Let v be a vector field on $R^2 \ (0, 0)$ of the form v(x, y) = f(r)(x, y), where $r =\sqrt{x^2 + y^2}$ and f : (0, ∞) → R is continuously differentiable.
Let $C_{R}$ be the circle of radius R centred at the origin. Evaluate the counterclockwise flux ¸ $$\oint_C v.n \,ds$$ in terms of f.
So far I have done: $For (x, y) ∈ C_{R}, x2 + y^2 = R^2$ and n(x, y) = $\frac{1}{R}$ (x, y). Therefore $\oint_C v.n \,ds$$ = $$\oint_C \frac{f(R)}{R}(x,y).(x,y) \,ds$ = $2\pi Rf(R)$.
But the solution is $2\pi R^2f(R)$. Where did the other R come from?
$\oint_C \frac{f(R)}{R}(x,y).(x,y) \,ds$ is correct.
But $(x,y).(x,y)=R^2$ , not $R$ (see dot product ) !
So : $\oint_C \frac{f(R)}{R}(x,y).(x,y) \,ds=\oint_C Rf(R) \,ds = Rf(R) \oint_C \,ds = Rf(R) \cdot 2\pi R =2\pi R^2f(R)$
PS : $\oint_C \,ds $ is of course just the circumference of the circle : $2 \pi R$.