Question

The divergence of \[ \left(2x^{2}z\,\hat{i}+xy^{2}z\,\hat{j}+3yz^{2}\,\hat{k}\right) \] at the point \((1,1,1)\) is

• A. \(-8\)
• B. \(0\)
• C. \(8\)
• D. \(12\)

Hint:

For \[ \vec{F}=P\hat{i}+Q\hat{j}+R\hat{k}, \] \[ \boxed{ \nabla\cdot\vec{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}. } \]

Answer 1

The Correct Option is D

Step 1: Use the divergence formula. For \[ \vec{F}=P\hat{i}+Q\hat{j}+R\hat{k}, \] \[ \nabla\cdot\vec{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}. \] Here, \[ P=2x^{2}z,\qquad Q=xy^{2}z,\qquad R=3yz^{2}. \]

Step 2: Differentiate and substitute the point. \[ \frac{\partial P}{\partial x}=4xz, \] \[ \frac{\partial Q}{\partial y}=2xyz, \] \[ \frac{\partial R}{\partial z}=6yz. \] Thus, \[ \nabla\cdot\vec{F} = 4xz+2xyz+6yz. \] At \[ (x,y,z)=(1,1,1), \] \[ \nabla\cdot\vec{F} = 4+2+6 = 12. \] Hence, \[ \boxed{12} \] Therefore, \[ \boxed{(D)} \] is the correct answer.