From Transcendental Functions (Smith and Minton) 14.3 #29
Evaluate $\int_CF \cdot dr $ for $F(x,y)= \langle {1\over y} -e^{2x}, 2x-{x \over y^2} \rangle$, $C$ is the circle $(x − 5)^2 + (y + 6)^2 = 16$, oriented counterclockwise
This vector field is not conservative, and since this is a circle $x$ and $y$ will have to be parametrized with $4 \cos t +5$ and $4 \sin t -6$ respectively. If this had to be done the regular way, it would be almost impossible.
Instead the book does this little maneuver:
$$\int_C \left( {1\over y} -e^{2x} \right)dx + \left( 2x-{x \over y^2}\right)dy$$
$$=\int_C \left( {1\over y}dx-{x \over y^2}dy \right) - \int_C e^{2x} dx+\int_C 2xdy$$
$$=\int_C d\left( x\over y \right) - \int_C e^{2x} dx+\int_C 2xdy$$
Then it says since this is a closed curve, both $\int_C d\left( x\over y \right)$ and $\int_C e^{2x} dx$ equal $0$, leaving us with only $\int_C 2xdy$ .
Why is this true? I learned that a closed curve is only $0$ when it is independent of path, but we already said that this vector field is not conservative? Also, why these terms specifically?
For the first term, the vector field $F = \langle \frac{1}{y}, -\frac{x}{y^2} \rangle$ is conservative since $\partial_y (\frac{1}{y}) = \partial_x (- \frac{x}{y^2}).$ The second term is solely dependent on $x$ so we can write $$\int_C e^{2x}dx = \int_9^1 e^{2x}dx + \int_1^9 e^{2x}dx$$ which is clearly $0.$