# Evaluating line integrals versus Green's Theorem

1. Apr 26, 2010

### theno1katzman

1. The problem statement, all variables and given/known data

Find the simple closed integral of (x+xy-y)(dx+dy) counterclockwise around the path of straight line segments from the origin to (0,1) to (1,0) to the origin...
a)as a line integral
b)using green's theorem

2. Relevant equations

Eq of line segment r(t)=(1-t)r0+tr1

Greens Theorem states that Integral of Pdx+Qdy on a closed simple path will equal the double integral or area of dQ/dx-dP/dy (see the work I did)

3. The attempt at a solution

See the pictures. The two were supposed to be equal, I just needed to show how to arrive at the same answer using both methods. The triangle shows the graph of the path, horizontal is x axis, vertical is y axis. The C2 on the left of the triangle should say C3, sorry. Also the first integral should read (1+y) not (1-y)

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Last edited: Apr 27, 2010
2. Apr 27, 2010

### HallsofIvy

Staff Emeritus
One obvious error is in saying that
$$\frac{\partial Q}{\partial x}= 1- y$$
Q= x+ xy- y so
$$\frac{\partial Q}{\partial x}= 1+ y$$

but it is written correctly in the next line so that's not the problem.

I notice you then integrate, in integrating over the area,
$$\int_{y=0}^{-x}$$
Why is that? The upper boundary is y= x, not y= -x. It looks to me like you should be integrating $2x+ x^2/2- x^2= 2x- x^2/2$ with respect to x. That would give $\left[x^2- x^3/6\right]_0^1= 5/6$ rather than -1/2.

Last edited: Apr 27, 2010
3. Apr 27, 2010

### theno1katzman

But that wouldn't give you the same answer for part a and b...
Why would that give y=x for the boundary. I looked it over again with my calculator, that upper boundary line (hypotenuse) is y=-x+1
however integrating that you end up with 2, which is still not the same as part a.

4. Apr 27, 2010

### HallsofIvy

Staff Emeritus