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We are asked to evaluate the iterated integral by converting to polar coordinates. The given integral is:

\[
\int_0^2 \int_0^{\sqrt{2x - x^2}} \frac{1}{5 \sqrt{x^2 + y^2}} \, dy \, dx
\]

### Step 1: Understanding the bounds and region
The given bounds suggest a region bounded by a circle, as we see \(\sqrt{2x - x^2}\) as an upper bound for \(y\). To confirm this, let's rewrite the bound expression:

\[
y = \sqrt{2x - x^2}
\]

Squaring both sides:

\[
y^2 = 2x - x^2 \implies x^2 + y^2 = 2
\]

This represents a circle with radius \(\sqrt{2}\) centered at \((1, 0)\), though we are only working with the positive quadrant (\(x \geq 0, y \geq 0\)).

### Step 2: Conversion to polar coordinates
In polar coordinates, we use the following transformations:
- \(x = r\cos(\theta)\)
- \(y = r\sin(\theta)\)
- \(x^2 + y^2 = r^2\)
- The differential element \(dx\,dy\) becomes \(r\,dr\,d\theta\)

### Step 3: Setting up the new bounds
In polar coordinates, the region we are integrating over is a quarter-circle (first quadrant) of radius \(\sqrt{2}\), so:
- \(r\) ranges from \(0\) to \(\sqrt{2}\)
- \(\theta\) ranges from \(0\) to \(\frac{\pi}{2}\)

### Step 4: Rewriting the integrand
The integrand \(\frac{1}{5 \sqrt{x^2 + y^2}}\) becomes \(\frac{1}{5r}\), since \(x^2 + y^2 = r^2\). Also, don't forget the Jacobian \(r\) from the differential transformation. Thus, the integral becomes:

\[
\int_0^{\frac{\pi}{2}} \int_0^{\sqrt{2}} \frac{1}{5r} r \, dr \, d\theta
\]

This simplifies to:

\[
\int_0^{\frac{\pi}{2}} \int_0^{\sqrt{2}} \frac{1}{5} \, dr \, d\theta
\]

### Step 5: Solving the integral
First, solve the inner integral with respect to \(r\):

\[
\int_0^{\sqrt{2}} \frac{1}{5} \, dr = \frac{1}{5} \times \left[r\right]_0^{\sqrt{2}} = \frac{1}{5} \times \sqrt{2}
\]

Next, solve the outer integral with respect to \(\theta\):

\[
\int_0^{\frac{\pi}{2}} \frac{\sqrt{2}}{5} \, d\theta = \frac{\sqrt{2}}{5} \times \left[\theta\right]_0^{\frac{\pi}{2}} = \frac{\sqrt{2}}{5} \times \frac{\pi}{2} = \frac{\pi \sqrt{2}}{10}
\]

### Final Answer:
The value of the iterated integral is:

\[
\boxed{\frac{\pi \sqrt{2}}{10}}
\]

Would you like more details on any of the steps? Here are five related questions you can explore:

1. How does the transformation to polar coordinates work in more complex regions?
2. What are the Jacobian determinants for different coordinate systems?
3. Why is the region of integration a quarter-circle, and how would this change in other coordinate systems?
4. How would this problem change if the region extended into different quadrants?
5. How can we apply the polar coordinate method to evaluate double integrals over ellipses?

**Tip**: When converting to polar coordinates, always check the shape of the region to properly adjust bounds for \(r\) and \(\theta\).

Evaluate the iterated integral by converting to polar coordinates: ∫₀² ∫₀√(2x - x²) (1 / 5√(x² + y²)) dy dx.

Solve the double integral by converting to polar coordinates. The problem involves evaluating a double integral over a quarter-circle region with key steps on setting bounds in polar coordinates, including Jacobian transformations. Suitable for college-level calculus students.

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