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Scalar line integral

  1. Aug 9, 2012 #1
    I have the vector:

    [tex]{\bf{u}}(x,y) = \frac{{x{\bf{i}} + y{\bf{j}}}}{{{x^2} + {y^2}}}[/tex]

    Where:

    [tex]x = a\cos t[/tex] [tex]y = a\sin t[/tex]

    I know I need to use the equation

    [tex]\int\limits_0^{2\pi } {{\bf{u}} \cdot d{\bf{r}}} [/tex]

    And the answer is

    [tex]\int\limits_0^{2\pi } {} ((a\cos t/{a^2})( - a\sin t) + (a\sin t/{a^2})(a\cos t)dt = 0[/tex]

    The trouble I have is finding that [tex]{d{\bf{r}}}[/tex] How is that done? Can someone help?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 9, 2012 #2

    uart

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    Start by actually stating the question correctly.

    For the given vector function [itex]{\bf u}(x,y)[/itex], evaluate the line integral,

    [tex]I = ~_\mathbb{C} \!\!\! \int {\bf u} \cdot d {\bf r} [/tex]

    Where [itex]\mathbb{C}[/itex] is a contour given parametrically by {[itex]x = a\cos t[/itex], [itex]y = a\sin t[/itex] : [itex]t = 0 \ldots 2\pi[/itex]}.
     
    Last edited: Aug 9, 2012
  4. Aug 9, 2012 #3

    uart

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    [itex]d {\bf r}[/itex] is the vector differential [itex]<dx, dy>[/itex]. So, using nothing but the definition of the dot product, the integral can be expanded as:

    [tex]I = ~_\mathbb{C} \!\!\! \int u_x(x,y) \, dx + u_y(x,y) dy \, [/tex]

    At this point it is still a line integral over a specified contour, but we can use the parametric equations of the contour to express it as a "normal" integral over 0 to 2 pi. That is:

    [tex]I = \int_0^{2 \pi} \left( u_x(t) \frac{dx}{dt} + u_y(t) \frac{dy}{dt} \right) \, dt [/tex]

    Does that help?
     
  5. Aug 9, 2012 #4

    uart

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    BTW. There's one other thing that you should notice about this question. It obvious that the vector function, [itex] x{\bf \, i} + y{\bf \, j}[/itex], is radially directed, and therefore at every point is perpendicular to the given contour. So we could actually state that the integral is zero by inspection.

    At this point however I'd guess that your teacher doesn't want you to do it that way. :smile:
     
  6. Aug 9, 2012 #5
    Thanks uart, that helps a lot. I was thinking along those lines but couldn't make it fit. What about that

    [tex]{x^2}[/tex]

    How does that become?

    [tex]{a^2}[/tex]

    Thanks.
     
  7. Aug 9, 2012 #6

    HallsofIvy

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    What "[itex]x^2[/itex]" are you talking about? The only [itex]x^2[/itex] in your integral is part of [itex]x^2+ y^2[/itex]. On the circle of radius a, [itex]x^2+ y^2=a^2cos^2(t)+ a^2 sin^2 t= a^2(cos^2(t)+ sin^2(t))= a^2[/itex].
     
  8. Aug 9, 2012 #7

    uart

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    It doesn't. However, [itex]x^2 + y^2[/itex] does become [itex]a^2[/itex]. That's a simple trig identity, can you see it?
     
  9. Aug 9, 2012 #8
    Did you find dr yet? Did you take the dot product?
     
  10. Aug 9, 2012 #9
    Yes thanks uart, I see how it's all working now.
     
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