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Question

Let F\mathbf{F} be a constant vector field. Show that

λSFndS=0\iint_{\lambda S} \mathbf{F} \cdot \mathbf{n} d S=0

for any "nice" solid S. What should we mean by "nice"?

Solution

Verified
Answered 2 years ago
Answered 2 years ago

Since F=(a,b,c){\bf F}=(a,b,c) is a constant vector field, then:

F=(a)x+(b)y+(c)z=0\nabla\cdot{\bf F}=\dfrac{\partial (a)}{\partial x}+\dfrac{\partial (b)}{\partial y}+\dfrac{\partial (c)}{\partial z}=0

Then, using the Gauss's Divergence Theorem we get:

δSFndS=VFdV=0\iint_{\delta S}{\bf F}\cdot{\bf n}dS=\iiint_V\nabla\cdot{\bf F}dV=0

A "nice" solid is a solid enclosed by a smooth and orientable closed surface.

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