Q: Let F= (-y/(x2+y2), x/(x2+y2), z) be a vector field and let U be the interior of the torus obtained by rotating the circle (x-2)2 + z2 = 1, y=0 about the z-axis. a) Show that curl F=0 but ∫ F . dx = 2pi where dx=(dx,dy,dz) C and C (contained in U) is the circle x2+y2=4, z=0. Therefore F is NOT conservative. b) Determine the possible values of (0,2,0) ∫ F . dx on a path in U. (2,0,0) =================== I am OK with part a. But part b is tricky!! Solution to part b: f(x,y,z)=arctan(y/x) + z2/2 is a function such that grad f = F => f(x,y,z)=arctan(rsin[tex]\theta[/tex]/rcos[tex]\theta[/tex]) (since z=0) => f=[tex]\theta[/tex] (0,2,0) ∫ F . dx = f([tex]\theta[/tex]=pi/2) - f([tex]\theta[/tex]=0) = pi/2 by F.T.C. for line integrals (2,0,0) Final Answer: (0,2,0) ∫ F . dx = pi/2 + (2n)pi (2,0,0) ============== Now, I don't understand the parts in red. Why is z=0? This is NOT true for everywhere in U Why + (2n)pi ? What if the path is not a circle? This is possible because the torus has some "width" and "height" Can someone please explain the tough part (part b)? I would really appreciate it!