Prove constant function using Gauss Theorem

Question

Given $\vec F = \nabla f$ and f is a smooth function from $\mathbb{R}^3$ to $\mathbb{R}$. And, $$\iiint_B f(\nabla \cdot \vec F) \,dV = \iint_{\partial B} (f\vec F) \cdot d\vec S $$ where $B$ is any open ball and $\partial B$ oriented with outward normal.

How can I prove $f$ is a constant function? Where should I start?

Answer 1

Using Gauß's theorem, we have $$ \int_{\partial B} fF\cdot n \, dS = \int_B \nabla \cdot (fF)\, dV $$ Note, that $$ \nabla \cdot (fF) = \nabla f \cdot F + f \nabla \cdot F $$ Hence, $$ \int_B \nabla f \cdot F \, dV = 0 $$ As $F = \nabla f$, we have $$ \int_B |\nabla f|^2\, dV = 0. $$ As $f$ is smooth, $\nabla f$ is continuous, hence the above implies (since $|\nabla f|^2 \ge 0$), that $\nabla f = 0$. Hence, $f$ is constant.

Answer 2

$$\iint_{\partial B} (f\vec F)\cdot d\vec S = \iiint_B \nabla\cdot (f\vec F)dV = \iiint_B \nabla f \cdot (\vec F) + f(\nabla \cdot \vec F)dV = \iiint_Bf(\nabla \cdot \vec F)dV$$ $$\Rightarrow \iiint_B\nabla f\cdot (\vec F)dV=0 \Rightarrow \iiint_B\nabla f\cdot (\nabla f)dV=0 \Rightarrow \iiint_B||\nabla f||^2dV=0 $$

since $||\nabla f||^2≥0 \text{ }\forall f,$

$$\iiint_B||\nabla f||^2dV=0 \Rightarrow ||\nabla f||^2=0 \Rightarrow \nabla f =\vec 0 \Rightarrow f=c$$