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The problem asks for the surface integral of the vector field \(\mathbf{F} = (y^2, z^2, x^2)\) over a surface \(S\), which is a part of the plane \(x + 2y + z = 4\) located in the positive quadrant, bounded by the planes \(x = 0\), \(y = 0\), and \(z = 0\).

### Steps to solve this:

1. **Parameterization of the Surface:**
   The surface \(S\) is the region where \(x + 2y + z = 4\). In this case, you can express \(z\) in terms of \(x\) and \(y\):
   \[
   z = 4 - x - 2y
   \]
   The limits of \(x\) and \(y\) are determined by the intersection of this plane with the boundaries \(x = 0\), \(y = 0\), and \(z = 0\).

2. **Bounds of Integration:**
   From the equation \(z = 4 - x - 2y\), the region in the \(xy\)-plane is bounded by:
   - \(x = 0\),
   - \(y = 0\), and
   - when \(z = 0\), we have \(x + 2y = 4\), or \(y = \frac{4 - x}{2}\).

   So the integration limits are:
   \[
   0 \leq x \leq 4, \quad 0 \leq y \leq \frac{4 - x}{2}.
   \]

3. **Normal Vector:**
   The normal vector \(\mathbf{n}\) to the surface can be found by taking the gradient of the plane equation \(x + 2y + z = 4\). This gives:
   \[
   \mathbf{n} = (1, 2, 1).
   \]
   To normalize, we calculate the magnitude:
   \[
   |\mathbf{n}| = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{6}.
   \]
   So the unit normal vector is:
   \[
   \hat{\mathbf{n}} = \frac{1}{\sqrt{6}}(1, 2, 1).
   \]

4. **Surface Integral Formula:**
   The surface integral of the vector field \(\mathbf{F}\) over the surface \(S\) is given by:
   \[
   \iint_S \mathbf{F} \cdot d\mathbf{S} = \iint_S \mathbf{F} \cdot \hat{\mathbf{n}} \, dA,
   \]
   where \(dA\) is the area element on the \(xy\)-plane.

5. **Compute the Dot Product \(\mathbf{F} \cdot \hat{\mathbf{n}}\):**
   On the surface, the vector field \(\mathbf{F} = (y^2, z^2, x^2)\) becomes:
   \[
   \mathbf{F} = \left(y^2, (4 - x - 2y)^2, x^2\right).
   \]
   Now, compute the dot product \(\mathbf{F} \cdot \hat{\mathbf{n}}\):
   \[
   \mathbf{F} \cdot \hat{\mathbf{n}} = \frac{1}{\sqrt{6}}(y^2 + 2(4 - x - 2y)^2 + x^2).
   \]

6. **Set Up the Integral:**
   The surface integral is then:
   \[
   \frac{1}{\sqrt{6}} \iint_S \left(y^2 + 2(4 - x - 2y)^2 + x^2\right) dA.
   \]
   In terms of \(x\) and \(y\), this becomes:
   \[
   \frac{1}{\sqrt{6}} \int_0^4 \int_0^{\frac{4 - x}{2}} \left(y^2 + 2(4 - x - 2y)^2 + x^2\right) dy \, dx.
   \]

Now, you can compute this double integral step by step to find the surface integral of the vector field \(\mathbf{F}\) over the surface \(S\).

Do you need more details on any part of this? Here are some relative questions to expand further:

1. How do we calculate the surface element \(d\mathbf{S}\)?
2. How can we simplify the bounds of integration using symmetry?
3. Can we use other coordinate systems (like cylindrical) for easier integration?
4. How do we verify the normal vector calculation?
5. What are some strategies for solving the double integral?

**Tip:** When calculating surface integrals, carefully examine the normal vector direction to ensure it is consistent with the orientation of the surface.

Diketahui medan vektor \(\mathbf{F} = (y^2, z^2, x^2)\) dan permukaan \(S\) adalah bagian dari bidang \(x + 2y + z = 4\) yang terletak di kuadran positif dan dibatasi oleh bidang \(x = 0\), \(y = 0\), dan \(z = 0\). Hitunglah integral permukaan dari medan vektor \(\mathbf{F}\) pada permukaan \(S\)!

Learn how to calculate the surface integral of the vector field \(\mathbf{F} = (y^2, z^2, x^2)\) over the plane \(x + 2y + z = 4\), located in the positive quadrant and bounded by the planes \(x = 0\), \(y = 0\), and \(z = 0\). This problem involves parameterization, finding the normal vector, and setting up the double integral for the surface integral.

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