Integrating a vector field along a curve is useful for computations such as work and circulation. Suppose [math]\textbf{F}[/math] is a vector field and the points of a curve [math]C[/math] lie in the domain of [math]\textbf{F}[/math]. Let [math]\textbf{T}[/math] be the unit tangent vector to [math]C[/math]. [i]The line integral of [/i][math]\textbf{F}[/math][i] along [/i][math]C[/math] is[br] [math]\int_C\textbf{F}\cdot\textbf{T}ds[/math].[br]If [math]C[/math] is parametrized by [math]\textbf{r}\left(t\right)[/math] for [math]a\le t\le b[/math], then it turns out that[br] [math]\int_C\textbf{F}\cdot\textbf{T}ds=\int_a^b\textbf{F}\cdot\frac{d\textbf{r}}{dt}dt[/math],[br]which is useful for actual computation.[br] In this interactive figure, carefully work down the controls in the right-hand pane from top to bottom, trying the sliders and checkboxes to see their effect. See how the parametrization of [math]C[/math] and the definition of [math]\textbf{F}[/math] are reflected in the graph. By examining [math]C[/math] and [math]\textbf{F}[/math], you should be able to see which points on [math]C[/math] make the integrand [math]\textbf{F}\cdot\textbf{T}[/math] positive, and which make it negative. From this you may gain some intuition about whether the line integral of [math]\textbf{F}[/math] along [math]C[/math] should be positive or negative.

[i]Developed for use with Thomas' Calculus and [url=https://www.pearson.com/en-us/subject-catalog/p/interactive-calculus-early-transcendentals-single-variable/P200000009666]Interactive Calculus[/url], published by Pearson.[/i]