3.6: Line Integrals
- Page ID
- 50414
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)The ingredients for line (also called path or contour) integrals are the following:
- A vector field \(F = (M, N)\)
- A curve \(\gamma (t) = (x(t), y(t))\) defined for \(a \le t \le b\)
Then the line integral of \(F\) along \(\gamma\) is defined by
\[\int_{\gamma} F \cdot dr = \int_a^b F(\gamma (t)) \cdot y'(t)dt = \int_{\gamma} M\ dx + N\ dy. \nonumber \]
Let \(F = (-y/r^2, x/r^2)\) and let \(\gamma\) be the unit circle. Compute line integral of \(F\) along \(\gamma\).
Solution
You should be able to supply the answer to this example
Properties of line integrals
1. Independent of parametrization.
2. Reverse direction on curve \(\Rightarrow\) change sign. That is,
\[\int_{-C} F \cdot dr = -\int_{C} F \cdot dr. \nonumber \]
(Here, \(-C\) means the same curve traversed in the opposite direction.)
3. If \(C\) is closed then we sometimes indicate this with the notation \(\oint_{C} F \cdot dr = \oint_{C} M\ dx + N\ dy\).
Fundamental theorem for gradient fields
If \(F = nabla f\) then \(\int_{\gamma} F \cdot d r = f(P) - f(Q)\), where \(Q, P\) are the beginning and endpoints respectively of \(\gamma\).
- Proof
-
By the chain rule we have
\[\dfrac{df(\gamma (t))}{dt} = \nabla f(\gamma (t)) \cdot \gamma '(t) = F(\gamma (t)) \cdot y'(t). \nonumber \]
The last equality follows from our assumption that \(F = \nabla f\). Now we can this when we compute the line integral:
\[\begin{array} {rcl} {\int_{\gamma} F \cdot dr} & = & {\int_a^b F (\gamma (t)) \cdot y' (t)\ dt} \\ {} & = & {\int_a^b \dfrac{df(\gamma (t))}{dt} dt} \\ {} & = & {f(\gamma (b)) - f(\gamma (a))} \\ {} & = & {f(P) - f(Q)} \end{array} \nonumber \]
Notice that the third equality follows from the fundamental theorem of calculus.
If a vector field \(F\) is a gradient field, with \(F = \nabla f\), then we call \(f\) a potential function for \(F\).
Note: the usual physics terminology would be to call \(f\) the potential function for \(F\).
independence and conservative functions
For a vector field \(F\), the line integral \(\int F \cdot dr\) is called path independent if, for any two points \(P\) and \(Q\), the line integral has the same value for \(every\) path between \(P\) and \(Q\).
\(\int_C F \cdot dr\) is path independent is equivalent to \(\oint_{C} F \cdot dr = 0\) for any closed path.
- Sketch of Proof
-
Draw two paths from \(Q\) to \(P\). Following one from \(Q\) to \(P\) and the reverse of the other back to \(P\) is a closed path. The equivalence follows easily. We refer you to the more detailed review of line integrals and Green’s theorem for more details.
A vector field with path independent line integrals, equivalently a field whose line integrals around any closed loop is 0 is called a conservative vector field.
We have the following equivalence: On a connected region, a gradient field is conservative and a conservative field is a gradient field.
- Proof
-
Again we refer you to the more detailed review for details. Essentially, if \(F\) is conservative then we can define a potential function \(f(x, y)\) as the line integral of \(F\) from some base point to \((x, y)\).