Work Done from (1,0,0) to (1,0,1) in Conservative Force Vector F

[Mindscrape]

I have a force vector that is F = (x^2 + y)i + (y^2 + x)j +ze^z k) and I am supposed to find the work done from (1,0,0) to (1,0,1). The question gives a bunch of paths to integrate, but I used the curl and found that the force was conservative (hence path independent), so I was going to make a function of x,y,z based on F.

\frac{\partial f}{\partial x} = x^2 + y

f(x,y,z) = (1/3)x^3 +xy+g(y,z)

\frac{\partial f}{\partial y}=x+\frac{\partial g}{\partial y} = y^2 +x

\frac{\partial g}{\partial y} = y^2

g(y,z)= (1/3)y^3 +h(z)

so now continue to build the function

f(x,y,z)=(1/3)x^3 +xy+(1/3)y^3 + h(z)

\frac{\partial f}{\partial z} = 0 + \frac{\partial h}{\partial z} = ze^z

solve by parts to get

h(z)=ze^z - \int e^z dz = e^z(z-1)

f(x,y,z)=(1/3)x^3 + xy+(1/3)y^3 + e^z(z-1)

Now I know that the last part is the part that I screwed up because f_z is supposed to equal ze^z, and it doesn't. What I can't figure out is what I screwed up. Anyone see it?

 

[Tomsk]

No I get the same answer as you, dx=dy=0, so the only term left in \int_{1,0,0}^{1,0,1}{F.dl} is \int_{0}^{1}{z e^{z} dz}, which gets you [e^z(z-1)]_{0}^{1} = -1. Why should it be ze^z? df is an element of work done, not dF.

 

[Mindscrape]

Don't you have to regain F with its respective partials? Taking the partial of the function with respect to x gives x^2 + y, which is the x (or M) term; with respect to y gives y^2 + x, which is the y (or N) term; yet, with repsect to z it is e^z+ze^z-e^z (errr... whoops I just realized it did work out - for some reason I assumed it didn't).

It would give +1 though, since evaluating gives 0--1=1.

 

[HallsofIvy]

Since you know the force is conservative, you know the work done is independent of the path. Integrate F along the straight line from (1, 0, 0) to (0, 0, 1): x= 1, y= 0, z= t. dx= 0, dy= 0, dz= dt so the integral is
\int_0^1 te^t dt.