# Path Integral equals zero on non conservative field

I was doing some excercises and I was asked to compute the line integral along certain path. I used greens formula to calculate the work. When computing the integral I had to divide the domain in two sections and one of those domains equals zero. Is this possible?

Here is what I have done so far (its question 1.2)

• I don't understand the question. It seems to me that $C_1$ is a semi-circle in the upper half plane, and $C_2$ is a parabolic arc running from $(0,-1)$ to $(1,0)$. Your picture doesn't look like this, and $C_1\cup C_2$ is not a closed curve. Perhaps the domain for $C_1$ is supposed to be $t\in[0,3\pi/2]$ – saulspatz Apr 23 at 15:20
• There is supposed to be a line from (0,0) to (0,-1), and thus closing the curve. Did I do something wrong? – Luis Guareschi Apr 23 at 15:28
• The semi circle is the upper half of the unit circle. It doesn't pass through $(0,0).$ It passes through the points $(1,0),(0,1),(-1,0).$ Also, there is nothing in the question about a line segment connecting $C_1$ and $C_2,$ and you aren't justified in drawing one. But the question makes no sense as written, so far as I can see. $C_1\cup C_2$ in not a closed curve, despite what the question says, and it's hard to know what to make of parts 2 and 3. Ask your teacher about it. If there's no way to do that, assume that $t$ is supposed to run from $0$ to $3\pi/2$ for $C_1.$ – saulspatz Apr 23 at 15:41
• @saulspatz is right. Unless your second parameter has values $t\in [-1,1]$ (or your first parameter has values $t \in [0, \frac{3\pi}{2}$] this is not a closed curve. Hence, Green's Theorem cannot be applied. I would assume this is most likely a typo though. – Hotdog Apr 23 at 22:53
• @Hotdog A different possible typo! I didn't think of that one. – saulspatz Apr 23 at 22:55

For a conservative field $$\vec{F}$$ the curve integral $$\oint_\gamma \vec{F} \cdot d\vec{x} = 0$$ for every closed path $$\gamma$$.