This is not a vector field. Did you mean ?

Assuming the the vector field to be integrated is indeed , on the path from (0,0) to (6, 0) we can take x= t ( ) and y= 0. Then dx= dt and y= 0dt so \math]\int \vec{u}\cdot\vec{ds}= \int (-ydx+ xdy}= \int (0dx+ x(0))= 0[/tex].find the line integral along the curve C from the origin along the x-axis to the point (6, 0) and then counterclockwise around the circumference of the circle x2 + y2 = 36 to the point \( (6)/sqrt(2), (6)/sqrt(2)\). Give an exact answer.

i think from x=0 to x=6 it is 0 but how do i get the second part? help please

Write the circle 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 .

Or course, if x= 6 cos(t) then x= when and that happens when . The integral is from t= 0 to .

(Blast! I was hoping I could get back and correct my silly mistake [I mixed up x and y] before anyone had looke at it!)