$U \in \mathbb{R}^2$ is a bounded region with a smooth edge, such that $0 \in U$, $f \in C_0^{\infty}(U)$$$V(x) = \begin{cases} \dfrac{x}{\|x\|^2} & \text{for } x \neq 0 \\ 0 & \text{for } x = 0 \end{cases}$$Prove that:$$\int_U (V, \nabla f) \ d\lambda_2 = -2 \pi f(0)$$
Hint: think about the region $U_t = U \ \setminus \ \{ \|x\| \leq \epsilon \}$
I found such exercise in the set of exercises preparing for an exam. I don't know how to start since I don't understand the idea of integrating over function $V$ on one dimension and gradient of $f$ on the other. Any help would be much appreciated.
As I understand it, $$\nabla f = \left( \frac{\partial f}{\partial x_1} \ , \ \frac{\partial f}{\partial x_2} \right)$$
We have:$$V(x) = \frac{x}{\|x\|^2} = \left( \frac{x_1}{x_1^2 + x_2^2} \ , \ \frac{x_2}{x_1^2 + x_2^2} \right)$$
The dot product is given by:$$(V, \nabla f) = \frac{\partial f}{\partial x_1} \frac{x_1}{x_1^2 + x_2^2} + \frac{\partial f}{\partial x_2} \frac{x_2}{x_1^2 + x_2^2} = \frac{\dfrac{\partial f}{\partial x_1} x_1 + \dfrac{\partial f}{\partial x_2} x_2}{x_1^2 + x_2^2}$$
Therefore we have:$$(V, \nabla f) = \begin{cases} \dfrac{\dfrac{\partial f}{\partial x_1} x_1 + \dfrac{\partial f}{\partial x_2} x_2}{x_1^2 + x_2^2} & \text{for } x \neq 0 \\ 0 & \text{for } x = 0 \end{cases}$$Right now, as I understand it, I would like to show that:$$\int_D \frac{\frac{\partial f}{\partial x_1} x_1 + \frac{\partial f}{\partial x_2} x_2}{x_1^2 + x_2^2} \lambda_2$$
Is equal to:$$−\int_D f \nabla \cdot V \ d \lambda_2 + \int_{\partial D} fV \cdot ndS$$
But while calculating $\int_D V \cdot \nabla f \ d \lambda_2$ I don't get why:$$\int_U (V, \nabla f) \ d\lambda_2 = \int_U (\nabla f, V) \ d\lambda_2$$
I guess dot product is commutative...