Evaluating line integrals versus Green's Theorem

[theno1katzman]

Homework Statement

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

Homework 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)

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)

 

[HallsofIvy]

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.

 

[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.

 

[HallsofIvy]

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.

Yes, you are right about that- I misread your problem.

But, still, the integral should be from 0 to 1- x, not to just -x.