Integrate the vector field $F(x,y)=(e^y+\frac{1}{y+3},xe^y-\frac{x+1}{(y+3)^2})$ over a curve that goes from $(-1,0)$ to $(-1,2)$ to $(0,1)$ to $(1,2)$ (in a linear fashion).

Now, I'm almost certain that this is easily solvable using Green's theorem, but I only know how to apply it for differential forms. Not vector functions.

Anyways, what I wanted to do was say that the integral is equal to the integral of the first segment, plus the integral over the triangle the other two segments would enclose if connected, minus the integral over the extra segment we had to add to close the triangle.

The numbers tell me I'm probably correct because both the first line integral (over a vertical line) and the one I have to subtract at the end (over a horizontal line) were trivial.

Now, as far as I understand, I need to do something like this for the integral over the triangle.

$\int_{\partial T}F=\int_TdF$

The problem is that I don't know what exactly $dF$ means. I've seen that a vector field like this corresponds to a 1-form $(e^y+\frac{1}{y+3})dx+(xe^y-\frac{x+1}{(y+3)^2})dy$ but I'm not sure if I can just simply integrate THAT over the triangle and get the same result. Could someone provide some insight?

Oh, and in case I'm right and I can just use the form instead of the vector function, $dF$ turns out to be $0$. Does that mean that the integral over the triangle is $0$?


Here is a clearer version. If $F(x,y) = (P(x,y),Q(x,y))$ then

$$ \int_C P\,dx + Q\,dy = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dy dx, $$

where $D$ is the region enclosed by $C$.

  • $\begingroup$ Ok. So that part in the brackets does turn out to be 0. Does that mean that the whole integral is 0? I can see how it makes sense because that form is actually exact as well as closed so integrating over any area should be 0 because it's surrounded by a closed curve. Just want to check if my logic holds. $\endgroup$ – Luka Horvat Jun 17 '14 at 19:29
  • $\begingroup$ @LukaHorvat: Do you know conservative fields? $\endgroup$ – Mhenni Benghorbal Jun 17 '14 at 19:42
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    $\begingroup$ Great. Thanks do the help. $\endgroup$ – Luka Horvat Jun 17 '14 at 20:23
  • $\begingroup$ @LukaHorvat: You are welcome. $\endgroup$ – Mhenni Benghorbal Jun 17 '14 at 20:29

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