To evaluate the double integral $\int \int_D \frac{4xy}{(x^2 + y^2 + 4)^3} dA$, where $D$ is the first quadrant, we can make the appropriate substitutions and evaluate the limits.

The integral is set up in the form of a Cartesian double integral over the first quadrant, but to simplify it, we can convert to polar coordinates because the integrand involves $x^2 + y^2$, which is naturally simplified using polar coordinates.

Step 1: Substituting Polar Coordinates

In polar coordinates:

• $x = r \cos \theta$
• $y = r \sin \theta$
• $x^2 + y^2 = r^2$
• The area element $dA = dx dy = r dr d\theta$.

We also need to account for the bounds. The region $D$ is the entire first quadrant, meaning $r$ runs from 0 to $\infty$, and $\theta$ runs from 0 to $\frac{\pi}{2}$ (as it covers the first quadrant).

The given integral becomes:

$\int_0^{\frac{\pi}{2}} \int_0^\infty \frac{4r^2 \cos \theta \sin \theta}{(r^2 + 4)^3} r \, dr \, d\theta$

Step 2: Simplifying the Integral

Notice that $\cos \theta \sin \theta = \frac{1}{2} \sin(2\theta)$. So, the integral becomes:

$\int_0^{\frac{\pi}{2}} \sin(2\theta) \, d\theta \int_0^\infty \frac{4r^3}{(r^2 + 4)^3} \, dr$

The angular integral:

$\int_0^{\frac{\pi}{2}} \sin(2\theta) \, d\theta = \frac{1}{2}$

So the integral becomes:

$\frac{1}{2} \int_0^\infty \frac{4r^3}{(r^2 + 4)^3} \, dr$

Step 3: Solving the Radial Integral

To solve the radial integral, let’s use the substitution $u = r^2 + 4$, which gives:

• $du = 2r \, dr$
• When $r = 0$, $u = 4$
• When $r = \infty$, $u = \infty$

This transforms the radial integral into:

$\frac{1}{2} \int_4^\infty \frac{4r^3}{u^3} \frac{du}{2r} = \frac{1}{2} \int_4^\infty \frac{2r^2}{u^3} \, du$

Since $r^2 = u - 4$, this becomes:

$\frac{1}{2} \int_4^\infty \frac{2(u - 4)}{u^3} \, du$

Breaking this up:

$\frac{1}{2} \left[ 2 \int_4^\infty \frac{u}{u^3} \, du - 2 \int_4^\infty \frac{4}{u^3} \, du \right]$

The first integral is:

$\int_4^\infty \frac{1}{u^2} \, du = \left[ -\frac{1}{u} \right]_4^\infty = 0 - \left( -\frac{1}{4} \right) = \frac{1}{4}$

The second integral is:

$4 \int_4^\infty \frac{1}{u^3} \, du = 4 \left[ -\frac{1}{2u^2} \right]_4^\infty = 4 \left( 0 - \left( -\frac{1}{2(4)^2} \right) \right) = 4 \times \frac{1}{32} = \frac{1}{8}$

Now substitute these back:

[ \frac{1}{2} \left[ 2 \times \frac{1}{4} - 2 \times \frac{1}{8} \right] = \frac{1}{2} \left[ \frac{1}{2} - \frac{1}{4} \right] = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8} ]

Final Answer:

The value of the double integral is $\boxed{\frac{1}{8}}$.

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Would you like more details or further clarification?

Here are five related questions:

1. How does the choice of polar coordinates simplify the integration process?
2. What would change if the region $D$ were different?
3. Can this method be applied to other types of integrals with similar structures?
4. How does the substitution $u = r^2 + 4$ work in simplifying the radial part?
5. What is the geometric interpretation of this integral?

Tip: When converting to polar coordinates, always remember to adjust both the limits and the area element (from $dx\,dy$ to $r\,dr\,d\theta$).