Integration along two separate paths.

1. Sep 12, 2010

petewil2009

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

The variation:
deltau = (6x^2y+3lny)dx + (2x^3+3x/y)dy
is an exact differential and u is given by
u = 2x^3y+3xlny
Demonstrate by integrating deltau along the two different paths from point A(1,1) to point B(3,2) (one path has a constant y from x=1 to x=3 and then the x is constant from y=1 to y=2, the second path is just the opposite, first the x is held constant until y=2 and then y is constant until x=3, essentially each path is two line segments that make up a rectangular looking graph) that the integral is path independent and equal to u(B)-u(A).

2. Relevant equations

3. The attempt at a solution

So far I have tried differentiate the equation twice. The first time setting y=0, x=2, dy=0 then doing it again with y=1, x=0, and dx=0. For the x=0 integration I got 10 and for y=0 I got 42.5 and then I added them together. I then substituted in y=x^2. That is my second path, however as I am writing this out I am realizing this is probably very wrong. Regardless I did the same thing integrating with the substituted y as I did earlier. This gave an answer of 341.5. My conclusion is that 52.5 does not equal 341.5 and since this is an exact differential both answers should be the same due to the path independence....
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 12, 2010

Dick

I have no idea what you are doing. You are just supposed to compute the path integral of du along the two different paths and compare the results. You don't have to differentiate du, and I don't see any part of the path where y=0. Start by computing the path integral along the first segment of the first path. From (1,1) to (3,1). What do you get?

3. Sep 12, 2010

LCKurtz

You don't mean differentiate. You are to do a line integral.

Where are you getting those numbers? Draw a picture of the path starting at (1,1) vertically to (1,2) then continuing horizontally to (3,2). That is the second path described. On the first segment of this path x =1, dx = 0, y = y, dy = dy, and y varies from 1 to 2. Similarly on the second segment. Adding those will complete the work for the second path you were given.
Neither of the paths described is along the parabola y = x2. See if you can understand the setup I gave above on the first segment and do a similar thing on the others.

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