= Solution
Substitute $x=r\cos\phi$, $y=r\sin\phi$ and include the <Jacobian determinant> factor $r$. The integrand times the area element becomes
$$
\big(\cos\phi+\sin\phi+r\cos\phi\sin\phi\big)dr\,d\phi.
$$
Both terms containing one factor $\sin\phi$ are odd in $\phi$, while the radial bounds and angular interval are even. They therefore integrate to zero. The remaining <double integral> is
$$
\begin{aligned}
I&=\int_{-\pi/3}^{\pi/3}\cos\phi\left(1+\cos\phi-\frac32\right)d\phi\\
&=\int_{-\pi/3}^{\pi/3}\left(\cos^2\phi-\frac12\cos\phi\right)d\phi
=\boxed{\frac\pi3-\frac{\sqrt3}{4}.}
\end{aligned}
$$
The domain stays away from $r=0$, so the denominator of the original integrand causes no singularity in this region.
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