We are given the integral:

$\int_{E} \frac{x \sin(xy)}{81 + z^2} \, dV$

where the region $E$ is defined by:

$E = \{(x, y, z) \mid 1 \leq x \leq 2, \, 0 \leq y \leq 3\pi, \, 0 \leq z \leq 9\}.$

This is a triple integral, where $dV = dx \, dy \, dz$.

Step-by-step solution:

The integral can be written explicitly as:

$\int_{0}^{9} \int_{0}^{3\pi} \int_{1}^{2} \frac{x \sin(xy)}{81 + z^2} \, dx \, dy \, dz.$

1. Factor the integrand:

The integrand $\frac{x \sin(xy)}{81 + z^2}$ is separable in terms of $x, y, z$ because $\frac{1}{81 + z^2}$ depends only on $z$, and $x \sin(xy)$ depends only on $x$ and $y$. So, the triple integral can be split into two parts:

$\int_{0}^{9} \frac{1}{81 + z^2} \, dz \cdot \int_{0}^{3\pi} \int_{1}^{2} x \sin(xy) \, dx \, dy.$

2. Evaluate the integral with respect to $z$:

We need to compute:

$\int_{0}^{9} \frac{1}{81 + z^2} \, dz.$

This is a standard integral, which is an arctangent:

$\int \frac{1}{a^2 + z^2} \, dz = \frac{1}{a} \tan^{-1}\left(\frac{z}{a}\right).$

Here $a = 9$, so:

$\int_{0}^{9} \frac{1}{81 + z^2} \, dz = \frac{1}{9} \left[ \tan^{-1}\left(\frac{z}{9}\right) \right]_{0}^{9} = \frac{1}{9} \left( \frac{\pi}{4} - 0 \right) = \frac{\pi}{36}.$

3. Evaluate the integral with respect to $x$ and $y$:

Now, we need to compute the double integral:

$\int_{0}^{3\pi} \int_{1}^{2} x \sin(xy) \, dx \, dy.$

First, evaluate the inner integral with respect to $x$:

$\int x \sin(xy) \, dx.$

This is solved using integration by parts. Let $u = x$ and $dv = \sin(xy) \, dx$. Then:

$u = x, \quad du = dx, \quad v = -\frac{1}{y} \cos(xy).$

So, the integration by parts formula gives:

$\int x \sin(xy) \, dx = -\frac{x}{y} \cos(xy) + \frac{1}{y} \int \cos(xy) \, dx.$

The integral of $\cos(xy)$ is $\frac{1}{y} \sin(xy)$, so:

$\int x \sin(xy) \, dx = -\frac{x}{y} \cos(xy) + \frac{1}{y^2} \sin(xy).$

Now evaluate this at the limits $x = 1$ and $x = 2$:

$\left[ -\frac{x}{y} \cos(xy) + \frac{1}{y^2} \sin(xy) \right]_{1}^{2} = \left( -\frac{2}{y} \cos(2y) + \frac{1}{y^2} \sin(2y) \right) - \left( -\frac{1}{y} \cos(y) + \frac{1}{y^2} \sin(y) \right).$

Thus, the double integral becomes:

$\int_{0}^{3\pi} \left( -\frac{2}{y} \cos(2y) + \frac{1}{y^2} \sin(2y) + \frac{1}{y} \cos(y) - \frac{1}{y^2} \sin(y) \right) \, dy.$

While this is a more complex integral, it can be solved numerically or using advanced techniques.

Final Result:

The final answer will be:

$\frac{\pi}{36} \times \text{(value of the double integral)}.$

Do you want further details on solving the double integral or a specific part?

---

Related Questions:

1. How do you perform integration by parts for products of functions?
2. What are the standard forms for integrals involving trigonometric functions?
3. How do you evaluate arctangent integrals in definite form?
4. What is the importance of separable integrals, and how do they simplify complex problems?
5. How do numerical methods help in evaluating complex integrals?

Tip: Always check if an integrand can be simplified or factored before attempting the full computation—it often reduces complexity.