A particle of mass $m$ moves along the quarter section of the circular path whose centre is at the origin. The radius of the circular path is $a$. A force $\vec{F}=y\hat{i}-x\hat{j}$ newton acts on the particle, where $x,y$ denote the coordinates of position of the particle. The work done by this force in taking the particle from point A ($a,0$) to point B ($0,a$) along the circular path is

$W=\int \mathbf{F}.\mathbf{d}\mathbf{r}=\int (y\hat{i}-x\hat{j}).(dx\hat{i}+dy\hat{j})$

$=\int (ydx-xdy)$         .........(1)

$\because x^2+y^2=a^2\therefore xdx+ydy=0$

$\Rightarrow W=\int \left(y\left(\frac{-ydy}{x}\right)-xdy\right)=-\int \frac{x^2+y^2}{x}dy$

$=-{\int }_0^a\frac{a^2}{\sqrt{a^2-y^2}}dy=-\frac{\pi a^2}{2}$

Alternate Method :
It can be observed that the force is tangent to the curve at
each point and the magnitude is constant. The direction of
force is opposite to the direction of motion of the particle.
$\therefore$  work done = (force) (distance)
$=-\sqrt{x^2+y^2}=\frac{\pi a}{2}=-a\times \frac{\pi a}{2}=-\frac{\pi a^2}{2}J$
$W=-\frac{\pi a^2}{2}J$