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Work of a non conservative field

  1. Jan 29, 2017 #1
    1. The problem statement, all variables and given/known data
    Compute the work of the vector field $$H: \mathbb{R^2} \setminus{(0,0}) \to \mathbb{R}$$

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

    in the path $$g(t) = (1-t^2, t^2+t-1)$ with $t\in[-1,1]$$

    2. Relevant equations
    3. The attempt at a solution


    So first I considered my vector field as a sum of 2 vector fields: $$H = F + G$$

    $$F(x,y)=\bigg(y^2,1+2xy\bigg)$$

    $$G(x,y)=\bigg(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2}\bigg)$$

    The vector field $$F$$ is conservative with one of many potentials $$A(x,y) = y^2x+y$$
    Then I worked out the work using the definition and fundamental theorem of calculus obtaining the value 2.

    So no problems at this point.

    But $$G$$ is not a conservative vector field (it doesn't have a potential, even though it's a closed field). How should I proceed? I tried the definition but we get to a very complicated integral... The path isn't closed so we can't apply Green's theorem... What should I do?
     
  2. jcsd
  3. Jan 29, 2017 #2

    Ray Vickson

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    What is preventing you from just going ahead and computing ##\int_{-1}^1 \vec{H}(t) \cdot \nabla g(t) \, dt?##
     
  4. Jan 29, 2017 #3
    You mean H(g(t)) (with the vector sign). Because the integral gets complicated.
     
  5. Jan 29, 2017 #4

    Ray Vickson

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    Yes, I know, but if you simplify the integrand first (to a single ratio of two polynomials) you DO get a tractable integral that is not too bad. (The integral is horrible if you do not simplify the integrand first.) For example, you could submit the integration to Wolfram Alpha and get a usable answer. Alternatively, you could use a numerical integration method to get a numerical answer to as much accuracy as you want.

    Basically, the integral is complicated, and that's life. Sometimes problems are not easy, and we just have to deal with them anyway.
     
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