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Find the value of integral ${\int}_{C}\text{F}\xb7d\text{r},$ where $C$ is the semicircle parameterized by $\text{r}\left(t\right)=\u27e8\text{cos}\phantom{\rule{0.2em}{0ex}}t+\pi ,\text{sin}\phantom{\rule{0.2em}{0ex}}t\u27e9,0\le t\le \pi $ and $\text{F}=\u27e8\text{\u2212}y,x\u27e9.$
Notice that this is the same problem as [link] , except the orientation of the curve has been traversed. In this example, the parameterization starts at $\text{r}\left(0\right)=\u27e8\pi ,0\u27e9$ and ends at $\text{r}\left(\pi \right)=\u27e80,0\u27e9.$ By [link] ,
Notice that this is the negative of the answer in [link] . It makes sense that this answer is negative because the orientation of the curve goes against the “flow” of the vector field.
Let C be an oriented curve and let − C denote the same curve but with the orientation reversed. Then, the previous two examples illustrate the following fact:
That is, reversing the orientation of a curve changes the sign of a line integral.
Let $\text{F}=x\text{i}+y\text{j}$ be a vector field and let C be the curve with parameterization $\u27e8t,{t}^{2}\u27e9$ for $0\le t\le 2.$ Which is greater: ${\int}_{C}\text{F}\xb7\text{T}ds$ or ${\int}_{\text{\u2212}C}\text{F}\xb7\text{T}ds}?$
${\int}_{C}\text{F}\xb7\text{T}}ds$
Another standard notation for integral $\int}_{C}\text{F}\xb7d\text{r$ is ${\int}_{C}Pdx+Qdy+Rdz.$ In this notation, P , Q , and R are functions, and we think of d r as vector $\u27e8dx,dy,dz\u27e9.$ To justify this convention, recall that $d\text{r}=\text{T}ds={r}^{\prime}\left(t\right)dt=\u27e8\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt}\u27e9dt.$ Therefore,
If $d\text{r}=\u27e8dx,dy,dz\u27e9,$ then $\frac{d\text{r}}{dt}=\u27e8\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt}\u27e9,$ which implies that $\frac{d\text{r}}{dt}=\u27e8\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt}\u27e9dt.$ Therefore
Find the value of integral ${\int}_{C}zdx+xdy+ydz,$ where C is the curve parameterized by $\text{r}\left(t\right)=\u27e8{t}^{2},\sqrt{t},t\u27e9,1\le t\le 4.$
As with our previous examples, to compute this line integral we should perform a change of variables to write everything in terms of t . In this case, [link] allows us to make this change:
Find the value of ${\int}_{C}4xdx+zdy+4{y}^{2}dz,$ where $C$ is the curve parameterized by $\text{r}\left(t\right)=\u27e84\phantom{\rule{0.2em}{0ex}}\text{cos}\left(2t\right),2\phantom{\rule{0.2em}{0ex}}\text{sin}\left(2t\right),3\u27e9,0\le t\le \frac{\pi}{4}.$
$\mathrm{-26}$
We have learned how to integrate smooth oriented curves. Now, suppose that C is an oriented curve that is not smooth, but can be written as the union of finitely many smooth curves. In this case, we say that C is a piecewise smooth curve . To be precise, curve C is piecewise smooth if C can be written as a union of n smooth curves ${C}_{1},{C}_{2}\text{,\u2026},{C}_{n}$ such that the endpoint of ${C}_{i}$ is the starting point of ${C}_{i+1}$ ( [link] ). When curves ${C}_{i}$ satisfy the condition that the endpoint of ${C}_{i}$ is the starting point of ${C}_{i+1},$ we write their union as ${C}_{1}+{C}_{2}+\cdots +{C}_{n}.$
The next theorem summarizes several key properties of vector line integrals.
Let F and G be continuous vector fields with domains that include the oriented smooth curve C . Then
Notice the similarities between these items and the properties of single-variable integrals. Properties i. and ii. say that line integrals are linear, which is true of single-variable integrals as well. Property iii. says that reversing the orientation of a curve changes the sign of the integral. If we think of the integral as computing the work done on a particle traveling along C , then this makes sense. If the particle moves backward rather than forward, then the value of the work done has the opposite sign. This is analogous to the equation $\int}_{a}^{b}f\left(x\right)dx=\text{\u2212}{\displaystyle {\int}_{a}^{b}f\left(x\right)dx.$ Finally, if $\left[{a}_{1},{a}_{2}\right],\left[{a}_{2},{a}_{3}\right]\text{,\u2026},\left[{a}_{n-1},{a}_{n}\right]$ are intervals, then
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