Divergence Theorem: Does Multiplying div F Multiply Volume?

[randomcat]

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

 

[Dick]

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

The divergence theorem always holds, the third equation doesn't hold. What is true is that if div(F)=1, then the volume integral is V. If div(F)=2 then the volume integral is 2V. You don't insert an extra constant into the divergence theorem.