume of 2π. In fact it's volume is 3π. Now move the curve out so that z=1 when x>9 and x<10, and z=0 elsewhere. The area under the curve is still 1, so your logic would say the volume of this curve revolved around the z-axis would still be 2π. In fact this volume is 19π.

All these have dimensions of area and not volume!
Thanks. Doesnt the first integral give area and the second integral - volume ? The expression dydx gives the infinitesimal footprint of each riemann column on the xy plane. The expression $e^-{x^2}$ gives the height of the column along the z axis. That should give us volume.

e−x2e−x2e^-{x^2}
that quantity is dimensionless.

tnich
Homework Helper
@phyzguy and @Let'sthink have both pointed out that you have not correctly written the integral for the volume of revolution. It should be $\int_{0}^{2\pi}\int_{0}^{\infty}xe^{-x^2}dx~d{\theta}$.
to keep it simple, the area under the curve $z=e^{-x^2}$
I think this next point is easier to see without the above simplification. If you try it with $z=\frac 1 {\sqrt{2\pi}}e^{-\frac {x^2} 2}$, you will see that the volume of revolution $\int_{0}^{2\pi}\int_{0}^{\infty}\frac 1 {\sqrt{2\pi}}xe^{-\frac {x^2} 2}dx~d{\theta}$ differs from the volume you obtain from $I^2=\int_{0}^{2\pi}\int_{0}^{\infty}\frac 1 {2\pi}e^{-\frac {r^2} 2}r~dr~d{\theta}$ by a factor of $\sqrt{2\pi}$. So your assumption that revolving half the normal pdf around the z-axis produces the bivariate normal pdf is not correct.

This calculation of the volume is simply not correct. Imagine replacing your half curve with the curve z=1 when x>1 and x<2, and z=0 elsewhere. The integral under this curve is 1. Now revolve it around the z-axis as before. By your logic, this cylinder would have a volume of 2π. In fact it's volume is 3π. Now move the curve out so that z=1 when x>9 and x<10, and z=0 elsewhere. The area under the curve is still 1, so your logic would say the volume of this curve revolved around the z-axis would still be 2π. In fact this volume is 19π.

Thanks.
Let me make that "revolution" around the z-axis more specific:-

$I^2 = \int_0^{2\pi}\int_0^\infty~e^{-x^2}dx~d\theta$

Orodruin
Staff Emeritus
Homework Helper
Gold Member
All these have dimensions of area and not volume!
There are no dimensions in this problem. If $r$ was a dimensionful quantity you would need a dimensionful constant to make the exponent dimensionless (you cannot exponentiate a dimensionful number!). You therefore should not expect to be able to apply dimensional analysis without introducing said constant. Even then, you can have ”volume” integrals where one of the dimensions does not have units of length. Consider the integral of velocity (dimension L/T) over time (dimension T). The result of this integration is distance (L), but you can still consider it the ”area” under the velocity curve. It is in that meaning ”volume” is being used here.

There are no dimensions in this problem. If $r$ was a dimensionful quantity you would need a dimensionful constant to make the exponent dimensionless (you cannot exponentiate a dimensionful number!). You therefore should not expect to be able to apply dimensional analysis without introducing said constant. Even then, you can have ”volume” integrals where one of the dimensions does not have units of length. Consider the integral of velocity (dimension L/T) over time (dimension T). The result of this integration is distance (L), but you can still consider it the ”area” under the velocity curve. It is in that meaning ”volume” is being used here.
Thanks for bringing out the dimension issue clearly.

phyzguy
Thanks.
Let me make that "revolution" around the z-axis more specific:-

$I^2 = \int_0^{2\pi}\int_0^\infty~e^{-x^2}dx~d\theta$
This is where you are making your mistake. This quantity is not the volume of revolving the curve $z = e^{-x^2}$ about the z-axis. The volume element in this case is $x \, dx \, d\theta$, not $dx \, d\theta$. When you add that x, it all works out.

@phyzguy and @Let'sthink have both pointed out that you have not correctly written the integral for the volume of revolution. It should be $\int_{0}^{2\pi}\int_{0}^{\infty}xe^{-x^2}dx~d{\theta}$.
I think this next point is easier to see without the above simplification. If you try it with $z=\frac 1 {\sqrt{2\pi}}e^{-\frac {x^2} 2}$, you will see that the volume of revolution $\int_{0}^{2\pi}\int_{0}^{\infty}\frac 1 {\sqrt{2\pi}}xe^{-\frac {x^2} 2}dx~d{\theta}$ differs from the volume you obtain from $I^2=\int_{0}^{2\pi}\int_{0}^{\infty}\frac 1 {2\pi}e^{-\frac {r^2} 2}r~dr~d{\theta}$ by a factor of $\sqrt{2\pi}$. So your assumption that revolving half the normal pdf around the z-axis produces the bivariate normal pdf is not correct.
Forget for a moment that we are dealing with a PDF. We are simply trying to compute the area under a curve.
There is no way of integrating exp(-x^2) so a smart way was established by
This is where you are making your mistake. This quantity is not the volume of revolving the curve $z = e^{-x^2}$ about the z-axis. The volume element in this case is $x \, dx \, d\theta$, not $dx \, d\theta$. When you add that x, it all works out.
Got it. Thanks.I must have been half asleep to have got perplexed. Developing a solid of revolution from a polar curve requires shell integration. The solid has to be formed out of myriad concentric shells of thickness dr. Revolution is performed in steps of dtheta yielding wedge shaped radial sections. . However the circumferential length of each element is r*dtheta.

Thanks folks.

Mark44
Mentor
The expression $e^-{x^2}$
@Ganesh Mahadevan, you have made great strides since your earlier post with all the images, but here's a tip: If an exponent is a single character, you don't need the braces. In the exponential above, the minus sign is treated as the exponent, and the part in braces is treated as ordinary text.

To fix this, use $e^{-x^2}$ (raw text is $e^{-x^2}$)
e−x2e−x2e^-{x^2}
that quantity is dimensionless.
@Let'sthink, you really mangled up what Ganesh wrote, making it essentially unreadable. Be more careful when you copy and paste things written using LaTeX. A better approach is to highlight the text you want to copy. On the right-click menu one option is "Show Math As" --> "TeX Commands". Then copy the TeX stuff and paste it, surrounding each end with a pair of # characters

@Let'sthink, you really mangled up what Ganesh wrote, making it essentially unreadable. Be more careful when you copy and paste things written using LaTeX. A better approach is to highlight the text you want to copy. On the right-click menu one option is "Show Math As" --> "TeX Commands". Then copy the TeX stuff and paste it, surrounding each end with a pair of # characters
Ok, I note it for future what I mean to say is that sin cosine and exponential functions even log functions are to be treated as dimensionless.

Mark44
Mentor
All these have dimensions of area and not volume!
Ok, I note it for future what I mean to say is that sin cosine and exponential functions even log functions are to be treated as dimensionless.
A definite integral is just a number, so no dimensions are involved. If we're doing an applied problem, which was not the case here, then we can attach dimensions to the number we get from the integral.