Integrate e^(-u^2) du: Polar Coords Explained

[Chronos000]

Homework Statement

I need to integrate e^(-u^2) du


I know I can write this as a double integral of exp(-x^2)dx . exp(-y^2)dy

I'm told I need to change this into polar coordinates but I'm unsure really what's going on

 

[funnyguy]

It may help if you describe what you are trying to accomplish. I don't follow how \int {e^{-{u^2}} du = \int\int e^{-{x^2}} e^{-{y^2}} dx dy. I think you are trying to say that it is the product of two integrals, \int {e^{-{u^2}} du = \int e^{-{x^2}} dx \int e^{-{y^2}} dyYou can't directly integrate e^{x^2}. I believe you can use a Maclaurin series to approximate a value. Look at the Maclaurin series for e^x and substitute x with x^2.

Assuming that your x and y are Cartesian, you can substitute their polar alternative. Just like on the unit circle, the x-horizontal axis component of a point is normally associated as the product of the magnitude of the point from the zero origin and the cosine of the angle, namely r \cos{\Theta}. Use this in your integral and you might get something that can be integrated, but without more about the problem, this may not be correct.

 

[RTW69]

Go to Wolframalpha.com and use their computational engine to integrate E^(-x^2) dx

Their solution is .5*Sqrt(Pi) erf(x) + Constant

 

[Chronos000]

the answer I am looking to get is SQRT(Pi).

The x and y were in cartesian yes. I have a part solution which states

"Change to circular polar coordinates (x, y) -> (rho,phi ) and evaluate the angular
integral (remember if you change variables correctly, the area of a ring should
enter as 2*Pi*r*dr)."

If r = (x,y) = (rho.cos(phi), rho.sin(phi) ), I am unsure of the limits or really what's going on. i don't see why you need cylindrical polars? (apart from the fact you can't do it in cartesian)