Lecture 49: Line Integrals with Respect to Coordinates
Graduate Entrance Examination Mathematics study notes: Lecture 49: Line Integrals with Respect to Coordinates. Original formulas, diagrams, and examples are retained.
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Graduate Entrance Examination MathematicsAdvanced MathematicsMultivariable Integral Calculus and Applications
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58.1 Curve Integration with Respect to Coordinates
Explanation: Method selection
Closed Section:
Green's formula
Non-closed interval:
Check: Check whether it is independent of the path -> Use partial derivatives to see if they are equal
It is path-independent
Change the path
Use a potential function.
The integral is path-dependent.
Direct calculation is convenient -> Direct calculation
Direct calculations are inconvenient -> Supplement Green
Description: the current parameter equation is $\begin{cases}x=\varphi(t)\\y=\psi(t)\end{cases}$, from start point A to endpoint B, t moves from α to β:$$\int_{L}P(x,y)dx+Q(x,y)dy = \int_{\alpha}^{\beta}[p(\psi(t),\psi(t))\psi(t)+Q(\varphi(t),\psi(t))\psi^{\prime}(t)]dt$$
Explanation
Concept:
Write out the parameter equations, carry them in, and convert them into definite integral calculations;
Note:
The upper and lower limits are calculated from the starting point parameter -> the endpoint parameters, rather than by size;
58.2.2 Method Two: Green's Formula
Introduction
On a double integral of a closed region D in a plane, can only the value difference on the boundary curve L be found without calculating the values of all points on the surface?
This function is achieved by Green's formula;
#####Definition: Singleconnectedarea
Description: A plane region $D$ is simply connected if every simple closed curve in $D$, together with its interior, lies entirely in $D$. Otherwise, the region is multiply connected.
Explanation
The region must be simply connected (it contains no holes).
#####Theorem: Greensformula
Description: Let the closed region $D$ be bounded by a piecewise smooth, positively oriented curve $L$. If $P(x,y)$ and $Q(x,y)$ have continuous first-order partial derivatives on $D$, then:
The range of Green's theorem used -> must be on a closed region: that is, the curve is closed;
The positive and negative directions are relative to the current area;
Where:
L is the positive boundary curve of region D;
Supplement: Closing an open path before applying Green's theorem
Green's theorem applies to a closed curve. For an open path, add a convenient auxiliary segment to form a closed curve, apply Green's theorem, and then subtract the integral over the auxiliary segment.
(a) Change path: First, change to a simpler path (usually along the coordinate axis).
(b) Using a potential function: $\int_{(x_{1},y_{1})}^{(x_{2},y_{2})}P\mathrm{d}x+Q\mathrm{d}y=F(x_{2},y_{2})-F(x_{1},y_{1})$
Methods for finding the potential function: 1. Integrate one partial derivative and determine the remaining function; 2. Complete the total differential.
28.3 Connections Between Two Types of Line Integrals
Example: $\text{ Let }L\text{ This is the column face }x^2+y^2=1\text{ and planes }y+z=0\text{ The intersection of the line, from }z\text{ The axis is steadfastly longing }z\text{ If the negative axis is viewed counterclockwise, then the curve integral }\int_{L}z\operatorname{d}x+y\operatorname{d}z$
Let $x=\cos t,y=\sin t,z=-\sin t$, substitute to get the definite integral: $I=\int_{0}^{2x}(\sin^{2}t-\sin t\cos t)dt$
