allsmiles
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



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allsmiles
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I got up to ∫∫[0 to 2pi] [0 to 2] cos(sqrt(r)) r dr dt

allsmiles
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but I can't solve for it! How do integrate r*cos(sqrt(r))

myko
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\[\int\limits\int\limits_{D}=\int\limits_{0}^{2\pi} d\theta\ \int\limits_{0}^{2}r \cos rdr\]

allsmiles
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omg you are correct.

allsmiles
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spent literally 30 mins trying to see what I did wrong. That was quick thanks a lot man

myko
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yw

myko
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by the way:int( rcos r) it's done by integration by parts

allsmiles
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can't I do substitution

myko
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you can try

allsmiles
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haha I just did and it didn't work, alright thanks again!


myko
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there exacly you example

myko
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your*

allsmiles
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Yeah I got it now, thanks a lot for your help!