´ 1
−1
´ √
1−x2
0
(x
2 + y
2
)
3/2dydx

The given integral appears to represent a double integral over a region in a Cartesian coordinate system. Here's how we can interpret and break it down:

\[
\int_{-1}^{1} \int_{0}^{\sqrt{1-x^2}} \frac{(x^2 + y^2)}{(x^2 + y^2)^{3/2}} \, dy \, dx
\]

### Step 1: Simplifying the integrand
The integrand is:

\[
\frac{x^2 + y^2}{(x^2 + y^2)^{3/2}}
\]

This expression simplifies to:

\[
\frac{1}{(x^2 + y^2)^{1/2}} = \frac{1}{\sqrt{x^2 + y^2}}
\]

Thus, the double integral becomes:

\[
\int_{-1}^{1} \int_{0}^{\sqrt{1-x^2}} \frac{1}{\sqrt{x^2 + y^2}} \, dy \, dx
\]

### Step 2: Changing to Polar Coordinates
This is a good candidate for converting to polar coordinates, since \(x^2 + y^2\) becomes \(r^2\) in polar coordinates. In polar form:

- \(x = r \cos \theta\)
- \(y = r \sin \theta\)
- The element of area \(dx \, dy\) becomes \(r \, dr \, d\theta\).

The limits also change in polar coordinates:
- For \(x\) from \(-1\) to \(1\) and \(y\) from \(0\) to \(\sqrt{1 - x^2}\), this describes the upper half of a circle with radius 1, so \(r\) will go from 0 to 1, and \(\theta\) will go from 0 to \(\pi\).

Now the integral is:

\[
\int_{0}^{\pi} \int_{0}^{1} \frac{1}{r} \cdot r \, dr \, d\theta
\]

### Step 3: Simplifying and Integrating
The \(r\) in the numerator and denominator cancel out, leaving:

\[
\int_{0}^{\pi} \int_{0}^{1} 1 \, dr \, d\theta
\]

First, integrate with respect to \(r\):

\[
\int_{0}^{1} 1 \, dr = 1
\]

Now, integrate with respect to \(\theta\):

\[
\int_{0}^{\pi} 1 \, d\theta = \pi
\

1 ∫ −1 √(1 − x²) 0 (x² + y²)^(3/2) dy dx

Solve a double integral over a circular region using polar coordinates. The integral covers key concepts such as double integration, polar transformation, and area calculation. Ideal for college-level calculus students.

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