To prove that a vector field $\mathbf{A} = xyz^2 \hat{i}$ is solenoidal, we need to show that the divergence of $\mathbf{A}$ is zero.

A vector field $\mathbf{A} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}$ is solenoidal if:

$\nabla \cdot \mathbf{A} = \frac{\partial A_x}{\partial x} + \frac{\partial A_y}{\partial y} + \frac{\partial A_z}{\partial z} = 0$

Given the vector field $\mathbf{A} = xyz^2 \hat{i}$, the components of $\mathbf{A}$ are:

• $A_x = xyz^2$
• $A_y = 0$
• $A_z = 0$

Now, let's compute the divergence $\nabla \cdot \mathbf{A}$:

$\nabla \cdot \mathbf{A} = \frac{\partial A_x}{\partial x} + \frac{\partial A_y}{\partial y} + \frac{\partial A_z}{\partial z}$

Substituting the components:

$\nabla \cdot \mathbf{A} = \frac{\partial}{\partial x} (xyz^2) + \frac{\partial}{\partial y} (0) + \frac{\partial}{\partial z} (0)$

$\nabla \cdot \mathbf{A} = \frac{\partial}{\partial x} (xyz^2)$

Since $y$ and $z^2$ are constants with respect to $x$, we differentiate $xyz^2$ with respect to $x$:

$\nabla \cdot \mathbf{A} = yz^2$

Given that $yz^2$ is not zero (unless $y$ or $z$ is zero), the divergence $\nabla \cdot \mathbf{A}$ is not zero. Hence, the vector field $\mathbf{A} = xyz^2 \hat{i}$ is not solenoidal.

Summary:

• The divergence of the given vector field $\mathbf{A} = xyz^2 \hat{i}$ is $yz^2$.
• Since the divergence is not zero, the vector field is not solenoidal.

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

Related Questions:

1. What is the physical significance of a solenoidal vector field?
2. How would you prove a vector field is irrotational?
3. Can a vector field be both solenoidal and irrotational? Explain.
4. What is the relationship between solenoidal vector fields and incompressible fluid flow?
5. How would you compute the divergence in spherical coordinates?

Tip:

For a vector field to be solenoidal, always remember that its divergence must be zero everywhere. This property is often related to the conservation of mass in fluid dynamics.