Question

Vector field F(x,y)=[2x+y,e^(-y)] what is the value of the integral Fdr (dotted) over a closed path C (C is a circle with radius r based in the center) ? ..... now i don't know why but they say that because the y component of the vector field is the total differential, the value of the integral over a closed path is zero ... why is that ? because i now that the x component is not zero when you calculate it over the path of the circle ...

Answer 1

I don't think this is zero for every closed path.

Answer 2

Are you familiar with Green's theorem?

Answer 3

Green:
\[\oint_C F dr=\int\int_{interior}\frac{\partial2x+y}{\partial y}-\frac{\partial e^{-y}}{\partial x}dxdy=\int\int_{interior}1dxdy\]