$\newcommand{\dis}{\displaystyle} \newcommand{\m}{\hspace{1em}} \newcommand{\mm}{\hspace{2em}} \newcommand{\x}{\vspace*{1ex}} \newcommand{\xx}{\vspace*{2ex}} \let\limm\lim \renewcommand{\lim}{\dis\limm} \let\fracc\frac \renewcommand{\frac}{\dis\fracc} \let\summ\sum \renewcommand{\sum}{\dis\summ} \let\intt\int \renewcommand{\int}{\dis\intt} \newcommand{\sech}{\text{sech}} \newcommand{\csch}{\text{csch}} \newcommand{\Ln}{\text{Ln}} \newcommand{\p}{\partial} \newcommand{\intd}[1]{\int\hspace{-0.7em}\int\limits_{\hspace{-0.7em}{#1}}} $

Lecture 28, 11/2/2022. This page is for Section 1 only.
ACMS 20550: Applied Mathematics Method I
Instructor: Bei Hu, b1hu@nd.edu, Hurley 174A

Chapter 6: Vector Analysis

  1. Line Integrals.
    1. For $\vec F = (F_1,F_2), \m d\vec r = (dx, dy)$, the line integral
      $\m\int \vec F\cdot d\vec r = \int (F_1,F_2)\cdot (dx, dy) = \int F_1 dx + F_2 dy$.
      Then find $\fcolorbox{white}{yellow}{the equation for the curve,}$ the upper and lower limit, and compute.
      The problem can be set up in 3D: $\mm\vec F = (F_1,F_2,F_3), \m d\vec r = (dx, dy, dz)$,
      $\m\int \vec F\cdot d\vec r = \int (F_1,F_2, F_3)\cdot (dx, dy, dz) = \int F_1 dx + F_2 dy+F_3 dz$.
    2. Conserved Fields:
      $\fcolorbox{white}{yellow}{Conserved Fields:}$ the value of the line integral is independent of the path.
      If $\vec F = \nabla W$, then $\vec F$ is a conserved field, and $\phi = - W$ is called the $\fcolorbox{white}{yellow}{potential.}$