= Solution
In <polar coordinates>, the circle is $r=2$, the line $y=\sqrt3x$ is $\theta=\pi/3$, and $y=1$ is $r\sin\theta=1$. The inequalities select
$$
\boxed{
\frac\pi6<\theta<\frac\pi3,\qquad
\csc\theta<r<2}.
$$
The surface height is
$$
z=\frac{xy}{x^2+y^2}
=\sin\theta\cos\theta.
$$
Thus the required volume is
$$
\begin{aligned}
V
&=\int_{\pi/6}^{\pi/3}
\int_{\csc\theta}^{2}
\sin\theta\cos\theta\,r\,dr\,d\theta\\
&=\int_{\pi/6}^{\pi/3}
\left(2\sin\theta\cos\theta-\frac12\cot\theta\right)d\theta\\
&=\boxed{\frac12-\frac14\log3}.
\end{aligned}
$$
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