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I'm currently studying vector calculus and have a question about the application of Gauss's Theorem (also known as the divergence theorem).

When calculating the surface integral in Gauss's Theorem, we use a normal vector. My question is about the normalization of this vector:

  1. Does the normal vector always need to be a unit vector (normalized) when applying Gauss's Theorem?

  2. If the normal vector is not normalized, I've noticed that the result of the surface integral changes. However, the result of Gauss's Theorem itself remains the same. Could someone explain why this is the case?

I'm trying to understand the underlying mathematics that allows Gauss's Theorem to be invariant to the normalization of the normal vector, while the surface integral is affected by it.

Sam12
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    Add some mathematical details to the questions. – Sillyasker Dec 21 '23 at 14:02
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    One has to guess a certain amount of intent (e.g., "the result of Gauss's theorem" refers to a volume integral of the divergence?), but your last sentence appears already to answer your question: Yes, it matters that we use (i) a unit normal field, and further (ii) that this field points outward from the solid bounded by a closed surface in three-space. (The surface integral is a flux, the integral of a normal component of a given vector field $F$; that normal component may be viewed as the dot product of $F$ with an outward unit normal field.) <> Is that what you're asking? – Andrew D. Hwang Dec 21 '23 at 14:55
  • This answer goes to the bottom of the question. What applies to Stokes also applies to Gauss. In short: when you use symmetry to bypass the integration you normalize. When you parametrize to carry out the surface integral you don't. Otherwise you could not change variables. – Kurt G. Dec 21 '23 at 17:08

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