The evaluation of  
∫
∫
D
4
x
y
(
x
2
+
y
2
+
4
)
3
d
A
 , where  
D
  is the entire first quadrant, can be accomplished by :

lim
b
→
∞
  
lim
d
→
∞
 
∫
b
0
∫
d
0
4
x
y
(
x
2
+
y
2
+
4
)
3
d
x
d
y
 

Work through the appropriate substitutions to complete the integrals, and evaluate the limits. What is the value? 
  

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 \( \boxed{\frac{1}{8}} \).

---

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 \( r\,dr\,d\theta \)).

The evaluation of ∫∫D 4xy / (x^2 + y^2 + 4)^3 dA, where D is the entire first quadrant, can be accomplished by lim b → ∞ lim d → ∞ ∫_0^b ∫_0^d 4xy / (x^2 + y^2 + 4)^3 dx dy. Work through the appropriate substitutions to complete the integrals, and evaluate the limits. What is the value?

Learn how to evaluate the double integral ∫∫_D (4xy) / (x^2 + y^2 + 4)^3 dA over the first quadrant using polar coordinates. This detailed guide covers key steps such as coordinate substitution, trigonometric simplifications, and solving the integral using Fubini's theorem. Ideal for calculus students at the undergraduate level.

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