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Exercise [12.10]

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Exercise [12.10]
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timray60 Joined: 12 Nov 2008, 02:55 Posts: 12 Location: Japan	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. Attachments: File comment: Problem 12.10 Excercise 12.10.doc [22.5 KiB] Downloaded 219 times
08 Jan 2009, 12:33

DimBulb Joined: 07 May 2009, 16:45 Posts: 62	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 why 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 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

DimBulb Joined: 07 May 2009, 16:45 Posts: 62	Re: Exercise [12.10] Correction, I missed this solution because it is in the Exercise discussion forum, not Solutions.
31 Jul 2009, 23:01

vampirbiber Joined: 22 May 2010, 14:58 Posts: 1	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 and $d\theta$ are orthogonal their wedge product is a "normal" product thanks vb
22 May 2010, 15:11

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