Convert to polar form and solve
$$\int^{1}_{0}\int_{0}^{\sqrt{2y-y^2}}(1-x^2-y^2)\text{ dx dy}$$
$x^2+y^2-2y=0$
$x^2+(y-1)^2=1$
$x=r\cos\theta$, $y=r\sin\theta+1$
$r^2=1$, $r=1$
$$\int^{\pi}_{0}\int^{1}_{0}(1-(r\cos\theta)^2-(r\sin\theta+1)^2) rdrd\theta$$ $$\int^{\pi}_{0}\int^{1}_{0}(-r^3-2r^2\sin\theta)drd\theta = -4/3-{\pi}/4$$
I cant figure out what I did wrong... answer should be 0.27396...
The problem is that $\theta$ doesn't go from $0$ to $\pi$. It goes from $\frac{3\pi}2$ to $2\pi$ instead, so that you get the lower right quarter of the circle centered at $(0,1)$ with radius $1$ (with $\theta$ going from $0$ to $\pi$, what you get is the top half of that circle). So, the answer is$$\frac23-\frac\pi8\approx0.273968.$$