Show that $\iiint_{\mathbb{R}^3}(\nabla f)\cdot(\operatorname{curl}\textbf{X})dV = 0$.

Question

Let $f:\mathbb{R}^3\rightarrow \mathbb{R}$ be a smooth function with compact support and let $\textbf{X}$ be a smooth vector field on $\mathbb{R}^3$. Show that

$$\iiint_{\mathbb{R}^3}(\nabla f)\cdot(\operatorname{curl}\textbf{X})dV = 0$$

So far, I understand that letting $F = \operatorname{curl}\textbf{X}$,

$$\nabla f\cdot F = \nabla \cdot (fF) - f(\nabla\cdot F) = \nabla \cdot (fF)$$

because the gradient of the curl is $0$. Hence, we have

$$\iiint_{\mathbb{R}^3}(\nabla f)\cdot(\operatorname{curl}\textbf{X})dV = \iiint_{\mathbb{R}^3}\nabla \cdot (fF)dV$$

I assume I am to use divergence theorem here, but I'm not sure exactly where to go with this.

Any help appreciated!!

Answer 1

In your argument it's not that the gradient of the curl is zero, but rather than the divergence of a curl is zero. With this in hand you use the fact that $f$ has compact support. This allows us to say that $f=0$ outside of $B(0,R)$ for some $R>0$. Then $$ \iiint_{\mathbb{R}^3} \nabla f \cdot \text{curl}X = \iiint_{B(0,R)} \nabla f \cdot \text{curl}X = \iiint_{B(0,R)} \nabla \cdot( f \text{curl}X) \\ = \iint_{\partial B(0,R)} f \text{curl}X \cdot \nu =0 $$ where the second-to-last equality is from the divergence theorem and the last equality is from the fact that $f=0$ on $\partial B(0,R)$.