Let $x = r\cos \theta ,\,y = r\sin \theta$, therefore ${x^2} + {y^2} = {r^2}$. In polar- coordinates ${V_r} = u\cos \theta  + v\sin \theta$ and ${V_\theta } =  – u\sin \theta  + v\cos \theta$

Hence, 

$u = \frac{{5y}}{{{x^2} + {y^2}}} = \frac{{5r\sin \theta }}{{{r^2}}} = \frac{{5\sin \theta }}{r}$ ,  $v =  – \frac{{5x}}{{{x^2} + {y^2}}} =  – \frac{{5r\cos \theta }}{{{r^2}}} =  – \frac{{5\cos \theta }}{r}$

${V_r} = \frac{{5\sin \theta }}{r}\left( {\cos \theta } \right) + \left( { – \frac{{5\cos \theta }}{r}} \right)\sin \theta  = 0$ , ${V_\theta } =  – \frac{{5\sin \theta }}{r}\sin \theta  + \left( { – \frac{{5\cos \theta }}{r}} \right)\cos \theta  =  – \frac{5}{r}$

$V.ds = \left( {{V_r}{e_r} + {V_\theta }{e_\theta }} \right).\left( {dr{e_r} + rd\theta {e_\theta }} \right)$$= \left( {{V_r}dr + r{V_\theta }d\theta } \right)$$= 0 + r\left( { – \frac{5}{r}} \right)d\theta$$=  – 5d\theta$

Therefore, $\tau  =  – \oint_c {V.ds}  =  – \int\limits_0^{2\pi } { – 5d\theta }  = 5\int\limits_0^{2\pi } {d\theta }$$= 5 \times 2\pi \ = \,10\,{m^2}/s$.

Here, value of circulation, is independent of diameter of circular path.