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\begin{align}
\!\!\!\!\!{\cal A}&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
\Theta\pars{4 - x^{2} - y^{2}}\Theta\pars{6x - x^{2} - y^{2}}\,\dd x\,\dd y
\\[3mm]&=\int_{0}^{\infty}\dd r\,r\int_{0}^{2\pi}\Theta\pars{4 - r^{2}}
\Theta\pars{6r\cos\pars{\theta} - r^{2}}
\\[3mm]&=\int_{0}^{\infty}\dd r\,r\int_{0}^{2\pi}\dd\theta\,\Theta\pars{2 - r}
\Theta\pars{6\cos\pars{\theta} - r}
=\int_{0}^{2}\dd r\,r\int_{0}^{2\pi}\dd\theta\,
\Theta\pars{\cos\pars{\theta} - {r \over 6}}\tag{1}
\end{align}
where $\Theta\pars{z}$ is the Heaviside Step Function.
\begin{align}
&\left.\int_{0}^{2\pi}
\Theta\pars{\cos\pars{\theta} - {r \over 6}}\,\dd\theta\right\vert_{0\ <\ r\ <\ 2}
=\left. 2\int_{0}^{\pi}
\Theta\pars{-\cos\pars{\theta} - {r \over 6}}\,\dd\theta\right\vert_{0\ <\ r\ <\ 2}
\\[3mm]&=\left. 2\int_{-\pi/2}^{\pi/2}
\Theta\pars{\sin\pars{\theta} - {r \over 6}}\,\dd\theta\right\vert_{0\ <\ r\ <\ 2}
=2\left. \int_{\arcsin\pars{r/6}}^{\pi/2}\,\dd\theta\right\vert_{0\ <\ r\ <\ 2}
\end{align}
$$
\left.\int_{0}^{2\pi}
\Theta\pars{\cos\pars{\theta} - {r \over 6}}\,\dd\theta\right\vert_{0\ <\ r\ <\ 2}
=\left.\pi - 2\arcsin\pars{r \over 6}\right\vert_{0\ <\ r\ <\ 2}
\tag{2}
$$
In replacing $\pars{2}$ in $\pars{1}$, we find:
\begin{align}
{\cal A}&=\int_{0}^{2}\bracks{\pi - 2\arcsin\pars{r \over 6}}r\,\dd r
=2\pi - 2\int_{0}^{2}\arcsin\pars{r \over 6}r\,\dd r
\end{align}
The last integral in the right member is an elementary one: It can be easily solved after integration by parts and the final result is
$$
\color{#00f}{\large{\cal A} = 2\pi - 4\root{2} + 14\arcsin\pars{1 \over 3}}
\approx 5.3841
$$
