Using Green's Theorem on an non-closed curve by adding or subtracting another curve/line?

Question

I understand Green's Theorem can only be used on curves that are simple and closed. However, taking a look at these two examples, it seems like you can add a line so that the curve becomes closed so that Green's Theorem can be used.

Example A: let C be the curve from points $(2,0)$ to $(0,0)$ along circle $(x-1)^2 + y^2 = 1$

For this example, I added the curve L, which is a line $y = 0,$ from $(0,0)$ to $(2,0).$ So now, the integral $$\int_{C}\vec{F} \cdot d\vec{r} + \int_{L}\vec{F} \cdot d\vec{r} = \oint_{C+L}\vec{F} \cdot d\vec{r} = \iint_{R}-y\,dA$$

In another example, B: consider the vector field, $\vec{F}(x,y) = \left\langle -x^3y,\sqrt{1+y^3}\right\rangle$ defined for all real x's and $y\geq-1$. Let $C$ be the curve from $(-1,1)$ to $(1,1)$ along the parabola $y=x^2$.

To compute the line integral, I first let $R$ (the area??) be the region enclosed by the lines $C-L$, where $L$ is the straight line from point $(-1,1)$ to $(1,1).$ Therefore: $$\int_{C}\vec{F} \cdot d\vec{r} - \int_{L}\vec{F} \cdot d\vec{r} = \oint_{C-L}\vec{F} \cdot d\vec{r} = \iint_{R}x^3\,dA$$

My question is why do we add in the first example, but subtract in the second example? My first assumption was to always add, as if like we're "adding the two lines" to create an enclosed curve.

Also: had I not been able to seen that I could have used Green's Theorem, say in Example B, could I have been able to parametrize this line by letting $x=t$ and letting $y=t^2$, then computing the line integral using $r(t)=\left\langle t,t^2 \right\rangle$?

Answer 1

One of the key ideas in vector calculus and in using Stokes' Theorem/Green's Theorem is the idea of orientation.

So in your first example, $C$ is a curve from $(-2, 0)$ to $(0, 0)$ along the circle centered at $(1, 0)$ with radius $1$. This description, in itself, is actually slightly ambiguous. $C$ could either be the line integral going counter-clockwise through the point $(1, -1)$, or clockwise through $(1, 1)$. But it ends up not mattering when you include the line $L$ because, by way of adding it, you're asserting that $L$ and $C$ have the same orientation.

On the other hand, in your second example of a curve along the parabola $y=x^2$ from $(-1, 1)$ to $(1, 1)$, there is only one well-defined curve, and so the orientation is given to you from the problem statement. This means that when you want to complete the curve with line $L$, in order for it to have the same orientation as the curve along $y=x^2$, it must be the straight line from $(1, 1)$ to $(-1, 1)$. However, you defined it as from $(-1, 1)$ to $(1, 1)$. This can be fixed by 'subtracting' the curve, which has the effect of reversing the orientation.

Edit: To answer your additional question (parameterizing the curve and taking the line integral instead) -- yes, that is absolutely the way you would solve that integral without using Green's Theorem.

Answer 2

The issue here has to do with the orientation of the curves themselves. Green's theorem specifies that the region R has to be on the left as one "traverses" the boundary curve(s).

In example A, as you move along both curve L and C, the region R will be on your left. Since this is the correct orientation, Green's theorem applies and one simply adds the line integrals around each curve to get the total closed line integral.

In example B, curve C exhibits the correct orientation, as the region R will be on the left as you traverse across C. However if you move along L from (-1,1) to (1,1), R will be on your right. Since the tangent vector, dr, points in the "wrong" direction, the sign of the line integral along L from (-1,1) to (1,1) will be the opposite sign of what it should be. Thus, you have to add a negative sign to the line integral in question and then add it to the line integral around C to get the correct closed line integral. You could also just reorientate L so that you are moving from (1,1) to (-1,1), which would be the correct orientation. Then, you would simply add the resulting line integrals as in example A. Both methods yield the same result.

And your parametrization of the curve y=x^2 is correct as it yields the correct orientation of C.