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Question Number 6174 by madscientist last updated on 17/Jun/16
prove the identity  ∫_∂D φ▽φ∙nds=∫∫_D (φ▽^(2 ) φ+▽φ∙▽φ)dA
$${prove}\:{the}\:{identity} \\ $$$$\int_{\partial{D}} \phi\bigtriangledown\phi\centerdot{nds}=\int\int_{{D}} \left(\phi\bigtriangledown^{\mathrm{2}\:} \phi+\bigtriangledown\phi\centerdot\bigtriangledown\phi\right){dA} \\ $$

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