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This problem asks us to use the **Gauss Divergence Theorem** to evaluate the surface integral of the function \( ax^2 + by^2 + cz^2 \) over the surface of the sphere given by \( x^2 + y^2 + z^2 = 1 \).

### Step 1: Recall the Gauss Divergence Theorem
The **Gauss Divergence Theorem** states:
\[
\iint_{S} \mathbf{F} \cdot \mathbf{n} \, dS = \iiint_{V} (\nabla \cdot \mathbf{F}) \, dV
\]
where:
- \( S \) is the closed surface (in this case, the surface of the sphere),
- \( V \) is the volume enclosed by \( S \) (the volume of the sphere),
- \( \mathbf{F} \) is a vector field, and
- \( \nabla \cdot \mathbf{F} \) is the divergence of \( \mathbf{F} \).

### Step 2: Define the Vector Field
In this case, the function inside the surface integral is \( ax^2 + by^2 + cz^2 \), and this can be interpreted as the dot product of a vector field \( \mathbf{F} \) with the normal vector \( \mathbf{n} \). To apply the divergence theorem, we need to define a vector field whose divergence can be computed. We assume the vector field \( \mathbf{F} \) is:
\[
\mathbf{F}(x, y, z) = \left( ax^2, by^2, cz^2 \right)
\]
This is a natural vector field based on the given function.

### Step 3: Compute the Divergence of \( \mathbf{F} \)
The divergence of a vector field \( \mathbf{F} = (F_1, F_2, F_3) \) is given by:
\[
\nabla \cdot \mathbf{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z}
\]
In this case:
\[
F_1 = ax^2, \quad F_2 = by^2, \quad F_3 = cz^2
\]
So, the divergence is:
\[
\nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(ax^2) + \frac{\partial}{\partial y}(by^2) + \frac{\partial}{\partial z}(cz^2)
\]
Each term can be computed as follows:
\[
\frac{\partial}{\partial x}(ax^2) = 2ax, \quad \frac{\partial}{\partial y}(by^2) = 2by, \quad \frac{\partial}{\partial z}(cz^2) = 2cz
\]
Thus, the divergence of \( \mathbf{F} \) is:
\[
\nabla \cdot \mathbf{F} = 2ax + 2by + 2cz
\]

### Step 4: Set up the Volume Integral
Using the divergence theorem, the surface integral over the sphere can be converted into a volume integral over the sphere:
\[
\iint_{S} (ax^2 + by^2 + cz^2) \, dS = \iiint_{V} (2ax + 2by + 2cz) \, dV
\]
Now we integrate the expression \( 2ax + 2by + 2cz \) over the volume of the sphere \( x^2 + y^2 + z^2 = 1 \).

### Step 5: Symmetry Considerations
By symmetry, the integrals of \( x \), \( y \), and \( z \) over the volume of a sphere are zero. This is because for any positive value of \( x \), there is a corresponding negative value of \( x \) with the same magnitude, and similarly for \( y \) and \( z \). Thus, we conclude that:
\[
\iiint_{V} 2ax \, dV = 0, \quad \iiint_{V} 2by \, dV = 0, \quad \iiint_{V} 2cz \, dV = 0
\]
Therefore, the entire volume integral evaluates to zero:
\[
\iiint_{V} (2ax + 2by + 2cz) \, dV = 0
\]

### Final Answer:
\[
\iint_{S} (ax^2 + by^2 + cz^2) \, dS = 0
\]

Would you like further details or have any questions on any step? Here are 5 follow-up questions you might be interested in:

1. How is the divergence of a vector field computed in general?
2. Why do symmetry considerations simplify the evaluation of integrals over a sphere?
3. Can you apply the divergence theorem to other types of surfaces, and how would the process differ?
4. How does the Gauss Divergence Theorem relate to the physical concept of flux?
5. What are some practical applications of the Gauss Divergence Theorem in physics or engineering?

**Tip:** When using symmetry to simplify integrals, always check the structure of the vector field and surface to ensure components will indeed cancel out as expected.

Use Gauss Divergence Theorem to evaluate ∬_S (ax² + by² + cz²) dS over the sphere x² + y² + z² = 1.

Explore how to apply the Gauss Divergence Theorem to evaluate the surface integral ∬_S (ax² + by² + cz²) dS over the sphere x² + y² + z² = 1. This solution walks through defining the vector field, computing the divergence, and utilizing symmetry arguments.

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