Confirming Vector Field is Conservative: ln(x^2 + y^2)

[gtfitzpatrick]

Homework Statement

confirm that the given function is apotential for the given vector field

$ln(x^{2} + y^{2}) for \frac{2x}{\sqrt{x^{2}+y^{2}}} \vec{i} + \frac{2y}{\sqrt{x^{2}+y^{2}}} \vec{j}$

Homework Equations

The Attempt at a Solution

the first thing i did was let my equation = $P\vec{i}+Q \vec{j}$

then if they are conservative$\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$

$\frac{\partial P}{\partial y} = \frac{-2xy}{\sqrt{x^{2}+y^{2}}}$

and

$\frac{\partial Q}{\partial x} = \frac{-2xy}{\sqrt{x^{2}+y^{2}}}$

so the vector field is conservative.

then

f(x,y) = $\int P(x,y) dx$ and f(x,y) = $\int Q(x,y) dy$


from tables i get f(x,y) = 2($\sqrt{x^2 + y^2}$

what am i doing wrong here? am i getting my integration wrong?

 

[DryRun]

The conservative vector field, $\vec F$, is:
$$\nabla \phi = P\hat i + Q \hat j$$
But,
$$\nabla \phi = \frac{\partial \phi}{\partial x}\hat i + \frac{\partial \phi}{\partial y} \hat j$$
So,
$$\frac{\partial \phi}{\partial x} = P \, ... equation \, (1)$$
and
$$\frac{\partial \phi}{\partial y} = Q \, ... equation \, (2)$$

Actually, i got $$\phi (x,y)=2\sqrt{x^2+y^2}+C$$
By definition, the potential of the vector field is $- \phi(x,y)$.

 

[gtfitzpatrick]

thats what i got, but i left out the constant of integration.
So do you think the question is wrong...

 

[DryRun]

The potential of the vector field $\vec F$ should be:
$$-\phi (x,y)=-2\sqrt{x^2+y^2}-C$$Unless there is a nice way to convert the above expression into:
$$\ln (x^2 + y^2)$$ then, i don't see how.