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Line Integral of grad(f)

  1. Jun 25, 2009 #1
    EDIT:
    SORRY, I didn't read the directions. It says, "Answer the following short questions: If true, justify, if false give a counterex-
    ample."

    I'm certain that this question is one of the "false" ones, which is why I was so confused. LOL

    1. The problem statement, all variables and given/known data
    Let f(x, y, z) = y - x. Then the line integral of grad(f) around the unit circle in the xy plane is [tex]\pi[/tex], the area of the circle.


    2. Relevant equations
    A line integral of a vector field which is the gradient of a scalar field is path independent.

    3. The attempt at a solution
    I had two ways of solving:

    Method 1
    Fundamental theorem of line integrals:
    [tex] \oint_C \nabla f dr = 0 [/tex] around a closed curve C.
    Path independence guarantees that the line integral of grad(f) depends only on the initial and final points, which are the same on the unit circle.


    Method 2
    [tex]grad(f) = <-1, 1, 0> [/tex]

    Parametrize the unit circle C by:
    [tex]x(t) = cos(t) 0 \leq t \leq 2\pi [/tex]
    [tex]y(t) = sin(t) 0 \leq t \leq 2\pi [/tex]
    [tex]z=0 [/tex]
    [tex]x'(t) = -sin(t)[/tex]
    [tex]y'(t) = cos(t)[/tex]

    [tex]\int_C grad(f) dr [/tex]
    [tex]= \int <-1, 1, 0><dx, dy, dz> [/tex]
    [tex]= \int -dx + \int dy [/tex]
    [tex]= \int_0^{2\pi} -\frac{dx}{dt}dt +\int_0^{2\pi} \frac{dy}{dt}dt [/tex]
    [tex]= \int_0^{2\pi} sin(t) dt + \int_0^{2\pi} cos(t) dt [/tex]

    [tex]= 0 [/tex]


    Using either method, I get 0 as the answer. The question suggests that the answer should be [tex]\pi[/tex]. Am I missing something?
     
  2. jcsd
  3. Jun 26, 2009 #2

    dx

    User Avatar
    Homework Helper
    Gold Member

    Your answer is correct. The line integral is zero.
     
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