Math Is Fun Forum

Talvon

Member

Registered: 2006-11-15

Posts: 16

Using Polar Coordinates to evaluate a double integral

I have this problem to do, and I've tried subbing in x=rcos and y=rsin, but I can't get it to do anything Any tips on how to get the ball rolling?

The problem is:

Use polar coordinates to evaluate
infinity
   ∫∫exp(-x²-y²)dydx
    0
and hence show that
  infinity
     ∫exp(-x²)dx=(√pi)/2
    0

Thanks

Last edited by Talvon (2007-03-03 12:25:37)

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Using Polar Coordinates to evaluate a double integral

The formula for converting from (x,y) to polar co-ordinates for double integration is given here:
http://ltcconline.net/greenl/courses/20 … ration.htm

So if you substitute r and θ into your double integral, the integrand becomes rexp(−r[sup]2[/sup]), which can be easily integrated with respect to r.

Last edited by JaneFairfax (2007-03-03 13:00:34)

Talvon

Member

Registered: 2006-11-15

Posts: 16

Re: Using Polar Coordinates to evaluate a double integral

I'm sorry but I don't get how -x² can equal -y² (From the page in your forum, which will be useful for the 2nd part of the question Half the range of the integral = Half the solution <_<) (http://z8.invisionfree.com/DYK/index.php?showtopic=136)

I can reproduce it to the stage of rexp(-r²), and I am having trouble getting to the next part (-1/2exp(-r²)). I have tried integration by parts, using u=exp(-r²) and dv/dr=r, but it's coming out nowhere near what it should be. What should I use? Is this the correct method?

Also, would I be correct in writing:

π      ∞
∫ dθ ∫rexp(-r²) dr?
0     0

Thanks

lightning

Real Member

Registered: 2007-02-26

Posts: 2,060

Re: Using Polar Coordinates to evaluate a double integral

huh

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JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Using Polar Coordinates to evaluate a double integral

I can reproduce it to the stage of rexp(-r²), and I am having trouble getting to the next part (-1/2exp(-r²)).

If you differentiate (−1/2exp(−r[sup]2[/sup])) with respect to r, you get rexp(−r[sup]2[/sup]), don’t you? Therefore, if you integrate rexp(−r[sup]2[/sup]) with respect to r, you should get (−1/2exp(−r[sup]2[/sup])). (Plus an arbitrary constant – but as you’re working with a definite integral, you don’t need to bother with arbitrary constants.) It’s that simple.

Also you’re wondering why

Why? Because it’s a definite integral, so the x and the y are just dummy variables. Indeed:

Last edited by JaneFairfax (2007-03-05 12:33:10)

Talvon

Member

Registered: 2006-11-15

Posts: 16

Re: Using Polar Coordinates to evaluate a double integral

Excellent, I got it done

Thanks a lot