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This problem asks us to calculate the outward flux of the given vector field through the surface \( S \) of a tetrahedron.

We are provided with:
- The vertices of the tetrahedron: \( (0,0,0) \), \( (1,0,0) \), \( (0,1,0) \), and \( (0,0,1) \).
- The vector field:
  \[
  \mathbf{F} = 3xy \mathbf{e_x} + y^2 \mathbf{e_y} - x^2 y^4 \mathbf{e_z}
  \]
  
### Approach: Divergence Theorem
The **Divergence Theorem** relates the flux of a vector field through a closed surface to the volume integral of the divergence of the vector field inside the surface.

The Divergence Theorem states:
\[
\iint_S \mathbf{F} \cdot d\mathbf{S} = \iiint_V (\nabla \cdot \mathbf{F}) dV
\]
where \( V \) is the volume enclosed by the surface \( S \).

#### Step 1: Compute the Divergence of \( \mathbf{F} \)
The divergence of the vector field \( \mathbf{F} = 3xy \mathbf{e_x} + y^2 \mathbf{e_y} - x^2 y^4 \mathbf{e_z} \) is given by:
\[
\nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(3xy) + \frac{\partial}{\partial y}(y^2) + \frac{\partial}{\partial z}(-x^2 y^4)
\]

Let's compute these partial derivatives:
1. \( \frac{\partial}{\partial x}(3xy) = 3y \)
2. \( \frac{\partial}{\partial y}(y^2) = 2y \)
3. \( \frac{\partial}{\partial z}(-x^2 y^4) = 0 \) (since there is no \( z \)-dependence in that term)

Thus, the divergence is:
\[
\nabla \cdot \mathbf{F} = 3y + 2y = 5y
\]

#### Step 2: Set up the Volume Integral
The volume \( V \) of the tetrahedron is bounded by the planes:
- \( x = 0 \)
- \( y = 0 \)
- \( z = 0 \)
- \( x + y + z = 1 \)

The volume integral becomes:
\[
\iiint_V 5y \, dV = 5 \iiint_V y \, dV
\]

We will integrate over the volume of the tetrahedron using the limits imposed by the vertices.

#### Step 3: Set up the Limits of Integration
For the given tetrahedron, the limits of integration are:
- \( z \) goes from 0 to \( 1 - x - y \)
- \( y \) goes from 0 to \( 1 - x \)
- \( x \) goes from 0 to 1

Thus, the volume integral is:
\[
5 \int_0^1 \int_0^{1-x} \int_0^{1-x-y} y \, dz \, dy \, dx
\]

#### Step 4: Compute the Integral
First, integrate with respect to \( z \):
\[
\int_0^{1-x-y} y \, dz = y(1 - x - y)
\]

Now, integrate with respect to \( y \):
\[
\int_0^{1-x} y(1 - x - y) \, dy
\]
This requires expanding the integrand:
\[
y(1 - x - y) = y(1 - x) - y^2
\]

So the integral becomes:
\[
\int_0^{1-x} \left[ y(1 - x) - y^2 \right] dy
\]

Compute this integral term by term:
1. \( \int_0^{1-x} y(1 - x) \, dy = (1 - x) \cdot \frac{(1-x)^2}{2} = \frac{(1-x)^3}{2} \)
2. \( \int_0^{1-x} y^2 \, dy = \frac{(1-x)^3}{3} \)

Thus, the result of the \( y \)-integration is:
\[
\frac{(1-x)^3}{2} - \frac{(1-x)^3}{3} = \frac{(1-x)^3}{6}
\]

Finally, integrate with respect to \( x \):
\[
5 \int_0^1 \frac{(1-x)^3}{6} dx = 5 \cdot \frac{1}{24} = \frac{5}{24}
\]

#### Step 5: Conclusion
The outward flux of the vector field through the surface \( S \) is \( \frac{5}{24} \).

Would you like more details on any step, or do you have any specific questions?

Here are 5 related questions you might explore:
1. How does the Divergence Theorem apply to different types of surfaces?
2. What is the physical interpretation of the divergence of a vector field?
3. How would the solution change if the vector field had a non-zero \( z \)-component?
4. Can the Divergence Theorem be applied in cylindrical or spherical coordinates?
5. How do you compute flux through non-planar surfaces?

**Tip**: Always check if the surface is closed when applying the Divergence Theorem. A non-closed surface might require a different approach.

Let S be the closed surface that consists of the four faces of the tetrahedron with vertices (0,0,0), (1,0,0), (0,1,0) and (0,0,1). Calculate the outward flux through S of the vector field F = 3xy e_x + y^2 e_y - x^2 y^4 e_z.

This problem explores how to calculate the outward flux of a vector field through the surface of a tetrahedron using the Divergence Theorem. The problem involves a vector field F = 3xy e_x + y^2 e_y - x^2 y^4 e_z and a step-by-step application of multivariable calculus techniques such as divergence and volume integration.

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