$\newcommand{\vF}{\mathbf{\vec F}}$ $\newcommand{\vN}{\mathbf{\vec N}}$ $\newcommand{\vT}{\mathbf{\vec T}}$ $\newcommand{\ve}{\mathbf{\vec e}{}}$ $\newcommand{\vx}{\mathbf{\vec x}{}}$ $\newcommand{\cC}{\mathcal{C}}$ $\newcommand{\cR}{\mathcal{R}}$

Math 234 — Final Exam Review

Line integrals

There are three kinds of line integrals: The work and flux integrals are special cases of the “plain” line integrals. In a work integral the quantity being integrated is $\vF\cdot\vT$, and in a flux integral it is $\vF\cdot\vN$.

The plain line integral of a function $f(\vx)$, which must be defined at least on the curve $\cC$: \[ \int_\cC f(\vx)\, ds = \int_{t=a}^b f(\vx(t))\, \|\vx{}'(t)\|dt \hspace{144px} ds=\|\vx'\|dt \] Work integrals There are several different notations, all of which are related: \begin{align*} \int_\cC \vF\cdot d\vx &= \int_\cC P(x, y)dx+Q(x, y)dy && \vF(x,y) = \begin{pmatrix} P(x,y) \\ Q(x,y) \end{pmatrix} \\ &= \int_{t=a}^b \vF\cdot\vx'(t)\,dt && d\vx = \vx'(t) dt\\ &= \int_\cC \vF\cdot\vT\, ds &&\vT=\frac{\vx'}{\|\vx'\|},\quad ds=\|\vx'\|dt\\ \end{align*} In the simplest cases you can figure out what $\vT$ is from a drawing, and figure out if $\vF\cdot\vT$ is a constant. If this is the case, then the last formula for the work integral is the best: \[ \vF\cdot\vT \text{ constant} \implies \int_\cC \vF\cdot\vT \,ds = (\vF\cdot\vT) \int_\cC \,ds = (\vF\cdot\vT)\,\cdot\, (\text{length of }\cC) . \] Flux integrals: from the given curve you must compute the unit normal $\vN$, usually by rotating the unit tangent $\vT$ (see the integral $L_3$ from the review problems). In some important special cases (where $\cC$ is a line segment, or a circle) you can get the unit normal $\vN$ from a drawing.

Conservative Vector Fields

A vector field $\vF$ is called conservative if the work integral \[ \int_\cC \vF\cdot\vT\, ds \] only depends on the endpoints of the curve $\cC$, and not on how $\cC$ goes from one end point to the other.

There is a Theorem that says that a vectorfield $\vF$ is conservative if and only if it is the gradient of some function, i.e. $\vF = \vec\nabla f$ for some function $f$.

If \[ \vF(x, y) = \vec\nabla f = \begin{pmatrix} f_x(x,y) \\ f_y(x,y) \end{pmatrix} \] then \[ \int_\cC \vF\cdot\vT\, ds = \int_\cC \vec\nabla f\cdot\vT\, ds = f(B) - f(A) \] where $A$ and $B$ are the initial and final point of the curve $\cC$.

Green's theorem

If $\cC$ is a closed curve which encloses a region $\cR$ in the plane, and if $\vF = \begin{pmatrix} P(x,y) \\ Q(x,y) \end{pmatrix}$ is a vectorfield then \[ \int_\cC \vF\cdot\vT \, ds = \iint_\cR \Bigl\{\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\Bigr\}\, dA \] Here $\vT$ must be the unit tangent for $\cC$ that points in the counterclockwise direction.

Since there are several different ways of writing the line integral in Green's theorem, there are several different ways of presenting the theorem: \[ \int_\cC \vF\cdot d\vx = \int_\cC P(x, y)dx+Q(x, y)dy = \iint_\cR \Bigl\{\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\Bigr\}\, dA \]