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}-CUnless there is a nice way to convert the above expression into:
\ln (x^2 + y^2) then, i don't see how.