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We are tasked with converting the given double integral from Cartesian coordinates to polar coordinates and evaluating it. The integral provided is:

\[
\int_0^{\sqrt{6}} \int_{-x}^{x} dy \, dx
\]

### Step 1: Sketch the region

The limits of integration suggest that \( y \) varies between \( -x \) and \( x \), and \( x \) varies between \( 0 \) and \( \sqrt{6} \). This region describes a triangular area in the first quadrant of the plane, bounded by the line \( y = -x \) and \( y = x \), and between \( x = 0 \) and \( x = \sqrt{6} \).

### Step 2: Convert to polar coordinates

In polar coordinates, the transformation between Cartesian and polar is given by:
\[
x = r \cos(t), \quad y = r \sin(t)
\]
The Jacobian determinant for polar coordinates is \( r \). The area element \( dx \, dy \) transforms to \( r \, dr \, dt \).

Now, we determine the bounds for \( r \) and \( t \). 

- \( r \) ranges from 0 to \( \sqrt{6} \), the radial distance from the origin to the outer boundary.
- \( t \) ranges from \( -\frac{\pi}{4} \) to \( \frac{\pi}{4} \), as the lines \( y = x \) and \( y = -x \) correspond to these angles in polar coordinates.

### Step 3: Set up the new integral

The integral in polar coordinates becomes:
\[
\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \int_0^{\sqrt{6}} r \, dr \, dt
\]

### Step 4: Evaluate the integral

1. First, integrate with respect to \( r \):
\[
\int_0^{\sqrt{6}} r \, dr = \left[ \frac{r^2}{2} \right]_0^{\sqrt{6}} = \frac{(\sqrt{6})^2}{2} = \frac{6}{2} = 3
\]

2. Next, integrate with respect to \( t \):
\[
\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 3 \, dt = 3 \left[ t \right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}} = 3 \left( \frac{\pi}{4} - \left( -\frac{\pi}{4} \right) \right) = 3 \times \frac{\pi}{2} = \frac{3\pi}{2}
\]

Thus, the value of the integral is:

\[
\boxed{\frac{3\pi}{2}}
\]

Would you like further details or clarifications?

Here are 5 related questions to expand your understanding:

1. How does the Jacobian determinant change the differential area element in polar coordinates?
2. Why is it necessary to change the limits of integration when converting to polar coordinates?
3. Can this region be described using a different coordinate system, like cylindrical or spherical?
4. How would the integral change if the region described a full circle instead of a triangular section?
5. What are some typical uses of polar coordinates in solving integrals in physics or engineering?

**Tip**: When converting integrals to polar coordinates, always consider the symmetry of the region, as polar coordinates are particularly useful for circular or radial symmetry.

Convert the integral \(\int_0^{\sqrt{6}} \int_{-x}^{x} dy \, dx\) to polar coordinates and evaluate it.

This page walks through the process of converting a double integral from Cartesian to polar coordinates and evaluating it. The integral \(\int_0^{\sqrt{6}} \int_{-x}^{x} dy \, dx\) is transformed and solved, covering polar coordinate transformations, the Jacobian, and area elements. Suitable for university-level calculus students.

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