To find the divergence of the vector field $\mathbf{F}(x, y) = 4x^2 \mathbf{i} + 2y^2 \mathbf{j}$, we use the formula for divergence in two dimensions:

$\text{div}(\mathbf{F}) = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y},$

where:

• $F_1 = 4x^2$ (the $i$-component of the vector field),
• $F_2 = 2y^2$ (the $j$-component of the vector field).

Step-by-Step Solution:

1. Differentiate $F_1$ with respect to $x$: $\frac{\partial}{\partial x}(4x^2) = 8x.$

2. Differentiate $F_2$ with respect to $y$: $\frac{\partial}{\partial y}(2y^2) = 4y.$

3. Add the results: $\text{div}(\mathbf{F}) = 8x + 4y.$

Final Answer:

$\text{div}(\mathbf{F}) = 8x + 4y.$

Let me know if you want further clarification or additional help!

Related Questions:

1. What does the divergence of a vector field represent physically?
2. How is divergence extended to three-dimensional vector fields?
3. Can you explain the significance of a zero-divergence field?
4. How would you calculate the curl of this vector field instead of the divergence?
5. What is the difference between divergence and gradient?

Tip:

When working with vector fields, always confirm whether you're calculating divergence, curl, or gradient to avoid confusion!