# Easy differentials question

1. Oct 2, 2006

### mewmew

Ok, I have having problems with the folloing:
http://www.physics.uc.edu/~simpson/pics/Desktop-Images/0.jpg
How exactly do I do this? I thought all I did, for example with the first du, was set up 2 integrals, one for the dx part, and one for the dy part. I then thought for (i) the dx integral would go from (a->x) and the dy integral would go from (b->y), then for (ii) I would integrate the dx from (b->y) and the dy from (a->x). To be honest though I am really pretty lost and any advice would be much appreciated.

2. Oct 3, 2006

### HallsofIvy

Staff Emeritus
You can't "set up 2 integrals, one for the dx part, and one for the dy part." because each contains the other variable.

How about doing what you were told to do: (i) First integrate along the line from (a,b) to (x,b) then from (x,b) to (x,y). (The use of (x,y) as a specific point is confusing. I'm going to call that point (u,v)) On the first line, take x= t(u-a)+ a, y= b with 0<t< 1. Then dx= (u-a)dt and dy= 0. Substitute those values into the two differentials and integrate. On the line from (u,b) to (u,v) let x= u and y= t(v-b)+ b. Then dx= 0, dy= (v-b)dt, 0< t< 1. Again, put those into the two differentials and integrate. Add the results to find the integrals on the entire path.

Now, do the same thing on the lines (a,b) to (a, v) to (u,v): let x= a, y= (v-b)t+ b so that dx= 0, dy= (v-b)dt, 0< t< 1, then let x= (u-a)t+ a, y= v so that dx= (u-a)dt, dy= 0, 0< t< 1.

You should have learned that integrals such as these are independent of the path if and only if the differentials are exact. For which of these do you get the same integral over both paths? (Of course, that doesn't prove that the integral will be the same over any path. That's why the problem says "could be".)

Do you remember the "cross derivative" test for exactness? If not, look it up!