VC3.21

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{\mathrm  {curl}}f={\begin{vmatrix}i&j&k\\{\frac  {\partial }{\partial x}}&{\frac  {\partial }{\partial y}}&{\frac  {\partial }{\partial z}}\\(x+y+1)&1&(-x-y)\end{vmatrix}}\,

=[{\frac  {\partial }{\partial y}}(-x-y)-{\frac  {\partial }{\partial z}}(1)]i-[{\frac  {\partial }{\partial x}}(-x-y)-{\frac  {\partial }{\partial z}}(x+y+1)]j+[{\frac  {\partial }{\partial x}}(1)-{\frac  {\partial }{\partial y}}(x+y+1)]k\,

=(-1-0)i-(-1-0)j+(0-1)k=-i+j-k\,

Now,f\cdot {\mathrm  {curl}}f=[(x+y+1)i+j+(-x-y)k]\cdot (-i+j-k)\,

=-(x+y+1)+1+(-1)(-x-y)=0\,

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