Show that if $C=\{(x,y) : (x-a)^2+y^2=b^2\}$ for some $b

Question

Show that if $C=\{(x,y) : (x-a)^2+y^2=b^2\}$ for some $b<a$ (in other words, $C$ is the circle in the plane centered at $(a,0)$ of radius $b<a$), then,$$\oint\limits_C \left[ \frac{x \hat x + y \hat y}{(x^2+y^2)^{3/2}} \right] ds = \frac{\hat x}{a}.$$Sidenote: yes, $ds$ is purposely a scalar and not the vector $d \vec s$, even though we're integrating a vector field.

anonymous · Answer

Just so this doesn't look so daunting and arbitrary, if we switch to polar coordinates for a second, the integrand becomes simply:$$\frac{x\hat x+y\hat y}{(x^2+y^2)^{3/2}}\equiv \frac{\hat r}{r^2}$$