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Path integral

  1. Nov 1, 2012 #1
    1. The problem statement, all variables and given/known data

    Find $$\int_{C} \frac{ds}{(2y^2+1)^{3/2}}$$
    where $$C$$ is the parabola $$ z^2=x^2+y^2 , x+z=1$$
    2. Relevant equations



    3. The attempt at a solution
    I tried to parametrize the C , s.t$$ x=t, z=1-t, y=\sqrt{2t-1}$$ ,
    but it seems to become a mess, and I don't know the bound of t,
    and I tried to let$$ x=sin^2ω , z=cos^2ω , y = \sqrt{cos^2ω-sin^2ω}$$
    Is $$ω $$from 0 to $$2\pi$$? , but it seems to not easy to compute it.
    So I want to know is there any nice method for that Q.?
     
  2. jcsd
  3. Nov 1, 2012 #2
    You got:

    [tex]z^2=x^2+y^2[/tex]
    [tex]z=1-x[/tex]
    or:
    [tex]x=1/2-y^2/2[/tex]
    so we can parameterize that in terms of y as:

    [tex]x(y)=1/2-y^2/2[/tex]
    [tex]y(y)=y[/tex]
    [tex]z=1-(1/2-y^2/2[/tex]
    or to make it look nicer:

    [tex]x(t)=1/2-t^2/2[/tex]
    [tex]y(t)=t[/tex]
    [tex]z(t)=1/2+t^2/2[/tex]
    and that last parametrization will track a parabolic curve on the surface of [itex] z^2=x^2+y^2[/itex] for how long? Well from one end of the parabola to the other end right? How long is that if I didn't give any end point? How about -100 to 100? Further? How about -1000 to 1000? Further.
     
  4. Nov 1, 2012 #3
    I got it , thank you very much :D
     
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