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.