Work done by a field along a circle

Question

I have the vector field

$$F(x, y) = (\frac{-y}{(x-1)^2+y^2} + \frac{y}{(x+1)^2+y^2}, \frac{x-1}{(x-1)^2+y^2} + \frac{-x-1}{(x+1)^2+y^2})$$

Using Green's Theorem I want to calculate the work done by $F$ along a circle of the form $(x-1)^2 + y^2 = 1$.

How should I do it since the denominator is zero at $(1, 0)$ and $(-1, 0)$? I know that theoretically I could divide the region in 2 parts, but how do I set it up analytically in order to compute the work?

Answer 1

You can't use Green's theorem because you don't have continuous partial derivatives.

Let $G = (\frac {-y}{(x-1)^2 + y^2},\frac {x-1}{(x-1)^2+y^2})$ and $H = (\frac {-y}{(x+1)^2 + y^2},\frac {x+1}{(x+1)^2+y^2})$

$F = G-H$

It is easier to check $G$ and $H$ independently to see that they are curl free.

$H$ does have continuous partials over the region, so the integral of a conservative field over a closed curve is zero.

Now we can parameterize the curve and integrate $G.$

$\int_0^{2\pi} (-\sin x,\cos x)\cdot (-\sin x. \cos x)\ dx = 2\pi$

Answer 2

The circle can be parametrized as

$ r = (1 + \cos(t) , \sin(t) ) \hspace{20pt} t \in [0, 2 \pi) $

and

$\mathbf{F} = \mathbf{F_1} + \mathbf{F_2} $

where

$\mathbf{F_1} = ( - \sin(t) , \cos(t) ) $

and

$ \mathbf{F_2} = ( \dfrac{\sin(t)}{ 5 + 4 \cos(t) } , - \dfrac{ 2 + \cos(t) }{ 5 + 4 \cos(t) } )$

The work is linear,

$W_1 = \displaystyle \int_0^{2\pi} \mathbf{F_1} \cdot (-\sin(t), \cos(t) ) \ dt = 2 \pi $

$W_2 = \displaystyle \int_0^{2 \pi} \mathbf{F_2} \cdot (-\sin(t), \cos(t) ) \ dt \\ = \displaystyle \int_0^{2 \pi} \dfrac{ - 1 - 2 \cos (t) }{5 + 4 \cos(t) } \ dt $

Let $z = \tan(\frac{t}{2})$ , then $dz = \frac{1}{2} \sec^2(\frac{t}{2}) \ dt = \frac{1}{2}(z^2 + 1) $, $\cos(t) = 2 \cos^2(\frac{t}{2}) - 1 = \dfrac{2}{z^2 + 1} - 1 = \dfrac{1 - z^2}{1 + z^2} $

Hence,

$W_2 = \displaystyle \dfrac{1}{2} \int \dfrac{- 1 - 2 (1 - z^2) }{5 (z^2 + 1) + 4(1 - z^2) } dz \\ = \displaystyle \dfrac{1}{2} \int \dfrac{2 z^2 - 3}{z^2 + 9} dz \\ = \displaystyle \dfrac{1}{2} \int \dfrac{2(z^2 + 9) - 21 }{z^2 + 9}dz \\ =\displaystyle \dfrac{1}{2} \bigg[ 2 z - 7 \tan^{-1}(\frac{z}{3}) \bigg] $

Plugging in $z$ and evaluating over $t \in [0, 2\pi) $ gives the value $ 2 \pi$

Therefore, the total work is

$ W = W_1 + W_2 = 2 \pi + 2 \pi = 4 \pi $

Answer 3

There is only one way to use Green's theorem. To perturb slightly the vector field defining as

$F_{1}=\dfrac{-y}{(x-1)^{2}+y^{2}+\epsilon}$+$\dfrac{y}{(x+1)^{2}+y^{2}}$

$F_{2}$=$\dfrac{x-1}{(x-1)^{2}+\epsilon+y^{2}}$$-$$\dfrac{x+1}{(x+1)^{2}+y^{2}}$

in order to avoid singularities. Set $F_{2}=Q,\,\,F_{1}=P$.

We get $Q_{x}-P_{y}=\dfrac{2\epsilon}{[(x-1)^{2}+y^{2}+\epsilon]^{2}}$

Changing to polar coordinates by $x=1+rcos\theta,\,\,y=sin\theta$

we obtain $\,\,\,2\epsilon\int_{0}^{1}\int_{0}^{2\pi}\dfrac{1}{(r^{2}+\epsilon)^{2}}rdrd\theta$

which is equal to : $2\pi\dfrac{1}{1+\epsilon}$ and taking $\epsilon\to 0$ we get $2\pi$.

It seems that it is like having a removable singularity of the field!!