To solve this integral, let us evaluate the given double integral over the positive quadrant of the circle $x^2 + y^2 = 4$. Here's a step-by-step explanation:

Step 1: Interpret the Problem

The integral is: $\iint_{D} x y \, dx \, dy$ where $D$ is the positive quadrant ($x \geq 0$, $y \geq 0$) of the circle $x^2 + y^2 = 4$.

Step 2: Convert to Polar Coordinates

We convert the integral to polar coordinates using the following relations: $x = r \cos\theta, \quad y = r \sin\theta, \quad dx\,dy = r\,dr\,d\theta.$ The equation $x^2 + y^2 = 4$ in polar coordinates becomes $r^2 = 4$, or $r = 2$.

For the positive quadrant, the limits of $r$ and $\theta$ are: $0 \leq r \leq 2, \quad 0 \leq \theta \leq \frac{\pi}{2}.$

The integral becomes: $\iint_{D} x y \, dx \, dy = \int_{0}^{\frac{\pi}{2}} \int_{0}^{2} (r\cos\theta)(r\sin\theta) \cdot r \, dr \, d\theta.$

Step 3: Simplify the Integral

Simplify the integrand: $(r\cos\theta)(r\sin\theta) \cdot r = r^3 \cos\theta \sin\theta.$

Thus, the integral becomes: $\int_{0}^{\frac{\pi}{2}} \int_{0}^{2} r^3 \cos\theta \sin\theta \, dr \, d\theta.$

Step 4: Separate the Variables

The integral separates as: $\int_{0}^{\frac{\pi}{2}} \cos\theta \sin\theta \, d\theta \cdot \int_{0}^{2} r^3 \, dr.$

Evaluate the $r$-Integral:

$\int_{0}^{2} r^3 \, dr = \left[ \frac{r^4}{4} \right]_{0}^{2} = \frac{2^4}{4} - \frac{0^4}{4} = \frac{16}{4} = 4.$

Evaluate the $\theta$-Integral:

$\int_{0}^{\frac{\pi}{2}} \cos\theta \sin\theta \, d\theta.$ Using the identity $\cos\theta \sin\theta = \frac{1}{2} \sin(2\theta)$, we rewrite the integral: $\int_{0}^{\frac{\pi}{2}} \cos\theta \sin\theta \, d\theta = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \sin(2\theta) \, d\theta.$

The integral of $\sin(2\theta)$ is: $\int \sin(2\theta) \, d\theta = -\frac{1}{2} \cos(2\theta).$

Thus: [ \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \sin(2\theta) , d\theta = \frac{1}{2} \left[ -\frac{1}{2} \cos(2\theta) \right]_{0}^{\frac{\pi}{2}} = \frac{1}{2} \left[ -\frac{1}{2}(\cos(\pi) - \cos(0)) \right]. ]

Simplify: $\cos(\pi) = -1, \quad \cos(0) = 1.$ [ \frac{1}{2} \left[ -\frac{1}{2}((-1) - 1) \right] = \frac{1}{2} \left[ -\frac{1}{2}(-2) \right] = \frac{1}{2} \cdot 1 = \frac{1}{2}. ]

Step 5: Combine the Results

Multiply the results of the two integrals: $\int_{0}^{\frac{\pi}{2}} \cos\theta \sin\theta \, d\theta \cdot \int_{0}^{2} r^3 \, dr = \frac{1}{2} \cdot 4 = 2.$

Final Answer:

$\boxed{2}$

Let me know if you'd like further explanation or details!

Related Questions

1. How do we convert between Cartesian and polar coordinates for integrals?
2. What are the typical bounds for integrals in polar coordinates when working with circles?
3. How does the Jacobian factor $r$ arise in polar coordinate transformations?
4. Can this problem be solved without using polar coordinates? Why or why not?
5. What changes if the integration region is the full circle instead of the positive quadrant?

Tip:

When dealing with circular or radial symmetry in double integrals, polar coordinates often simplify the computation significantly.