Section 5.2 Line Integral

5.2 Line Integral

A line integral gives us the ability to integrate multivariable functions and vector fields over arbitrary curves in a plane or in space. There are two types of line integrals: scalar line integrals and vector line integrals. Scalar line integrals are integrals of a scalar function over a curve in a plane or in space. Vector line integrals are integrals of a vector field over a curve in a plane or in space. Let’s look at scalar line integrals first. 

Definition 

Let [latex]f[/latex] be a function with a domain that includes the smooth curve [latex]C[/latex] that is parameterized by [latex]\overrightarrow{r(t)}=\lt x(t),y(t),z(t)\gt[/latex], [latex]a\le t\le b[/latex]. The scalar line integral of [latex]f[/latex] along [latex]C[/latex] is 

\[\int_{C}f(x,y,z)ds=\lim_{n\rightarrow\infty}\sum_{i=1}^{n}f(P_{i}^{*})\triangle s_{i}\]

if this limit exists. 

If [latex]C[/latex] is a planar curve, then [latex]C[/latex] can be represented by the parametric equations [latex]x=x(t),y=y(t)[/latex], and [latex]a\le t\le b[/latex]. If [latex]C[/latex] is smooth and [latex]f(x,y)[/latex] is a function of two variables, then the scalar line integral of [latex]f[/latex] along [latex]C[/latex] is defined similarly.

Theorem: Evaluating a Scalar Line Integral 

Let [latex]f[/latex] be a continuous function with a domain that includes the smooth curve [latex]C[/latex] with parameterization [latex]\overrightarrow{r(t)}[/latex], [latex]a\le t\le b.[/latex] Then

\[\int_{C}f(x,y,z)ds=\int_{a}^{b}f(\overrightarrow{r(t)})||\overrightarrow{r'(t)}||dt.\]

Theorem: Scalar Line Integral Calculation 

Let [latex]f[/latex] be a continuous function with a domain that includes the smooth curve [latex]C[/latex] with parameterization [latex]\overrightarrow{r(t)}=\lt x(t),y(t),z(t)\gt[/latex], [latex]a\le t\le b[/latex]. Then

\[ \int_{C}f(x,y,z)ds=\int_{a}^{b}f(\overrightarrow{r(t)})\sqrt{(x'(t))^{2}+(y'(t))^{2}+(z'(t))^{2}}dt.\]

Similarly, 

\[\int_{C}f(x,y)ds=\int_{a}^{b}f(\overrightarrow{r(t)})\sqrt{(x'(t))^{2}+(y'(t))^{2}}dt \]

if [latex]C[/latex] is a planar curve and [latex]f[/latex] is a function of two variables. 

Example 1: Evaluate [latex]\int_{C}yds[/latex] where [latex]C[/latex] is the curve [latex]\overrightarrow{r(t)}=\lt t^{2},2t\gt[/latex] with [latex]0\leq t\leq2[/latex].

Exercise 1: Evaluate [latex]\int_{C}xds[/latex] where [latex]C[/latex] is the curve [latex]\overrightarrow{r(t)}=\lt 3t,t^{2}\gt[/latex] with [latex]0\leq t\leq1[/latex]. 

Example 2: Evaluate [latex]\int_{C}e^{y}ds[/latex] where [latex]C[/latex] is the line segment from [latex](1,2)[/latex] to [latex](3,4)[/latex].

Exercise 2: Evaluate [latex]\int_{C}e^{x}ds[/latex] where [latex]C[/latex] is the line segment from [latex](0,1)[/latex] to [latex](-1,2)[/latex]. 

Example 3: Evaluate [latex]\int_{C}x^{2}+y^{2}-zds[/latex] where [latex]C[/latex] is the part of the helix [latex]\overrightarrow{r(t)}=\lt \text{cos}(3t),\text{sin}(3t),t\gt[/latex], [latex]0\leq t\leq\pi[/latex].

Exercise 3: Evaluate [latex]\int_{C}x^{2}+z^{2}+yds[/latex] where [latex]C[/latex] is the part of the helix [latex]\overrightarrow{r(t)}=\lt \text{cos}(2t),t,\text{sin}(2t)\gt[/latex], [latex]0\leq t\leq\frac{\pi}{2}[/latex].

