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So I'm required to find a non-closed path $C$ so that the line integral

$$\displaystyle \int_C \mathbf{F} \cdot \mathrm d \mathbf{r}$$ equals $0$, and another non-closed path $C$ so that the line integral equals $2$.

It's given that $$\mathbf{F}=\nabla f$$ where $f=\sin(x-2y)$ (I guess this mean $\mathbf{F}$ Is conservative).

So far what I have managed to have is

$\nabla f=\cos(x-2y)\vec{i},-2\cos(x-2y)\vec{j}$,

The fundamental Theorem of Line Integral: $\displaystyle \int_C \mathbf{F} \cdot \mathrm d \mathbf{r} = f(b)-f(a)$

and the definition of a non-closed path, which is a path where the position vector $\mathbf{r}(t)$ describes $C$ so that $\mathbf{r}(b) \neq \mathbf{r}(a)$. Other than that, I'm pretty stuck.

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You should use the Fundamental Theorem of Line Integrals: $$ \int_C\mathbf{F}\cdot d\mathbf{r}=f(\mathbf{b})-f(\mathbf{a}),$$ where $\mathbf{a}$ and $\mathbf{b}$ are the initial and ending points of the path $C,$ respectively.

So, if you need the last integral to take some value $c$ you can need a path $C$ from some point $\mathbf{a}=(x_1,y_1)$ to some point $\mathbf{b}=(x_2,y_2)$ such that $f(\mathbf{b})-f(\mathbf{a})=c.$ Since $f(x,y)=\sin(x-2y),$ if $c=0$ then you need that $\sin(x_1-2y_1)=\sin(x_2-2y_2)$ which you can obtain if $x_2-2y_2=x_1-2y_1+2\pi.$ Thus, for example, you can take any path from $(x_1,y_1)=(0,0)$ to $(x_2,y_2)=(2\pi,0),$ say a straigh line path: $$ C_1=\{\mathbf{r}(t)=(2\pi t,0) :t\in [0,1]\}.$$ Then $$ \int_{C_1}\mathbf{F}\cdot d\mathbf{r}=f(2\pi,0)-f(0,0)=\sin(2\pi)-\sin(0)=0.$$

Similarly, if you need $f(\mathbf{b})-f(\mathbf{a})=2,$ you note that this can be obtained only if $\sin(x_2-2y_2)=1$ and $\sin(x_1-2y_1)=-1.$ You can take for example the points $\mathbf{a}=(x_1,y_1)=(-\pi/2,0)$ and $\mathbf{b}=(x_2,y_2)=(\pi/2,0)$ and take any path between them, the easiest being a straight line path: $$ C_2=\{\mathbf{r}(t)=(1-t)(-\frac{\pi}{2},0)+t(\frac{\pi}{2},0) :t\in [0,1]\}.$$ Thus $$ \int_{C_2}\mathbf{F}\cdot d\mathbf{r}=f(\frac{\pi}{2},0)-f(-\frac{\pi}{2},0)=\sin(\frac{\pi}{2})-\sin(-\frac{\pi}{2})=2.$$

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  • $\begingroup$ can I also obtain $\sin(x_1+2y_1)=\sin(x_2+2y_2)$ if $x_1+2y_1=\pi-(x_2+2y_2)?$ $\endgroup$ – user98937 Mar 31 '18 at 2:29
  • $\begingroup$ Yes, that also works. $\endgroup$ – positrón0802 Mar 31 '18 at 17:55

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