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Set up the Surface area Double integration f(x,y)=cos(x^2+y^2) R={(x,y), x^2+y^2<=pi/2}
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|dw:1280356924796:dw|
\[\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi/2}rcos(r^2)drd \theta\]
correct?
yes
actually it'd be pi/4 right? the diameter is pi/4
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pi/2
|dw:1280357053887:dw|
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