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For VII.4.1 I get the bottom side of the rectangle tends to pi and the sides tend to 0. But when I try and evaluate the top side by letting <math>z (t) = 2at - a + \sqrt b i</math>, my integrand becomes a mess: <math>2a/((4a^2 t^2-4a^2 t+a^2-b+1)+i(4a\sqrt b t-2a\sqrt b))</math>. I don't see how that simplifies to what the integrand in the problem gives. Just keep messing around with the algebra until something happens? Or did I do something wrong
 
For VII.4.1 I get the bottom side of the rectangle tends to pi and the sides tend to 0. But when I try and evaluate the top side by letting <math>z (t) = 2at - a + \sqrt b i</math>, my integrand becomes a mess: <math>2a/((4a^2 t^2-4a^2 t+a^2-b+1)+i(4a\sqrt b t-2a\sqrt b))</math>. I don't see how that simplifies to what the integrand in the problem gives. Just keep messing around with the algebra until something happens? Or did I do something wrong
 
Edit: Never mind, I figured it out by changing the parametrization so <math>-a<=t<=a</math>--[[User:Aata|Aata]]
 
Edit: Never mind, I figured it out by changing the parametrization so <math>-a<=t<=a</math>--[[User:Aata|Aata]]
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Just work with the algebra and then take the real part and you should get it
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--[[User:Kfernan|Kfernan]] 20:11, 29 October 2009 (UTC)

Revision as of 15:11, 29 October 2009


Homework 7

HWK 7 problems

For VII.4.1 I get the bottom side of the rectangle tends to pi and the sides tend to 0. But when I try and evaluate the top side by letting $ z (t) = 2at - a + \sqrt b i $, my integrand becomes a mess: $ 2a/((4a^2 t^2-4a^2 t+a^2-b+1)+i(4a\sqrt b t-2a\sqrt b)) $. I don't see how that simplifies to what the integrand in the problem gives. Just keep messing around with the algebra until something happens? Or did I do something wrong Edit: Never mind, I figured it out by changing the parametrization so $ -a<=t<=a $--Aata


Just work with the algebra and then take the real part and you should get it

--Kfernan 20:11, 29 October 2009 (UTC)

Alumni Liaison

Recent Math PhD now doing a post-doctorate at UC Riverside.

Kuei-Nuan Lin