# An inconsistent conservative vector field

1. Jan 22, 2016

### riemannsigma

< y^2, 2xy+ e^(3z), 3ye^(3z)> is the vector field.

the above vector field is inside an open simply connected domain.

the parametric equations all have a continuous first order derivative inside the domain.

Lastly, the curl of the vector field is <0, 0, 0>

Thus, the vector field is conservative and is path independent.

HOWEVER....

If i take the

x axis, y axis, and z axis,
Y axis, x axis, and z axis, OR
Y axis, z axis, and x axis

pathways, then I get the scalar field equation to be

[f(x,y,z)] = xy^2 + y + ye^(3z)

This is NOT CORRECT!!!!

If I take the three axis pathways in any other order than the order listed above, then I get the correct answer.

Correct function is
[f(x,y,z)] = xy^2 + ye^(3z)

There is not doubt that the vector field is conservative.

If a vector field is conservative, then the path integral MUST BE path independent.

Why do I get two different answers?...

2. Jan 23, 2016

### Samy_A

Probably because you made a computation error. How can we tell without seeing the computation?
Can you show how you got $xy²+y+ye^{3z}$ in one case and $xy²+ye^{3z}$ in an other case?

3. Jan 23, 2016

### riemannsigma

X axis- y axis - z axis pathway.

1. x axis
y,z,dy,dz=0

Interval is from x=0 to x=x1

Integrate <0, 1, 0> • <dx, 0, 0> to get 0

2. Y-axis

x=x1, dx,dz,z=0

Interval is from y=0 to y=y1

Integrate <y^2, 2x1y + 1, 3y> • <0, dy, 0> to get
x1y^2 + y

3. Z-axis
x=x1, y=y1, dx=dy=0

Interval is from z = 0 to z = z1
Integrate <(y1)^2, 2x1y1 + e^(3z1), 3y1e^(3z)> • <0,0,dz> to get

y1e^(3z1)

The total path integral is NOT CORRECT!!!

[f(x1, y1, z1)] = Constant + x1y^2 + y + y1e^(3z1)

Where did I go wrong?!?! Help

4. Jan 23, 2016

### riemannsigma

Also... The same inconsistency occurs for this conservative vector field.

<e^y , xe^y>

Having e in the parametric equation seems to make the line integral path DEPENDENT.

WHATS GOING ON?!?!

5. Jan 23, 2016

### Samy_A

In the last integral in the z-direction you miss a $-y_1$ term.
$\displaystyle \int_0^{z_1} 3y_1e^{3z} \, dz = \left. y_1e^{3z} \right|_0^{z_1} = y_1e^{3z_1}-y_1$

Last edited: Jan 23, 2016
6. Jan 23, 2016

### riemannsigma

Yes! Lol. Thanks