The divergence theorem states that the volume integral of the divergence of a vector field F over a volume V is equal to the surface integral of F over the boundary surface S of V. Specifically, if div F = 1, then ∫∫∫V div F dV equals ∫∫S F(dot)Ndσ, resulting in the volume V. However, if div F = 2, the equation ∫∫∫V div F dV does not equal 2∫∫S F(dot)Ndσ; rather, the volume integral becomes 2V, confirming that an extra constant cannot be inserted into the divergence theorem.
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randomcat said:Homework Statement
The divergence theorem states that
∫∫∫V div F dV = ∫∫S F(dot)Ndσ
Suppose that div F = 1, then
∫∫∫V div F dV = ∫∫S F(dot)Ndσ
If divF = 2, does the following hold true?
∫∫∫V div F dV = 2∫∫S F(dot)Ndσ
Homework Equations
Since the divergence theorem computes the volume, if div F is a constant, then the volume formed by the closed surface would just be multiplied by that constant?
The Attempt at a Solution