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Green's Theorem vs Fundamental

  1. Apr 28, 2015 #1
    1. The problem statement, all variables and given/known data
    1) How do I know when to use Green's Theorem, the Fundamental Theorem for Line Integrals or the regular method of using parametrization?
    2) Assuming that the three methods above are all used to solve line integrals, why do the Fundamental Theorem and Green give different answers? My textbook solves ##\int_C F \cdot dr## given ##F(x,y,z) = yz \vec i + xz \vec j + (xy + 2z) \vec k## and C is the line segment fro (1, 0, -2) to (4, 6, 3). It is can be derived that the vector field is conservative. If it's conservative, solving with Green's Theorem gives zero, but using the Fund Thm, the answer is 77.
    3) What is the difference between a vector field and a line integral?
    4) Intuitively, what is the difference between solving ##\int_C y^2dx + xdy ## and ##\int_C 2+x^y ds ##? If ##dx## and ##dy## signify integrating the line integral along the x and y direction, what is ##ds##? Is there an s-axis somewhere?

    2. Relevant equations
    n/a

    3. The attempt at a solution
    My textbook's chapters don't seem to be related to one another, which is why I am confused.
     
  2. jcsd
  3. Apr 28, 2015 #2

    Zondrina

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    1) The fundamental theorem for line integrals:

    $$\oint_C \vec{\nabla f} \cdot d \vec r = f(\vec r(b)) - f(\vec r(a))$$

    Says we can evaluate the line integral of a conservative vector field ##\vec F = \vec{\nabla f}## by knowing the value of ##f## at the endpoints of ##C##. The line integral of ##\vec{\nabla f}## is the net change in ##f##. This should be used when the function ##f## is given with the endpoints of the curve.

    Green's theorem:

    $$\oint_C \vec F \cdot d \vec r = \iint_D \left[ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right] \space dA$$

    Should be used in the two-dimensional case when given a positively oriented, piecewise smooth, simple closed curve ##C## that encloses a region ##D##. It's particularly useful for annular regions.

    Simply parametrizing a curve ##C## is the most standard way to do any line integral, but it may prove to be tedious, which is why you have so many useful theorems.

    2) You can't use Green's theorem in three dimensions.

    3) A vector field is all of the vectors produced from ##\vec F(P)## for any point ##P \in \mathbb{R}^n##. A line integral ##\oint_C## is very similar to a regular integral ##\int_a^b##, except you integrate the area along a curve as opposed to an interval.

    4) There is really no difference between ##\oint_C P \space dx + Q \space dy## and ##\oint_C f \space ds##. It's just a matter of how to parametrize the curve.

    As for what ##ds## actually means, suppose a curve ##C## is given by a vector equation ##\vec r(t) = x(t) \hat i + y(t) \hat j, \space t \in [a, b]##. Suppose further ##\vec r(t)## is smooth, and ##s(t)## is the length of ##C## between ##\vec r(a)## and ##\vec r(t)##. Then:

    $$\frac{ds}{dt} = \sqrt{\left(\frac{dx}{dt} \right)^2 + \left(\frac{dy}{dt} \right)^2} = |\vec r'(t)|$$
    $$ds = \sqrt{\left(\frac{dx}{dt} \right)^2 + \left(\frac{dy}{dt} \right)^2} \space dt = |\vec r'(t)| dt$$
     
    Last edited: Apr 28, 2015
  4. Apr 28, 2015 #3

    LCKurtz

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    Green's theorem applies when the path is a closed curve enclosing an area, not just a line segment.

    How would you even apply Green's theorem? There is no enclosed area for which to do a double integral.

    A vector field might be, for example, a force field. It just sits there. A line integral might be used to calculate the work done in moving around in a force field. They are completely different things.

    The first integral is of the form ##\int_C \vec F\cdot d\vec R## which might, as above, be calculating work. The second integral as an arc length type integral. ##ds## is the differential element of are length and such an integral might represent and integral of the type ##\int_C \delta(s)~ds## where ##\delta(s)## is a mass density and the integral represents the mass of a wire. Different types of integrals represent solutions to different types of problems.
     
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