Conservative vector field problem

[Gianmarco]

Homework Statement

Determine for which real values of the parameter $\alpha$ the vector field given by
$F(x,y) = (\frac{2xy}{y-\alpha}, 2 - \frac{4x^2}{(y-\alpha)^2})$
is conservative. For those values of $\alpha$, calculate the work done along the curve of polar equation:
$\rho = \frac{\theta}{\pi}$

Homework Equations

If F is a conservative vector field, then:
$rotF=0$
we can find a potential function $\phi s.t. \ F=\nabla\phi$

The Attempt at a Solution

To find the parameter, I calculated the derivative of the first component of the field with respect to y and set it equal to the derivative of the second component with respect to x to show that the vector field is irrotational (I don't know if I can do that since the domain of the field is not simply connected and so schwarz' theorem for second derivatives doesn't apply). Either way, I found the plausible result that $\alpha = 4$.
Since the vector field is conservative, I tried to find the potential function by integrating the second component $2 - \frac{4x^2}{(y-\alpha)^2}$ with respect to y.
This gives $\phi(x,y) = 2y + \frac{4x^2}{y-4} + g(x)$
By deriving it with respect to x and equating it with the first component of the field, I got $\frac{8x}{y-4} + g'(x) = \frac{2xy}{y-4}$ which leads to $g'(x) = \frac{2xy+8}{y-4}$.
This makes no sense because g(x) should be a function of x only and since I've proven that the field is conservative for that value of $\alpha$, I should be able to find a potential function. Am I wrong?

 

[Samy_A]

Homework Statement

Determine for which real values of the parameter $\alpha$ the vector field given by
$F(x,y) = (\frac{2xy}{y-\alpha}, 2 - \frac{4x^2}{(y-\alpha)^2})$
is conservative. For those values of $\alpha$, calculate the work done along the curve of polar equation:
$\rho = \frac{\theta}{\pi}$

Homework Equations

If F is a conservative vector field, then:
$rotF=0$
we can find a potential function $\phi s.t. \ F=\nabla\phi$

The Attempt at a Solution

To find the parameter, I calculated the derivative of the first component of the field with respect to y and set it equal to the derivative of the second component with respect to x to show that the vector field is irrotational (I don't know if I can do that since the domain of the field is not simply connected and so schwarz' theorem for second derivatives doesn't apply). Either way, I found the plausible result that $\alpha = 4$.
Since the vector field is conservative, I tried to find the potential function by integrating the second component $2 - \frac{4x^2}{(y-\alpha)^2}$ with respect to y.
This gives $\phi(x,y) = 2y + \frac{4x^2}{y-4} + g(x)$
By deriving it with respect to x and equating it with the first component of the field, I got $\frac{8x}{y-4} + g'(x) = \frac{2xy}{y-4}$ which leads to $g'(x) = \frac{2xy+8}{y-4}$.
This makes no sense because g(x) should be a function of x only and since I've proven that the field is conservative for that value of $\alpha$, I should be able to find a potential function. Am I wrong?

Error in the very last step: $\frac{8x}{y-4} + g'(x) = \frac{2xy}{y-4}$ doesn't lead to $g'(x) = \frac{2xy+8}{y-4}$.

 

[Gianmarco]

Error in the very last step: $\frac{8x}{y-4} + g'(x) = \frac{2xy}{y-4}$ doesn't lead to $g'(x) = \frac{2xy+8}{y-4}$.

Thank you.