If my vector field is:
$F=(1-\frac{x}{x^2+y^2})i-(\frac{y}{x^2+y^2})j$
How would I go about finding the work done between A(3,2) and B(4, -3)?
I have proved that the vector field is conservative:
$\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=\frac{2xy}{(x^2+y^2)^2}-\frac{2xy}{(x^2+y^2)^2}=0$
Can Green's theorem be used in this case?
Best Answer
I'll elaborate the idea of @RobBland .
Let us find $f$, so that $\nabla f=\mathbf{F}$. We will do that as follows:
$$ f(x,y)=\int P(x,y)dx=\int\left(1-\frac{x}{x^2+y^2}\right)dx=x-\frac{1}{2}\ln(x^2+y^2)+c(y).\tag 1 $$ We assume $y$ to be constant while performing integration $(1)$. $f(x,y)$ satisfies $\frac{\partial f}{\partial x}=P(x,y)$ condition if and only if it has the form $(1)$. Here $c(y)$ is some arbitrary (smooth enough) function; it depends only on $y$.
Now we should find $c(y)$ using $\frac{\partial f}{\partial y}=Q(x,y)$ condition. We will differentiate $(1)$ by $y$: $$ \frac{\partial f}{\partial y}=\frac{\partial}{\partial y}\left(x-\frac{1}{2}\ln(x^2+y^2)+c(y)\right)=-\frac{y}{x^2+y^2}+c^\prime(y).\tag2 $$
But $\frac{\partial f}{\partial y}$ should equal $Q(x,y)=-\frac{y}{x^2+y^2}$: $$ -\frac{y}{x^2+y^2}+c^\prime(y)=-\frac{y}{x^2+y^2}. $$ Thus we obtain $c^\prime=0$, so $c=\mathrm{const}$. Now we got the potential: $$ f(x,y)=x-\frac{1}{2}\ln(x^2+y^2)+c, $$ here $c$ is an arbitrary constant, that may be omitted.
Now we obtain readily: $$ \mathrm{Work}_{A\to B}=f(B)-f(A)=f(4,-3)-f(3,2)=\left[x-\frac{1}{2}\ln(x^2+y^2)\right]\Bigg|^{(4,-3)}_{(3,2)}=\\ =1-\frac{1}{2}\ln\frac{25}{13} $$
REMARK
$f(x,y)$ may be found easier, if we recall, that $\left(-\frac{x}{x^2+y^2}, -\frac{y}{x^2+y^2}\right)=-\frac{\mathbf{r}}{r^2}$ is the field of a charged straight wire (the wire is along $z$ axis). Its potential $f$ is $\ln r$ (up to some factor).