VC3.40

From Exampleproblems

We have the identity, $\nabla\times(f\times g)=f(\nabla\cdot g)-g(\nabla\cdot f)+(g\cdot\nabla)f-(f\cdot\nabla)g\,$ --(1)

By putting f=F and g=r in (1),we have

$\nabla\times(F\times r)=F(\nabla\cdot r)-r(\nabla\cdot F)+(r\cdot\nabla)F-(F\cdot\nabla)r\,$ --(2)

Now, $\nabla\cdot r=\nabla\cdot(xi+yj+zk)=\frac{\partial}{\partial x}(x)+\frac{\partial}{\partial y}(y)+\frac{\partial}{\partial z}(z)=3\,$ --(3)

Let $F=F_1i+F_2j+F_3k\,$ --(4)

Therefore, $F\cdot\nabla=(F_1i+F_2j+F_3k)\cdot(i\frac{\partial}{\partial x}+j\frac{\partial}{\partial y}+k\frac{\partial}{\partial z})=F_1\frac{\partial}{\partial x}+F_2\frac{\partial}{\partial y}+\frac{\partial}{\partial z}\,$ --(5)

and so $(F\cdot\nabla)r=(F_1\frac{\partial}{\partial x}+F_2\frac{\partial}{\partial y}+\frac{\partial}{\partial z})(xi+yj+zk)\,$

= $F_1i+F_2j+F_3k\,$ ,by (4) --(6)

Using (3) and (6),(2) becomes $\nabla\times(F\times r)=3F-r(\nabla\cdot f)+(r\cdot\nabla)F-F=2F-(\nabla\cdot F)r+(r\cdot\nabla)F\,$

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