coordinate transformation double integral

nha · 2010-10-09 23:31:07

Hey everyone, I am stuck on this question, I just can't get anymore done. Any help would be great.

Evaluate

by making the coordinate transformation and .

My working so far:

So:

But I don't get how to go from here, with the limits of one of the integrals having terminals with u and v in it.

If anyone can help that would be fantastic!

Thanks

bobbym · 2010-10-09 23:45:23

Hi nha;

Is there some reason why you are not algebraically simplifying the inner integrand?

nha · 2010-10-09 23:53:24

bobbym wrote:

Hi nha;
Is there some reason why you are not algebraically simplifying the inner integrand?

Hi bobbym, yes I know it simplifies down to a rather simple function but I am confused about the terminals uv and 1/uv after I integrate.

Last edited by nha (2010-10-09 23:54:45)

bobbym · 2010-10-09 23:57:27

Hi;

What I am saying is that the inner integral can be done by parts or by a table lookup. The intervals of integration having variables is no problem at all. Is that what you mean by terminals the intervals?

nha · 2010-10-10 00:03:12

bobbym wrote:

Hi;
What I am saying is that the inner integral can be done by parts or by a table lookup. The intervals of integration having variables is no problem at all. Is that what you mean by terminals the intervals?

If it is no trouble, could you show me some working out for what you mean, I don't understand. What I mean is how do I integrate the definite integral with terminals uv and 1/uv with respect to either du or dv?

bobbym · 2010-10-10 00:10:04

Hi nha;

To show you what the answer to the inner integral looks like:

Before we work on getting this answer. I am not sure about your substitutions.
We might be working on the wrong integrals!

nha · 2010-10-10 00:40:27

bobbym wrote:

Hi nha;
To show you what the answer to the inner integral looks like:
Before we work on getting this answer. I am not sure about your substitutions.
We might be working on the wrong integrals!

I think my substitutions were wrong, and I think the double integral is:

as I forgot to look at the domains of x and y and change u and v after considering the domains. I think this is right.

Last edited by nha (2010-10-10 00:41:34)

bobbym · 2010-10-10 00:49:30

Assuming that everything is okay up to here do you want this integral done.

Something is still wrong here you have 2 integrals with different variables but only one variable in the integrand. Are you sure about your subs?

nha · 2010-10-10 01:06:03

bobbym wrote:

Assuming that everything is okay up to here do you want this integral done.
Something is still wrong here you have 2 integrals with different variables but only one variable in the integrand. Are you sure about your subs?

I think this is right:

nha · 2010-10-10 01:09:36

The limits of integration, in terms of x and y are: y= 1, y= 2, and x= 1/y, x= y.

Since x= u/v and y= uv, those become uv= 1, uv= 2, and u/v= 1/uv, u/v= uv. The first two give v= 1/u and and v= 2/u while the last two reduce to u^2=1 and v^2=1.

Drawing those lines on a uv- graph, you can see that u= 1 and v= 1 intersect at (1, 1) on the graph of v= 1/u and intersect the graph of y= 2/v at u= 1, v= 2 and u= 2, v= 1. Essentially, then, the region you want to integrate over, in the uv-plane, is bounded by u= 1, v= 1, and v= 2/u. If u ranges from 1 to 2 then, for each u, v ranges from 1 to 2/u.

bobbym · 2010-10-10 01:10:19

That is not correct you did not do the first integration right. You integrated ok but your putting in of the limits of integration is wrong.

nha · 2010-10-10 01:13:58

bobbym wrote:

That is not correct you did not do the first integration right. You integrated ok but your putting in of the limits of integration is wrong.

I don't see where I went wrong.

bobbym · 2010-10-10 01:19:58

nha · 2010-10-10 01:23:14

bobbym wrote:

Oh yeah, but I think you mean

as the u's cancel out on the first one. But thanks again bobbym.

bobbym · 2010-10-10 01:27:38

Yes, that is what I mean, sorry a typo. I adjusted the above post.
You will now get a different answer for the next integration.

You have an error in your integration in post #9.
You should get 2e^2 - 4e as the answer instead of -2e

nha · 2010-10-10 01:34:18

bobbym wrote:

Yes, that is what I mean, sorry a typo. I adjusted the above post.
You will now get a different answer for the next integration.
You should get 2e^2 - 4e as the answer instead of -2e

Yes, thanks again bobbym.

bobbym · 2010-10-10 08:06:47

Hi nha;

It was fun to work on, hope you thought so too.

Math Is Fun Forum

#1 2010-10-09 23:31:07

coordinate transformation double integral

#2 2010-10-09 23:45:23

Re: coordinate transformation double integral

#3 2010-10-09 23:53:24

Re: coordinate transformation double integral

#4 2010-10-09 23:57:27

Re: coordinate transformation double integral

#5 2010-10-10 00:03:12

Re: coordinate transformation double integral

#6 2010-10-10 00:10:04

Re: coordinate transformation double integral

#7 2010-10-10 00:40:27

Re: coordinate transformation double integral

#8 2010-10-10 00:49:30

Re: coordinate transformation double integral

#9 2010-10-10 01:06:03

Re: coordinate transformation double integral

#10 2010-10-10 01:09:36

Re: coordinate transformation double integral

#11 2010-10-10 01:10:19

Re: coordinate transformation double integral

#12 2010-10-10 01:13:58

Re: coordinate transformation double integral

#13 2010-10-10 01:19:58

Re: coordinate transformation double integral

#14 2010-10-10 01:23:14

Re: coordinate transformation double integral

#15 2010-10-10 01:27:38

Re: coordinate transformation double integral

#16 2010-10-10 01:34:18

Re: coordinate transformation double integral

#17 2010-10-10 08:06:47

Re: coordinate transformation double integral

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