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?
Best Answer
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 $0.$