Divergence of vector field \[\overrightarrow {\rm{V}} \left( {{\rm{x}},\,{\rm{y}},\,{\rm{z}}} \right) = – \left( {{\rm{x}}\cos {\rm{xy}} + {\rm{y}}} \right){\rm{\hat i}} + \left( {{\rm{y}}\cos {\rm{xy}}} \right){\rm{\hat j}} + \left[ {\left( {\sin {{\rm{z}}^2}} \right) + {{\rm{x}}^2} + {{\rm{y}}^2}} \right]{\rm{\hat k}}\] is A. 2z cos z2 B. sin xy + 2z cos z2 C. x sin xy – cos z D. None of these

The divergence of a vector field $\overrightarrow{V}(x, y, z)$ is defined as:

$$\nabla \cdot \overrightarrow{V} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}$$

where $P$, $Q$, and $R$ are the $x$, $y$, and $z$ components of $\overrightarrow{V}$, respectively.

In this case, we have:

$$\overrightarrow{V}(x, y, z) = -(x \cos xy + y) \hat{\imath} + (y \cos xy) \hat{\jmath} + \left[ (\sin z^2) + x^2 + y^2 \right] \hat{k}$$

Therefore, the divergence of $\overrightarrow{V}$ is:

$$\begin{align}
\nabla \cdot \overrightarrow{V} &= \frac{\partial}{\partial x} \left[ -(x \cos xy + y) \right] + \frac{\partial}{\partial y} \left[ (y \cos xy) \right] + \frac{\partial}{\partial z} \left[ (\sin z^2) + x^2 + y^2 \right] \
&= -\cos xy – \frac{\partial}{\partial y} \left[ y \cos xy \right] + 2z \cos z^2 \
&= -\cos xy – y \sin xy + 2z \cos z^2 \
&= \boxed{-\cos xy + y \sin xy + 2z \cos z^2}.
\end{align}$$

Here is a brief explanation of each option:

• Option A: $2z \cos z^2$. This is the divergence of the vector field $\overrightarrow{V}(x, y, z) = 2z \cos z^2 \hat{k}$. However, the given vector field is $\overrightarrow{V}(x, y, z) = -(x \cos xy + y) \hat{\imath} + (y \cos xy) \hat{\jmath} + \left[ (\sin z^2) + x^2 + y^2 \right] \hat{k}$, which is different from $\overrightarrow{V}(x, y, z) = 2z \cos z^2 \hat{k}$. Therefore, option A is incorrect.
• Option B: $\sin xy + 2z \cos z^2$. This is the divergence of the vector field $\overrightarrow{V}(x, y, z) = \sin xy + 2z \cos z^2 \hat{k}$. However, the given vector field is $\overrightarrow{V}(x, y, z) = -(x \cos xy + y) \hat{\imath} + (y \cos xy) \hat{\jmath} + \left[ (\sin z^2) + x^2 + y^2 \right] \hat{k}$, which is different from $\overrightarrow{V}(x, y, z) = \sin xy + 2z \cos z^2 \hat{k}$. Therefore, option B is incorrect.
• Option C: $x \sin xy – \cos z$. This is the divergence of the vector field $\overrightarrow{V}(x, y, z) = x \sin xy – \cos z \hat{k}$. However, the given vector field is $\overrightarrow{V}(x, y, z) = -(x \cos xy + y) \hat{\imath} + (y \cos xy) \hat{\jmath} + \left[ (\sin z^2) + x^2 + y^2 \right] \hat{k}$, which is different from $\overrightarrow{V}(x, y, z) = x \sin xy – \cos z \hat{k}$. Therefore, option C is incorrect.
• Option D: None of these. This is the correct answer, as none of the other options are the divergence of the given vector field.