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allsmiles
Group Title
Evaluate ∫∫ over D (cos(sqrt(x^2+y^2)) dA by changing polar coordinates where the disk is with center of origin and radius 2
 one year ago
 one year ago
allsmiles Group Title
Evaluate ∫∫ over D (cos(sqrt(x^2+y^2)) dA by changing polar coordinates where the disk is with center of origin and radius 2
 one year ago
 one year ago

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allsmiles Group TitleBest ResponseYou've already chosen the best response.0
I got up to ∫∫[0 to 2pi] [0 to 2] cos(sqrt(r)) r dr dt
 one year ago

allsmiles Group TitleBest ResponseYou've already chosen the best response.0
but I can't solve for it! How do integrate r*cos(sqrt(r))
 one year ago

myko Group TitleBest ResponseYou've already chosen the best response.1
\[\int\limits\int\limits_{D}=\int\limits_{0}^{2\pi} d\theta\ \int\limits_{0}^{2}r \cos rdr\]
 one year ago

allsmiles Group TitleBest ResponseYou've already chosen the best response.0
omg you are correct.
 one year ago

allsmiles Group TitleBest ResponseYou've already chosen the best response.0
spent literally 30 mins trying to see what I did wrong. That was quick thanks a lot man
 one year ago

myko Group TitleBest ResponseYou've already chosen the best response.1
by the way:int( rcos r) it's done by integration by parts
 one year ago

allsmiles Group TitleBest ResponseYou've already chosen the best response.0
can't I do substitution
 one year ago

allsmiles Group TitleBest ResponseYou've already chosen the best response.0
haha I just did and it didn't work, alright thanks again!
 one year ago

myko Group TitleBest ResponseYou've already chosen the best response.1
if you have problems look here, :) http://en.wikipedia.org/wiki/Integration_by_parts
 one year ago

myko Group TitleBest ResponseYou've already chosen the best response.1
there exacly you example
 one year ago

allsmiles Group TitleBest ResponseYou've already chosen the best response.0
Yeah I got it now, thanks a lot for your help!
 one year ago
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