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The problem asks you to evaluate the surface integral

\[
\iint_S (\hat{n} \cdot \vec{F}) \, dS
\]

where \(\vec{F}\) is given as:

\[
\vec{F} = (3x + z) \hat{i} + (y^2 - \sin(x^2)z) \hat{j} + (xz + y e^{x^5}) \hat{k}
\]

and \(\hat{n}\) is the unit normal vector to the surface \(S\), which is described with bounds \(0 \leq x \leq 1\), \(0 \leq y \leq 3\), and \(0 \leq z \leq 2\).

### Strategy:
This is a vector surface integral of the form \(\iint_S (\hat{n} \cdot \vec{F}) \, dS\). Typically, this type of problem suggests using the **Divergence Theorem** if the surface \(S\) is closed, as the divergence theorem relates a surface integral to a volume integral:

\[
\iint_S (\hat{n} \cdot \vec{F}) \, dS = \iiint_V (\nabla \cdot \vec{F}) \, dV
\]

We will first calculate the divergence of the vector field \(\vec{F}\).

### 1. Compute the divergence \(\nabla \cdot \vec{F}\):
The divergence of a vector field \(\vec{F} = F_1 \hat{i} + F_2 \hat{j} + F_3 \hat{k}\) is given by:

\[
\nabla \cdot \vec{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z}
\]

where:
- \(F_1 = 3x + z\)
- \(F_2 = y^2 - \sin(x^2)z\)
- \(F_3 = xz + y e^{x^5}\)

### 2. Boundaries:
The problem provides the following bounds for the surface:
\[
0 \leq x \leq 1, \quad 0 \leq y \leq 3, \quad 0 \leq z \leq 2
\]

Now, I'll proceed to compute the divergence of \(\vec{F}\) and then integrate it over the volume described by these bounds.The divergence of the vector field \(\vec{F}\) is:

\[
\nabla \cdot \vec{F} = x + 2y + 3
\]

### 3. Set up the volume integral:

Now, we need to integrate this divergence over the volume \(V\) described by the bounds:

\[
0 \leq x \leq 1, \quad 0 \leq y \leq 3, \quad 0 \leq z \leq 2
\]

The volume integral becomes:

\[
\iiint_V (x + 2y + 3) \, dV = \int_0^1 \int_0^3 \int_0^2 (x + 2y + 3) \, dz \, dy \, dx
\]

I'll compute this next.It seems like I can’t do more advanced data analysis right now. Please try again later.

However, you can manually complete the volume integral as follows:

1. **Integrate over \( z \)**:
   \[
   \int_0^2 (x + 2y + 3) \, dz = (x + 2y + 3) \cdot 2 = 2(x + 2y + 3)
   \]

2. **Integrate over \( y \)**:
   \[
   \int_0^3 2(x + 2y + 3) \, dy = 2 \left(3x + 6 + 18\right) = 6x + 48
   \]

3. **Integrate over \( x \)**:
   \[
   \int_0^1 (6x + 48) \, dx = 3 + 48 = 51
   \]

So, the final result of the integral is \( 51 \).

If you have more questions or would like further clarification, feel free to ask! Here are some related questions to explore this concept:

1. How does the Divergence Theorem simplify surface integrals?
2. What is the physical interpretation of the vector field \(\vec{F}\) in this problem?
3. Can you apply the Divergence Theorem in cases where the surface is not closed?
4. How would the result change if the vector field had a different divergence?
5. How can you calculate the normal vector \(\hat{n}\) for surfaces in general?

**Tip**: In vector calculus, recognizing opportunities to apply the Divergence Theorem can often simplify surface integrals into easier volume integrals.

Determine the surface integral of (n̂ · F) over the surface S, where F is given as (3x + z)î + (y^2 - sin(x^2)z)ĵ + (xz + ye^x^5)k̂, and the bounds are 0 ≤ x ≤ 1, 0 ≤ y ≤ 3, 0 ≤ z ≤ 2.

This problem involves calculating the surface integral of a vector field F over a given surface S using the Divergence Theorem. Learn how to compute the divergence of the vector field and evaluate the resulting volume integral, with a detailed step-by-step solution. Suitable for students at the college level studying vector calculus.

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