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.

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 $\theta$.