Math Is Fun Forum

Identity

Member

Registered: 2007-04-18

Posts: 934

Double integral

Can someone please show me why the following is possible?

If

then

I don't understand this last step. In general are you allowed to multiply two integrals together to get a double integral?

Also, another question. The next step it says this is equivalent to

Is this a mistake? I thought you would integrate

from 0 to infinity, and integrate

from 0 to

?

Last edited by Identity (2009-08-29 23:17:07)

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Double integral

Hi identity;

Thought you might find this interesting:

http://betterexplained.com/articles/a-c … plication/

For this particular type problem there is a formula:

Last edited by bobbym (2009-08-30 01:34:03)

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Double integral

The theorem is known as Fubini's theorem.  You may ignore the hypotheses, they are all satisfied by your integral.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Double integral

For your 2nd question, if that's the order they have it, then yes it is a mistake.  Note that you must integrate over theta first.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

rzaidan

Member

Registered: 2009-08-13

Posts: 59

Re: Double integral

Hi Identity
I think that the proff is as follows:
If we can seperate h(x,y) as  h(x,y)=f(x)g(y) and x and y are independent variables  then
∫∫h(x,y)dxdy=∫∫f(x)g(y)dxdy=∫(g(y)(∫f(x)dx)) dy   by considering g(y) as a constant w.r.t  x
Also ∫f(x)dx can be considered as a constant w.r.t y  so   
∫∫f(x)g(y)dxdy=∫(g(y)(∫f(x)dx) )dy ∫f(x)dx)(∫g(y) dy) as requiered
I hope that I am right
Best regards
Riad Zaidan

Identity

Member

Registered: 2007-04-18

Posts: 934

Re: Double integral

Thanks everyone, in the end I think rzaidan's proof was easiest for me to understand.