The given double integral can be expressed as:

$\int_1^3 \int_0^{2\pi} \sqrt{2r} \, dr \, d\theta$

Step-by-step Solution:

1. Separate the integral: Since the limits for $r$ and $\theta$ are independent, we can separate the integral as: $\int_1^3 \int_0^{2\pi} \sqrt{2r} \, dr \, d\theta = \left( \int_0^{2\pi} 1 \, d\theta \right) \left( \int_1^3 \sqrt{2r} \, dr \right)$

2. Evaluate the $\theta$-integral: $\int_0^{2\pi} 1 \, d\theta = \theta \Big|_0^{2\pi} = 2\pi$

3. Simplify the $r$-integral: Rewrite $\sqrt{2r}$ as $\sqrt{2} \cdot \sqrt{r}$, a constant factor can be taken out: $\int_1^3 \sqrt{2r} \, dr = \sqrt{2} \int_1^3 \sqrt{r} \, dr$

4. Use the power rule for integration: Recall $\int r^{n} \, dr = \frac{r^{n+1}}{n+1}$: [ \int_1^3 \sqrt{r} , dr = \int_1^3 r^{1/2} , dr = \left[ \frac{r^{3/2}}{3/2} \right]_1^3 = \frac{2}{3} \left[ r^{3/2} \right]_1^3 ]

5. Evaluate the bounds: Substitute $r = 3$ and $r = 1$: $\frac{2}{3} \left[ 3^{3/2} - 1^{3/2} \right] = \frac{2}{3} \left( \sqrt{27} - 1 \right) = \frac{2}{3} \left( 3\sqrt{3} - 1 \right)$

6. Combine the results: Multiply the results of the $\theta$-integral and the $r$-integral: $\int_1^3 \int_0^{2\pi} \sqrt{2r} \, dr \, d\theta = 2\pi \cdot \sqrt{2} \cdot \frac{2}{3} \left( 3\sqrt{3} - 1 \right)$

7. Simplify: $\frac{4\pi \sqrt{2}}{3} \left( 3\sqrt{3} - 1 \right)$

Final Answer:

$\boxed{\frac{4\pi \sqrt{2}}{3} \left( 3\sqrt{3} - 1 \right)}$

Do you have questions or need further clarification? Here are some related questions:

1. What is the significance of separating variables in double integrals?
2. How do you integrate functions with square roots, such as $\sqrt{r}$?
3. Can you explain why the limits $0$ to $2\pi$ apply only to $\theta$?
4. How would this integral change if the limits of $r$ were dependent on $\theta$?
5. What does this integral represent geometrically in polar coordinates?

Tip: Always check for symmetry in polar integrals—it can simplify calculations significantly!