Question

If \( \vec{\mathbf{F}} (x, y, z) = (2x + 3yz) \hat{i} + (3xz + 2y) \hat{j} + (3xy + 2z) \hat{k} \) for \( (x, y, z) \in \mathbb{R}^3 \), then which among the following is (are) TRUE? 
 

• A. \( \nabla \times \vec{\mathbf{F}} = \vec{\mathbf{0}}\)
• B. \( \oint_C \vec{\mathbf{F}} \cdot d\vec{\mathbf{r}} = 0 \) along any simple closed curve \( C \)
• C. There exists a scalar function \( \phi : \mathbb{R}^3 \to \mathbb{R} \) such that \( \nabla \cdot \vec{\mathbf{F}} = \phi_{xx} + \phi_{yy} + \phi_{zz} \)
• D. \( \nabla \cdot \vec{\mathbf{F}} = 0 \)

Hint:

For conservative vector fields, the curl is zero, and the line integral over any closed curve is zero.

Answer 1

The Correct Option is A, B, C

\[\vec F=(2x+3yz, 3xz+2y, 3xy+2z)\]

Curl
\[\nabla\times\vec F=\begin{vmatrix} \mathbf i&\mathbf j&\mathbf k \\ \partial_x&\partial_y&\partial_z \\ 2x+3yz&3xz+2y&3xy+2z \end{vmatrix} =(0,0,0)\]
So (A) true.

Since \(\mathbb R^3\) is simply connected and \(\nabla\times\vec F=0\), \(\vec F\) is conservative, hence
\(\displaystyle \oint_C \vec F\cdot d\vec r=0\).
So (B) true.

Divergence
\[\nabla\cdot\vec F=2+2+2=6\neq0\]
So (D) false.

For (C), since \(\nabla\cdot\vec F=6\), take \(\phi=x^2+y^2+z^2\), then
\[\phi_{xx}+\phi_{yy}+\phi_{zz}=6=\nabla\cdot\vec F\]
So (C) true.

Correct options: A, B, C