Calculate the surface integral ∬s f · n dμ where
f = (x³e³ᶻ, x²y² cos z, xy² sin z)
S = {(x, y, z) | x²e²ᶻ + y² + z² = 1}
and n is the outward normal vector to the surface.

To calculate the surface integral \(\iint_S \mathbf{f} \cdot \mathbf{n} \, d\mu\), where \(\mathbf{f} = (x^3 e^{3z}, x^2 y^2 \cos z, xy^2 \sin z)\) and \(S\) is defined by the surface \(x^2 e^{2z} + y^2 + z^2 = 1\), with \(\mathbf{n}\) being the outward normal vector, we can proceed by using the **divergence theorem**.

### Step 1: Check the Divergence Theorem Applicability
The divergence theorem states:
\[
\iint_S \mathbf{f} \cdot \mathbf{n} \, d\mu = \iiint_V (\nabla \cdot \mathbf{f}) \, dV
\]
where \(V\) is the volume enclosed by the surface \(S\), and \(\nabla \cdot \mathbf{f}\) is the divergence of the vector field \(\mathbf{f}\).

### Step 2: Calculate the Divergence of \(\mathbf{f}\)
Given \(\mathbf{f} = (f_1, f_2, f_3) = (x^3 e^{3z}, x^2 y^2 \cos z, xy^2 \sin z)\), the divergence \(\nabla \cdot \mathbf{f}\) is:
\[
\nabla \cdot \mathbf{f} = \frac{\partial f_1}{\partial x} + \frac{\partial f_2}{\partial y} + \frac{\partial f_3}{\partial z}
\]

Let's calculate each partial derivative:

1. \(\frac{\partial f_1}{\partial x} = \frac{\partial}{\partial x} \left( x^3 e^{3z} \right) = 3x^2 e^{3z}\)
2. \(\frac{\partial f_2}{\partial y} = \frac{\partial}{\partial y} \left( x^2 y^2 \cos z \right) = 2x^2 y \cos z\)
3. \(\frac{\partial f_3}{\partial z} = \frac{\partial}{\partial z} \left( xy^2 \sin z \right) = xy^2 \cos z\)

Thus, the divergence is:
\[
\nabla \cdot \mathbf{f} = 3x^2 e^{3z} + 2x^2 y \cos z + xy^2 \cos z
\]

### Step 3: Set Up the Volume Integral
Now, we integrate the divergence over the volume \(V\) enclosed by the surface \(S\):
\[
\iiint_V (3x^2 e^{3z} + 2x^2 y \cos z + xy^2 \cos z) \, dV
\]
The region \(V\) is the volume inside the surface \(x^2 e^{2z} + y^2 + z^2 = 1\).

However, directly evaluating this integral could be complex due to the nature of the surface equation. The problem might require further simplifications or numerical methods depending on the context. Typically, if a direct evaluation is not feasible, this integral might be solved using specific symmetries or approximations.

### Final Answer
Thus, the surface integral is given by:
\[
\iint_S \mathbf{f} \cdot \mathbf{n} \, d\mu = \iiint_V \left(3x^2 e^{3z} + 2x^2 y \cos z + xy^2 \cos z\right) \, dV
\]
where \(V\) is the volume enclosed by the surface \(x^2 e^{2z} + y^2 + z^2 = 1\).

Would you like to explore the next steps, such as numerical approximation or applying symmetry to evaluate the integral?

### 5 Relative Questions:
1. What are some common methods for evaluating complex volume integrals?
2. How does the divergence theorem simplify surface integrals?
3. What role do symmetries play in simplifying triple integrals?
4. Can you apply similar techniques to other non-spherical surfaces?
5. How might this problem change if the surface equation were a simple sphere?

### Tip:
When dealing with complex integrals, always check for symmetries in the region of integration to potentially simplify the problem.

Learn how to calculate the surface integral ∬s f · n dμ where f = (x³e³ᶻ, x²y² cos z, xy² sin z) and S = {(x, y, z) | x²e²ᶻ + y² + z² = 1}. Detailed step-by-step solution using the divergence theorem and formulas for divergence and surface integrals.

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