Double integral of $-yf_x+xf_y$ on a rotating disc has zero value

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Question:

$u(x,y)=-yf_x+xf_y,f\in C^1(\mathbb{R^2})$ ,

$I(\alpha)=\iint_{D_\alpha}u(x,y)dxdy,D_\alpha\colon(x-2\cos\alpha)^2+(y-2\sin\alpha)^2\leq 1$ .

Prove:$\exists \alpha,I(\alpha)=0$ .

Attempt:

Applying Green's formula we get

$I(\alpha)=2\int_0^{2\pi}(yf(x,y)\cos\alpha-xf(x,y)\sin\alpha)dt \\=2\int_0^{2\pi}(f(x,y)\cos\alpha\sin t-f(x,y)\sin\alpha\cos t)dt,\\(x,y)=(2\cos\alpha+\cos t,2\sin\alpha+\sin t)$

$I(\alpha)=0\iff \cos\alpha\int_0^{2\pi}f(x,y)\sin tdt=\sin\alpha\int_0^{2\pi}f(x,y)\cos tdt$