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 Exercise [12.10] 
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Joined: 12 Nov 2008, 02:55
Posts: 12
Location: Japan
Post Exercise [12.10]
Sameed
Here is the problem. I can't type the question adequately so I will add it to the attached.
Thanks for any help.


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08 Jan 2009, 12:33

Joined: 07 May 2009, 16:45
Posts: 62
Post Re: Exercise [12.10]
timray60,
I did not see your solution to this exercise before I commenced my own. (I sorted the posts by subject, and because the "paper clip" sorts at the bottom I missed it)

Thanks for your explanation whydxdy is equivalent to dx \wedge dy.

I have a problem with your substitution of \theta limits of -\pi to \pi. I see how it is correct, because you have this ray or half line from the (0,0) to \infty that you are integrating over 2\pi to cover the whole plane. However, I don't see how you are getting there from your math. Could you look at my solution and tell me if you think I am wrong the way I do it?


31 Jul 2009, 17:03

Joined: 07 May 2009, 16:45
Posts: 62
Post Re: Exercise [12.10]
Correction, I missed this solution because it is in the Exercise discussion forum, not Solutions.


31 Jul 2009, 23:01

Joined: 22 May 2010, 14:58
Posts: 1
Post Re: Exercise [12.10]
Hi,

I do have a question to this exercise, more precisely is this correct?
dx\wedge dy with x = r cos(\theta) and y = r sin(\theta) follows: dx = dr cos(\theta)-r sin(\theta)d\theta and dy = dr sin(\theta)+r cos(\theta)d\theta
now inserting this into the wedge product:
(dr cos(\theta)-r sin(\theta)d\theta)\wedge(dr sin(\theta)+r cos(\theta)d\theta) =(cos(\theta) r cos(\theta)-(-r sin(\theta) sin(\theta))) dr\wedge d\theta = r dr d\theta
because dr and d\theta are orthogonal their wedge product is a "normal" product

thanks
vb


22 May 2010, 15:11
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