1. **State the problem:** We are given a vector field \( \vec{F} = (6xy + z^3)\mathbf{i} + 3x^2(-z)\mathbf{j} + 3xz^2(-y)\mathbf{k} \). We need to prove that \( \vec{F} \) is irrotational and find its scalar potential function \( \phi \) such that \( \vec{F} = \nabla \phi \).
2. **Recall the definition of irrotational:** A vector field \( \vec{F} = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k} \) is irrotational if its curl is zero:
$$
\nabla \times \vec{F} = \mathbf{0}
$$
where
$$
\nabla \times \vec{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right)\mathbf{i} - \left( \frac{\partial R}{\partial x} - \frac{\partial P}{\partial z} \right)\mathbf{j} + \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right)\mathbf{k}
$$
3. **Identify components:**
$$
P = 6xy + z^3, \quad Q = 3x^2(-z) = -3x^2 z, \quad R = 3xz^2(-y) = -3xyz^2
$$
4. **Compute partial derivatives:**
$$
\frac{\partial R}{\partial y} = \frac{\partial}{\partial y}(-3xyz^2) = -3xz^2
$$
$$
\frac{\partial Q}{\partial z} = \frac{\partial}{\partial z}(-3x^2 z) = -3x^2
$$
$$
\frac{\partial R}{\partial x} = \frac{\partial}{\partial x}(-3xyz^2) = -3yz^2
$$
$$
\frac{\partial P}{\partial z} = \frac{\partial}{\partial z}(6xy + z^3) = 3z^2
$$
$$
\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(-3x^2 z) = -6xz
$$
$$
\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(6xy + z^3) = 6x
$$
5. **Calculate curl components:**
$$
\left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right) = (-3xz^2) - (-3x^2) = -3xz^2 + 3x^2
$$
$$
- \left( \frac{\partial R}{\partial x} - \frac{\partial P}{\partial z} \right) = - \left( -3yz^2 - 3z^2 \right) = -(-3yz^2 - 3z^2) = 3yz^2 + 3z^2
$$
$$
\left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) = (-6xz) - 6x = -6xz - 6x
$$
6. **Check if curl is zero:**
The curl is zero if all components are zero:
- First component: \( -3xz^2 + 3x^2 \) is not generally zero.
- Second component: \( 3yz^2 + 3z^2 \) is not generally zero.
- Third component: \( -6xz - 6x \) is not generally zero.
Since the curl is not zero, \( \vec{F} \) is **not** irrotational.
7. **Conclusion:** The vector field \( \vec{F} \) is not irrotational, so it does not have a scalar potential.
**Final answer:** \( \vec{F} \) is not irrotational and has no scalar potential.
Irrotational Check C4161F
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