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To evaluate the given double integral using a polar transformation, let's first rewrite the given integral in polar coordinates.

The given double integral is:

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

### Step-by-Step Process:

#### 1. Converting to Polar Coordinates:
In polar coordinates, we know that:
- \(x = r \cos \theta\)
- \(y = r \sin \theta\)
- \(x^2 + y^2 = r^2\)
- The differential \(dx \, dy = r \, dr \, d\theta\)

The limits of integration need to be adjusted to fit the polar coordinates. The region of integration is a sector of the unit circle in the first quadrant.

#### 2. Limits of Integration:
For the region described:
- The angle \(\theta\) will range from \(0\) to \(\frac{\pi}{4}\) (since the upper limit \(y = \sqrt{1 - x^2}\) describes a quarter of the unit circle).
- The radius \(r\) will range from \(0\) to \(1\) (the radius of the unit circle).

#### 3. Substituting the Integral in Polar Coordinates:
In polar coordinates, the integrand \(\frac{x}{x^2 + y^2}\) becomes:

\[
\frac{r \cos \theta}{r^2} = \frac{\cos \theta}{r}
\]

Thus, the integral becomes:

\[
\int_0^{\frac{\pi}{4}} \int_0^1 \frac{\cos \theta}{r} r \, dr \, d\theta
\]

The \(r\) terms cancel out, simplifying to:

\[
\int_0^{\frac{\pi}{4}} \cos \theta \int_0^1 \, dr \, d\theta
\]

#### 4. Evaluating the Integral:
The inner integral with respect to \(r\) is:

\[
\int_0^1 dr = 1
\]

So, the integral reduces to:

\[
\int_0^{\frac{\pi}{4}} \cos \theta \, d\theta
\]

The integral of \(\cos \theta\) is:

\[
\sin \theta
\]

Evaluating this from \(0\) to \(\frac{\pi}{4}\):

\[
\sin\left(\frac{\pi}{4}\right) - \sin(0) = \frac{\sqrt{2}}{2} - 0 = \frac{\sqrt{2}}{2}
\]

### Final Answer:
The value of the double integral is \(\frac{\sqrt{2}}{2}\).

Would you like further details or explanations on any step?

Here are some related questions to consider:
1. What are the general steps to convert a Cartesian integral into polar coordinates?
2. How do you determine the limits of integration when switching to polar coordinates?
3. Why do the \(dx \, dy\) terms convert to \(r \, dr \, d\theta\) in polar coordinates?
4. What is the geometric interpretation of the region of integration for this problem?
5. How does symmetry in integrals simplify evaluation in polar coordinates?

**Tip:** Always check if a given region of integration can be simplified using symmetry when converting to polar coordinates.

Evaluate the following double integrals through the polar transformation: ∫(0 to 1) ∫(x to sqrt(1-x^2)) (x / (x^2 + y^2)) dy dx

This page explains how to evaluate a given double integral using a polar transformation. Learn how to convert Cartesian coordinates into polar coordinates, adjust the limits of integration, and evaluate the integral. This solution involves key concepts in calculus, such as double integrals, polar coordinates, and integral calculus theorems. Suitable for college-level calculus students.

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