Answer
The total charge on the disk is $ \frac{2 \pi}{3} \mathrm{C} $.
Work Step by Step
The total charge $Q$ is given by
$$
\begin{aligned} Q &=\iint_{D} \sigma(x, y) d A \\
&=\iint_{D} \sqrt{x^{2}+y^{2}} d A \\
&=\int_{0}^{2 \pi} \int_{0}^{1} \sqrt{r^{2}} r d r d \theta \\
&=\int_{0}^{2 \pi} d \theta \int_{0}^{1} r^{2} d r\\
&=[\theta]_{0}^{2 \pi}\left[\frac{1}{3} r^{3}\right]_{0}^{1}\\
&=2 \pi \cdot \frac{1}{3} \\
&=\frac{2 \pi}{3} \mathrm{C} \end{aligned}
$$
Thus, the total charge is $ \frac{2 \pi}{3} \mathrm{C} $.
