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The problem asks to compute the double integral of \( 4x^3 + y \) over the region \( D \), where the boundaries are defined as:

\[
D = \{x = 9, x = y^2\}
\]

The given integral is:

\[
\iint_D (4x^3 + y) \, dx \, dy
\]

The region \( D \) is bounded by \( x = 9 \) and \( x = y^2 \), so \( y \) ranges from \(-3\) to \(3\) (since \( x = y^2 \)), and for each \( y \), \( x \) ranges from \( y^2 \) to \( 9 \).

Now, we set up the double integral with these limits:

\[
\int_{y=-3}^{3} \int_{x=y^2}^{9} (4x^3 + y) \, dx \, dy
\]

### Step-by-Step Solution:

1. **Integrate with respect to \( x \)** first:
   \[
   \int_{y=-3}^{3} \left[ \int_{x=y^2}^{9} (4x^3 + y) \, dx \right] dy
   \]

The integral of \( 4x^3 \) with respect to \( x \) is:

\[
\int 4x^3 \, dx = x^4
\]

And the integral of \( y \) with respect to \( x \) is:

\[
\int y \, dx = yx
\]

So, the result of integrating \( 4x^3 + y \) with respect to \( x \) over the bounds \( x = y^2 \) to \( x = 9 \) is:

\[
\left[ x^4 + yx \right]_{x=y^2}^{x=9}
\]

2. **Substitute the limits**:

At \( x = 9 \):

\[
9^4 + y \cdot 9 = 6561 + 9y
\]

At \( x = y^2 \):

\[
(y^2)^4 + y \cdot y^2 = y^8 + y^3
\]

Thus, the expression becomes:

\[
\left( 6561 + 9y \right) - \left( y^8 + y^3 \right)
\]

This simplifies to:

\[
6561 + 9y - y^8 - y^3
\]

3. **Integrate with respect to \( y \)**:

Now, we integrate:

\[
\int_{y=-3}^{3} \left( 6561 + 9y - y^8 - y^3 \right) \, dy
\]

This can be split into separate integrals:

\[
6561 \int_{y=-3}^{3} 1 \, dy + 9 \int_{y=-3}^{3} y \, dy - \int_{y=-3}^{3} y^8 \, dy - \int_{y=-3}^{3} y^3 \, dy
\]

### Calculation of each term:

- \( 6561 \int_{y=-3}^{3} 1 \, dy = 6561 \times 6 = 39366 \)
- \( 9 \int_{y=-3}^{3} y \, dy = 0 \) (since it's an odd function over a symmetric interval)
- \( \int_{y=-3}^{3} y^8 \, dy = 2 \int_{0}^{3} y^8 \, dy \) (due to symmetry)
- \( \int_{y=-3}^{3} y^3 \, dy = 0 \) (since it's an odd function over a symmetric interval)

Let me now compute the value of the remaining term \( \int_{0}^{3} y^8 \, dy \).The value of the integral \( \int_0^3 y^8 \, dy \) is \( 2187 \).

Now, summarizing the integrals:

\[
\int_{y=-3}^{3} \left( 6561 + 9y - y^8 - y^3 \right) \, dy = 39366 + 0 - 2 \times 2187 + 0
\]

This simplifies to:

\[
39366 - 4374 = 34992
\]

Thus, the value of the double integral is:

\[
\boxed{34992}
\]

Would you like further details or clarifications on any step?

Here are 5 related questions you could explore:
1. How would the bounds change if the region was rotated or translated?
2. What if the integrand was more complex, such as involving trigonometric functions?
3. How do odd and even functions affect the computation of symmetric integrals?
4. What is the physical significance of computing double integrals in applications?
5. How would this integral be set up in polar coordinates?

**Tip:** For symmetric intervals, always check if the integrand is odd, as it can simplify calculations.

Compute the double integral of (4x^3 + y) over the region D, where the boundaries are defined as x = 9 and x = y^2.

Compute the double integral of (4x^3 + y) over the region bounded by x = 9 and x = y^2. This detailed solution involves integration with respect to x and y, applying concepts of bounded regions and symmetry in integration.

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