Example 4: A wire has a shape that can be modeled with the parameterization [latex]\overrightarrow{r(t)}=\lt \text{cos}(t),\text{sin}(t),t\gt[/latex], [latex]0\le t\le4\pi[/latex]. Find the length of the wire.

Exercise 4: A wire has a shape that can be modeled with the parameterization [latex]\overrightarrow{r(t)}=\lt \text{sin}(t),2t,\text{cos}(t)\gt[/latex], [latex]0\le t\le\pi[/latex]. Find the length of the wire.

Definition: line integral with respect to arc length

Let [latex]C[/latex] be the curve [latex]\overrightarrow{r(t)}=\lt x(t),y(t),z(t)\gt[/latex] then we can define the line integral with resopect to each component of [latex]\overrightarrow{r(t)}[/latex]. 

\[\int_{C}f(x,y,z)dx=\int_{a}^{b}f(\overrightarrow{r(t)})x'(t)dt\]

\[\int_{C}f(x,y,z)dy=\int_{a}^{b}f(\overrightarrow{r(t)})y'(t)dt\]

\[\int_{C}f(x,y,z)dz=\int_{a}^{b}f(\overrightarrow{r(t)})z'(t)dt.\]

Example 5: Evaluate [latex]\int_{C}xy+\text{cos}(x)dx[/latex] where [latex]C[/latex] is the arc of [latex]x=y^{2}[/latex] from [latex](0,0)[/latex] to [latex](\pi^{2},\pi)[/latex]. 

Exercise 5: Evaluate [latex]\int_{C}x^{2}y+\text{sin}(x)dy[/latex] where [latex]C[/latex] is the arc of [latex]y=x^{2}[/latex] from [latex](0,0)[/latex] to [latex](\pi,\pi^{2})[/latex]. 

Definition: We define the integral of [latex]F[/latex] along [latex]C[/latex] as the sum of the integrals of [latex]f[/latex] along each of the smooth pieces of [latex]C[/latex]:

\[\int_{C}f(x,y,z)ds=\int_{C_{1}}f(x,y,z)ds+\int_{C_{2}}f(x,y,z)ds+\cdots+\int_{C_{n}}f(x,y,z)ds.\]

Example 6: Evaluate [latex]\int_{C}x^{2}dx+y^{2}dy[/latex] where [latex]C[/latex] is the arc of circle [latex]x^{2}+y^{2}=9[/latex] from [latex](0,3)[/latex] to [latex](-3,0)[/latex] followed by the line segment from [latex](-3,0)[/latex] to [latex](-4,-5)[/latex]. 

Exercise 6: Evaluate [latex]\int_{C}ydx+xdy[/latex] where [latex]C[/latex] is the arc of circle [latex]x^{2}+y^{2}=1[/latex] from [latex](1,0)[/latex] to [latex](0,1)[/latex] followed by the line segment from [latex](0,1)[/latex] to [latex](-1,2)[/latex]. 

Example 7: Evaluate [latex]\int_{C}z^{2}dx+x^{2}dy+y^{2}dz[/latex] where [latex]C[/latex] is the line segments from [latex](1,0,0)[/latex] to [latex](1,0,2)[/latex] and from [latex](1,0,2)[/latex] to [latex](2,3,4)[/latex].

Group work:

1. Evaluate [latex]\int_{C}x^{2}+yzds[/latex] where [latex]C[/latex] is the curve [latex]\overrightarrow{r(t)}=\lt 2t,5t,-t\gt[/latex], [latex]0\leq t\leq4[/latex].

2. Evaluate [latex]\int_{C}xyds[/latex] where [latex]C[/latex] is the first half of the circle [latex]x^{2}+y^{2}=9[/latex] and is traverse in the clockwise direction.

3. Evaluate [latex]\int_{C}y^{2}dx+xdy[/latex] where [latex]C[/latex] is the arc of [latex]x=4-y^{2}[/latex] from [latex](4,0)[/latex] to [latex](0,2)[/latex].