# Integration of Gaussian equation

1. Aug 24, 2004

### kekly

Hi,

Need help desperately!

I am trying to figure out the area under a gaussian cone by finding the integral of

2PIArEXP[-(r^2)/(2sigma^2)] dr

My supervisor thought it is

2PI [A sigma^2 EXP(-(r^2)/(2 sigma^2)] Between 0 and infinity

and he came up with the answer

2 PI A sigma^2

I hope you can understand that!

I don't think that his integration is right to get to the second step there. But I'm not sure why. I don't think sigma will come down like that. Please help me. I haven't done maths like this for a few years and I'm very rusty at it!

Thanks

Kek

2. Aug 24, 2004

### TenaliRaman

I think he has taken
2PIArEXP[-(r^2)/(2sigma^2)] dr

-- AI

3. Aug 24, 2004

### HallsofIvy

Staff Emeritus
To find $2\pi {A} \int_0^{\infinity}{r e^{\frac{-r^2}{2\sigma^2}}dr}$, let $u= \frac{r^2}{2\sigma^2}$. Then $du= \frac{r}{\sigma^2}$ so $rdr= \sigma^2du$. When r=0, u= 0, when r= infinity, u= infinity so the integral becomes $2\pi {A} \sigma\int_{o}^{\infinity}{e^{-u} du}$.
That IS $-2\pi {A} \sigma e^{-u}$ evaluated between u=0 and u= infinity:
$2\pi {A} \sigma$.

Last edited: Aug 24, 2004
4. Aug 24, 2004

### TenaliRaman

rdr = sigma^2 du
the final answer is 2*pi*A*sigma^2

-- AI

5. Aug 24, 2004

### kekly

I see what he has done now!

Thanks so much. You guys are lifesavers!

Kek

6. Aug 24, 2004

### kekly

Thanks for your help. I am still a little confused. As I said I haven't done integration like this for a long time!

I can't see how you substituted $rdr= \sigma^2du$ back in to get $2\pi {A} \sigma\int_{o}^{\infinity}{e^{-u} du}$. I can see that when r=0, u= 0, when r= infinity, u= infinity. That is fine but where does the sigma come from and where does the r go?

Thanks
kek

7. Aug 24, 2004

### TenaliRaman

kekly,
Look at the substitution hurkyl makes ....
u = r^2/(2*sigma^2)
find du/dr

-- AI

8. Aug 24, 2004

### kekly

I understand the substitution that is made and I can find

du/dr = r/sigma^2

That can then be rearranged to

rdr = sigma^2du

How then do I get to this!

$2\pi {A} \sigma\int_{o}^{\infinity}{e^{-u} du}$

Forgive me I think it is part of the substiution method that I don't understand. I haven't done it for many years.

9. Aug 24, 2004

### kekly

Thanks for all your hlep. I have been playing around with it and I understand it now!

It was not remembering how to use the substitution method that was causing me problems!

Thansk again for all your help!

kek

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