Suppose $f$ is a real function over $\mathbb{R}^3$ and satisfies $f(tx)=t^kf(x)$ where $k$ is a positive integer and $x\in \mathbb{R}^3$. Let $B(0,1)$ be a unit ball and $\partial B(0,1)$ being its surface. Proove: $$ \int_{B(0,1)} \Delta f\ dxdydz = k \int_{\partial B(0,1)} f \mu_{\partial B(0,1)} $$ where $\mu_{\partial B(0,1)}$ is the volume-form over the surface.
I tried $$ \int_{B(0,1)} \Delta f\ dxdydz = \int_{B(0,1)} \nabla \cdot \nabla f\ dxdydz = \int_{\partial B(0,1)} \nabla f \cdot \vec{n}\ \mu_{\partial B(0,1)} $$ But I don't know how to get $\nabla f \cdot \vec{n} = kf$ if possible. As I know, $\vec{n}=\frac{(x,y,z)}{\sqrt{x^2+y^2+z^2}}$, but I don't know how to proceed. Any help is appreciated!