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**HallsofIvy** This is not a vector field. Did you mean $\displaystyle -y\vec{i}+ x\vec{j}$?

Assuming the the vector field to be integrated is indeed $\displaystyle -y\vec{i}+ x\vec{j}$, on the path from (0,0) to (6, 0) we can take x= t ($\displaystyle 0\le t\le 6$) and y= 0. Then dx= dt and y= 0 dt so $\displaystyle \int \vec{u}\cdot\vec{ds}= \int (-ydx+ xdy)= \int (0dx+ x(0))= 0$.

Write the circle $\displaystyle x^2+ y^2= 36$ as x= 6 cos(t), y= 6 sin(t) (that's a standard parameterization for a circle) so that dx= -6 sin(t)dt and dy= 6 cos(t) dt. The integral becomes $\displaystyle \int (6 cos(t)\vec{i}+ 6 sin(t)\vec{j})\cdot(-6 sin(t)dt\vec{i}+ 6 cos(t)dt\vec{j})$$\displaystyle = \int(-36sin(t)cos(t)dt+ 36cos(t)sin(t))dt= 0$!