# Help with conservative line integrals.

1. Dec 3, 2005

### seang

I'm really having trouble with this, I can't even understand the example in my textbook. The example problem is:

Show that ydx + xdy + 4dz is exact and then evaluate it over some limits.

I can show that its exact, and once I find the function, I can evaluate it, so I really just need help getting the function. So the book says:

df/dx = y; df/dy = x; df/dz = 4; I understand this part, then the next step is:

f(x,y,z) = xy + g(y,z); Got this too, but now the fun begins. The next line is:

df/dy = x + dg/dy = x or dg/dy = 0; I have no idea what they're talking about here. Where is the dg/dy from? And why is 'x' what the equation is supposed to equal? The next step is:

f(x,y,z) = xy + h(z); followed by:

df/dz = 0 + dh/dz = 4 or h(z) = 4z + C; Again, I have no idea where the dh/dz if from. Also, above df/dz = 4, now it is 0? I understand that if you differentiate 4, you get 0, but what's with the discrepancy?

Any help is appreciated! I know this is long so thanks for your time

2. Dec 3, 2005

### BobG

That seems like a really confusing way to explain the procedure, at least if I've gotten what it's trying to get at straight.

If you have the following function:

$$xy+4z$$

and find the partial derivatives, you'll get:

$$\frac{\partial (xy + 4z)}{\partial x}=y + 0$$

$$\frac{\partial xy + 4z}{\partial y}=x + 0$$

$$\frac{\partial xy + 4z}{\partial z}=0 + 4$$

What you're doing is the inverse. Integrate the partial derivative with respect to x and you should get the original function -- with the exception of any constants. With respect to x, 4z is just a constant.

You do the same with the partial derivative with respect to y and should also get the original function, with the exception of any constants (4z in this case).

You do the same with the partial derivative with respect to z and get the original function, with the exception of any constants (xy in this case).

Any unique terms are combined into one equation to get the original function. When you integrated with respect to x and with respect to y, you got xy for both (plus some constant). You only include the xy once in your final function, since you didn't get a new unique solution. When you integrated with respect to z, you got a new unique solution - 4z. You combine the unique solutions to get the potential function of xy + 4z.

Of course, since xy + 4z + 6 gets you the same partial derivatives, when you're going in reverse, there will be no way to know that the 6 should have been in the potential function, so your potential function should be xy + 4z + K to cover any constants that had a partial derivative of 0 for all three variables.

In your example, your first integration with respect to x yielded xy + g(y,z), since the partial derivative of any function consisting of just y, z, or yz, yielded a partial derivative with respect to x of 0. And so on.