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QUESTION

Show that

(a)\ \ $\nabla\times(\nabla\phi)=\textbf{0}$\ \ (b)\ \
$\nabla\cdot(\nabla\times \textbf{a})=0$

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ANSWER
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\item[(a)]
 $\textbf{a}=(x^2y,-2xz,2yz)\\
 \nabla\times \textbf{a}=(2z-2x,0,-2z-x^2)\\
 \nabla\times (\nabla\times
 \textbf{a})=(0,2+2x,0)$

\item[(b)]
 Same vector $\textbf{a},\ \
 \nabla\cdot\textbf{a}=2xy+2y\\
 \nabla(\nabla\cdot\textbf{a})=(2y,2x+2,0)\\
 \nabla^2\textbf{a}=(2y,0,0)\\
 \nabla(\nabla\cdot\textbf{a})-\nabla^2\textbf{a}=(0,2x+2,0)$\\
 This should agree with 2(a) by a general formula!

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