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Double Integration in Polar and its Applications
#3
For the conversion I was going off an example from class that I thought was similar, though your method makes more sense. Still a little bit confused as to what you would use for the u-sub (or v, in your case), but I guess that won't have any relevance since it's wrong to begin with. Also confused as to why the theta integral only goes from 0 to pi/2. :\

Was able to get 32/3 as the answer, ha. Looks like I'm missing a half somewhere and I think it might come from a trig property, though I'm not sure.

Subbed 1+sin^2(theta) instead of 1 - sin^2(theta) for cos^2(theta). Got 16/3.

I'll go ahead and give number 1 another try and take it slow to see if I can actually get to 54 this time. I got the polynomial but I might be simplifying wrong.

Got 54 after giving it another go. Blah.

I'll give number 2 a go without converting.

Got 16/5. Guess I was too anxious to try converting to polar that I made it harder than it should have been.


I'll report shortly.
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Double Integration in Polar and its Applications - by Panacea - 2012-03-27, 03:45 PM

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