VC3.19

From Exampleproblems

Given, $f=xy^2i+2x^2yzj-3yz^2k\,$

Therefore $\mathrm{div}f=\frac{\partial}{\partial x}(xy^2)+\frac{\partial}{\partial y}(2x^2yz)+\frac{\partial}{\partial z}(-3yz^2)\,$

= $y^2+2x^2z-6yz\,$

Therefore, $\mathrm{div}f\,$ at (1,-1,1)= $(-1)^2+2(1)^2(1)-6(-1)(1)=9\,$

Now, $\mathrm{curl}f=\begin{vmatrix} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} & \\ xy^2 & 2x^2yz & -3yz^2 \end{vmatrix}\,$

= $[\frac{\partial}{\partial y}(-3yz^2)-\frac{\partial}{\partial z}(2x^2yz)]i-[\frac{\partial}{\partial x}(-3yz^2)-\frac{\partial}{\partial z}(xy^2)]j+[\frac{\partial}{\partial x}(2x^2yz)-\frac{\partial}{\partial y}(xy^2)]k\,$