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I realize i need to use polar coordinates. First thing i do is calculate $z=1-y$ and then include that in $z=1-x^2-y^2$ that turns out be circle centered at $ (0,\frac 12) $. I know that $ \theta$ limits of integration should be from 0 to $\pi$ but what should be limits of integration for $r$. Also then the double integral should be $$ \int_0^\pi\int_?^?[(1-r^2)-(1-rsin\theta)]rdrd\theta $$ Any help is appreciated.

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We know $z=1-x^2-y^2=1-y$, that is $x^2+y^2=y$. Then in polar coordinates we have $$r^2=r\sin \theta.$$ So these surfaces intersect for $r=\sin \theta.$ The other limit is obviously $r=0.$

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  • $\begingroup$ Is my double integral correct? Should it be like i wrote in post? $\endgroup$ – D M Jan 27 '17 at 14:21
  • $\begingroup$ yes, it is correct. $\endgroup$ – Domates Jan 27 '17 at 15:59

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