I will just quote what my book has written down; however, I cannot put the picture so I am sorry on that note.
"Proof Let A(a,b) be a fixed point in D. We construct the desired potential function f by defining
f(x,y) = \int^{(x,y)}_{(a,b)} F \cdot dr
for any point (x,y) in D. Since \int_c F \cdot dr is independent of path, it does not matter which path C from (a,b) to (x,y) is used to evaluate f(x,y). Since D is open, there exists a disk contained in D with center (x,y). Choose any point (x_1,y) in the disk with x_1 < x and let C consist of any path C_1 from (a,b) to (x_1,y) followed by the horizontal line segement C_2 from (x_1,y) to (x,y). (see figure 4.) Then
f(x,y)= \int_{C_1} F \cdot dr + \int_{C_2} F \cdot dr = \int^{(x_1,y)}_{(a,b)} F \cdot dr + \int_{C_2} F \cdot dr
Notice that the first integral of these integrals does not depend on x, so
\frac {\partial}{\partial x}f(x,y) = 0 + \frac {\partial}{\partial x} \int_{C_2} F \cdot dr
If we write F =Pi +Qj , then
\int_{C_2} F \cdot dr = \int_{C_2} Pdx + Qdy
On C_2 y is constant, so dy=0. Using t as the parameter, where x_1 <= t >= x, we have
\frac {\partial}{\partial x}f(x,y)= \frac {\partial}{\partial x} \int_{C_2} Pdx+Qdy= \frac {\partial}{\partial x} \int^x_{x_1} P(t,y)dt = P(x,y)
by part 1 of the fundamental theorem of calculus (see section 5.4). A similar argument, using a verticle line segment (see figure 5), shows that
\frac {\partial}{\partial y}f(x,y)= \frac {\partial}{\partial y} \int_{C_2} Pdx+Qdy= \frac {\partial}{\partial y} \int^y_{y_1} Q(x,t)dt = Q(x,y)
Thus F= Pi +Qj = \frac {\partial f}{\partial x}i +\frac {\partial f}{\partial y}j "-end quote
Whew that's a lot of latex.
Now for the questions. I think Krab cleared up my first question about the f(x,y) = \int^{(x,y)}_{(a,b)} F \cdot dr
buisness.
Second Question, when they do the derivative with respect to x, and eliminate the integral of curve C_1. I've never seen that done before. I thought that when you take a derivative, it is with respect to the variables in the function, but here they are taking a derivative with respect to variables in the integrand. Also, they say the first curve does not depend on x, but in a way it does. You want the line segment from the end of the first curve to the end point of the second curve to be horizontal, so depending on where x is in the region D, determines where the curve C1 will terminate in order to keep the second line segment horizontal.
Third and final question, in the last step of this:
\frac {\partial}{\partial x}f(x,y)= \frac {\partial}{\partial x} \int_{C_2} Pdx+Qdy= \frac {\partial}{\partial x} \int^x_{x_1} P(t,y)dt = P(x,y)
they seem to be puling that same stunt that they did with f(a,b). What ever happened to the lower limit? Why isint it written as P(x,y)-P(x1,y). We origionally defined the potential to be zero at (a,b), so the value of P(x1,y) cannot be zero. What is justifiying them in doing this? Or is it becuase they are starting off with a new curve, (the horizontal one), that they are allowed to redefine where the potential is zero, in this case, the start of the second curve. (It seems that in solving this, they made two important assumptions they never bothered to mention, and that is where they defined the zero potential point in the evalutation of each line segment.
