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I question in a old exam is: Given the vector field $\vec{K}$ with the domain of $R^2\setminus {(0,0)}$, calculate the potential for $\vec K$

$$\vec{K}(x,y) = \left(\frac{-2xy}{(x^2+y^2)^2}, \frac{x^2-y^2}{(x^2+y^2)^2}\right)$$

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I solved this exercise by myself. Any correction or improvements of the solution would be highly welcome! – leo Jan 10 '13 at 12:02
up vote 2 down vote accepted

Given $\vec K$, we define $P(x,y)=\frac{-2xy}{(x^2+y^2)^2}$ and $Q(x,y) = \frac{x^2-y^2}{(x^2+y^2)^2}$.


$$P_y = \frac{\partial P}{\partial y} (x,y) = \frac{\partial}{\partial y} \frac{-2xy}{(x^2+y^2)^2} = -2x \frac{\partial}{\partial y} \frac{y}{(x^2+y^2)^2}= -2x \frac{(x^2+y^2)^2-y 2(x^2+y^2) 2y}{(x^2+y^2)^4} = -2x \frac{x^2+y^2-4y^2}{(x^2+y^2)^3}= -2x \frac{x^2-3y^2}{(x^2+y^2)^3}$$


$$Q_x = \frac{\partial Q}{\partial x} (x,y) = \frac{\partial}{\partial x} \frac{x^2-y^2}{(x^2+y^2)^2} = \frac{2x(x^2+y^2)^2-(x^2-y^2)2(x^2+y^2) 2x}{(x^2+y^2)^4} = -2x \frac{-(x^2+y^2) + 2(x^2-y^2)}{(x^2+y^2)^3} = -2x \frac{x^2-3y^2}{(x^2+y^2)^3}$$

$\Rightarrow P_y = Q_x \Rightarrow \vec K$ has a potential $f$.

  1. $f(x,y) = \int P(x,y)\,dx + C(y) = \int\frac{-2xy}{(x^2+y^2)^2}\,dx + C(y) =-y\int\frac{2x}{(x^2+y^2)^2}\,dx + C(y)=-y\int\frac{2x}{u^2}\,\frac{du}{2x} + C(y)=-2y\int\frac{1}{u^2}\,du + C(y) = \frac{y}{u} + C(y) = \frac{y}{x^2+y^2} + C(y)$
  2. $\frac{\partial f}{\partial y} (x,y) = \frac{\partial}{\partial y} \frac{y}{x^2+y^2} + C'(y) = \frac{(x^2+y^2)-2y^2}{(x^2+y^2)^2} + C'(y) = \frac{x^2-y^2}{(x^2+y^2)^2} + C'(y)$
  3. $Q(x,y) = \frac{x^2-y^2}{(x^2+y^2)^2} + C'(y) \Rightarrow \frac{x^2-y^2}{(x^2+y^2)^2} = \frac{x^2-y^2}{(x^2+y^2)^2} + C'(y) \Rightarrow C'(y) = 0$
  4. $\int C'(y) \,dy = \int 0 \,dy = c$
  5. $f(x,y) = \frac{y}{x^2+y^2} + C(y) = \frac{y}{x^2+y^2} + c$
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