VC3.19

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

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

=y^{2}+2x^{2}z-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^{2}yz&-3yz^{2}\end{vmatrix}}\,

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

=(-3z^{2}-2x^{2}y)i-(0-0)j+(4xyz-2xy)k\,

=-(3z^{2}+2x^{2}y)i+2xy(2z-1)k\,

{\mathrm  {curl}}f\, at (1,-1,1)

=-(3-2)i-2(2-1)k=-i-2k\,

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