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To convert the given integral to polar coordinates, let's analyze and solve it step-by-step.

### Original Integral
The given integral is:
\[
\int_0^3 \int_{-\sqrt{9 - x^2}}^{\sqrt{9 - x^2}} \frac{1}{\sqrt{x^2 + y^2}} \, dy \, dx
\]

### Step 1: Recognize the Region
The region of integration in rectangular coordinates is bounded by:
1. \( x \) goes from \( 0 \) to \( 3 \),
2. \( y \) goes from \( -\sqrt{9 - x^2} \) to \( \sqrt{9 - x^2} \).

This region represents the upper right quarter of a circle with radius 3, as \( x^2 + y^2 \leq 9 \) and \( x \geq 0 \).

### Step 2: Convert to Polar Coordinates
In polar coordinates:
- \( x = r \cos \theta \),
- \( y = r \sin \theta \),
- \( dx \, dy = r \, dr \, d\theta \).

Additionally, \( \sqrt{x^2 + y^2} = r \), so the integrand \( \frac{1}{\sqrt{x^2 + y^2}} \) becomes \( \frac{1}{r} \) in polar coordinates.

### Step 3: Set up the Limits
For the region defined:
1. \( r \) ranges from \( 0 \) to \( 3 \),
2. \( \theta \) ranges from \( 0 \) to \( \frac{\pi}{2} \), as it covers the first quadrant.

### Step 4: Rewrite the Integral
Now, rewriting the integral in polar coordinates:
\[
\int_0^{\pi/2} \int_0^3 \frac{1}{r} \cdot r \, dr \, d\theta = \int_0^{\pi/2} \int_0^3 1 \, dr \, d\theta
\]

This simplifies to:
\[
\int_0^{\pi/2} \int_0^3 \, dr \, d\theta
\]

### Step 5: Find the Correct Option
The correct answer is:
\[
\int_0^{\pi/2} \int_0^3 \, dr \, d\theta
\]

This corresponds to the third option in the list provided.

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Would you like further explanation on any part of this process? 

Here are some related questions for additional practice:
1. How would the limits change if the region were a full circle?
2. How does the Jacobian \( r \) affect the transformation of integrals?
3. Why does \( \frac{1}{r} \) simplify to \( 1 \) when multiplied by the \( r \) in \( dx \, dy \)?
4. How would this integral change if it included \( x^2 + y^2 \) in the integrand?
5. What if the limits were from \( -3 \) to \( 3 \) for both \( x \) and \( y \)?

**Tip:** Always identify the region of integration carefully before converting to polar coordinates, as this influences the limits for \( r \) and \( \theta \).

Change the integral to polar coordinates: ∫₀³ ∫_{-√(9−x²)}^{√(9−x²)} 1 / √(x² + y²) dy dx

This problem demonstrates how to convert a double integral in Cartesian coordinates to polar coordinates over a quarter-circle region. The key steps involve identifying the region, transforming the limits, and rewriting the integrand using polar coordinates.

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