Answer
$$
\iint_{D} x y \sqrt{1+x^{2}+y^{2}} d A
$$
where $D$ is the portion of the disk $x^{2}+y^{2}\leq 1 $ that lies in the first quadrant.
The required integral is given by:
$$
\begin{aligned}
& \iint_{D} x y \sqrt{1+x^{2}+y^{2}} d A \approx 0.1609
\end{aligned}
$$
Work Step by Step
$$
\iint_{D} x y \sqrt{1+x^{2}+y^{2}} d A
$$
where $D$ is the portion of the disk $x^{2}+y^{2}\leq 1 $ that lies in the first quadrant.
The region $D$ can be described as:
$$ D=\left\{(x, y ) | \quad x \geq 0,\, y \geq 0 , \quad x^{2}+y^{2}\leq 1 \right\} $$
In polar coordinates, we have $x^{2}+y^{2}=r^{2} $, $ x=r \cos \theta ,\,\, y= r \sin \theta $
Thus the disk D is given by:
$$ D=\left\{(r, \theta) | 0 \leq r \leq 1, \quad 0 \leq \theta \leq \frac{\pi}{2} \right\} $$
and the required integral is given by:
$$
\begin{aligned}
& \iint_{D} x y \sqrt{1+x^{2}+y^{2}} d A=\\
&= \int_{0}^{\pi / 2} \int_{0}^{1}(r \cos \theta)(r \sin \theta) \sqrt{1+r^{2}} r d r d \theta \\
&=\int_{0}^{\pi / 2} \sin \theta \cos \theta d \theta \int_{0}^{1} r^{3} \sqrt{1+r^{2}} d r\\
&=\left[\frac{1}{2} \sin ^{2} \theta\right]_{0}^{\pi / 2} \int_{0}^{1} r^{3} \sqrt{1+r^{2}} d r \\
&=\frac{1}{2} \int_{0}^{1} r^{3} \sqrt{1+r^{2}} d r \\
&\approx 0.1609
\end{aligned}
$$
