Multivariable Calculus: Line Integral I have this math problem. It states:

Calculate the given line integral $\oint _c {M dx+Ndy}$ where $C$ is
  the triangle with vertices $P_0=(0, 1)$, $P_1=(2, 1)$, $P_2=(3, 4)$
  with counterclockwise rotation.

Here is the problem: 
$\oint _c {x dx+ dy}$

I'm not sure as to what the "counterclockwise rotation" means. I'm assuming I have to start by finding 3 different line integrals. So I find three parametric lines...
$\vec{P_0} + \vec{P_0P_1}t=<0, 1> + <2t, 0> = <2t, 1>$
$\vec{P_1} + \vec{P_1P_2}t=<2, 1> + <t, 3t> = <2+t, 1+3t>$
$\vec{P_2} + \vec{P_2P_0}t=<3, 4> + <-3t, -3t> = <3-3t, 4-3t>$
That's pretty much where I got. Thanks for the help.
 A: Without appealing to Green's Theorem we proceed.
Inasmuch as this is a closed-path integral, 
$$\oint_C (xdy+dy) =\left(\frac12 x^2 +y\right)|_{P_i}^{P_i}=0$$
where $P_i$ is any starting point of the path.

Now let's show this explicitly by evaluating each contribution of the line integral.
First segment:  
Start at $P_0$ and end at $P_1$.  Parameterize the curve as $x=t$, $y=1$, with $0\le t\le 2$.  Then, $dx=dt$ and $dy=0$.  Thus, we have
$$\int_{C_{1}} (xdx+dy) = \int_0^2 tdt = 2$$
Second segment:  
Start at $P_1$ and end at $P_2$.  Parameterize the curve as $x=t+2$, $y=3t+1$, with $0\le t\le 1$.  Then, $dx=dt$ and $dy=3dt$.  Thus, we have
$$\int_{C_{2}} (xdx+dy) = \int_0^1 ((t+2)+3)dt = 11/2$$
Third segment:  
Start at $P_2$ and end at $P_3$.  Parameterize the curve as $x=3-3t$, $y=4-3t$, with $0\le t\le 1$.  Then, $dx=-3dt$ and $dy=-3dt$.  Thus, we have
$$\int_{C_{3}} (xdx+dy) = \int_0^1 (3-3t+1)(-3)dt = -15/2$$
Thus, summing the contributions reveals that
$$\oint_C (xdx+dy) = 2+11/2-15/2=0$$
as expected!
A: I think that "counterclockwise rotation" just means do the line integral starting from point 1, going in order which would take you counterclockwise around the triangle. So I think what you're doing is right
A: By Green's Theorem,
$$\int_{C} x\,dx+1\,dy=\iint_{T}\left(\frac{\partial 1}{\partial x}-\frac{\partial x}{\partial y}\right)d A=\iint_T 0\,dA=0.$$
A: Observe that your field is a conservative one as $\;\dfrac{\partial P}{\partial y}=0=\dfrac{\partial Q}{\partial x}\;$ , with $\;P=x\;,\;\;Q=1\;$ , and thus the integral equals zero on any simple, "nice" closed path.
You can also observe that 
$$\;(x,1)=\nabla\left(\frac12x^2\,,\,y\right)$$
