Applications of Line Integrals

Scalar line integrals have many applications. They can be used to calculate the length or mass of a wire, the surface area of a sheet of a given height, or the electric potential of a charged wire given a linear charge density. Vector line integrals are extremely useful in physics. They can be used to calculate the work done on a particle as it moves through a force field, or the flow rate of a fluid across a curve. Here, we calculate the mass of a wire using a scalar line integral and the work done by a force using a vector line integral.

Suppose that a piece of wire is modeled by curve [latex]C[/latex] in space. The mass per unit length (the linear density) of the wire is a continuous function [latex]{\rho}{({x}, {y}, {z})}[/latex]. We can calculate the total mass of the wire using the scalar line integral [latex]{\displaystyle\int_{C}}{\rho}{({x}, {y}, {z})}{ds}[/latex]. The reason is that mass is density multiplied by length, and therefore the density of a small piece of the wire can be approximated by [latex]{\rho}{({x}^{*}, {y}^{*}, {z}^{*})}{\Delta}{s}[/latex] for some point [latex]({x}^{*}, {y}^{*}, {z}^{*})[/latex] in the piece. Letting the length of the pieces shrink to zero with a limit yields the line integral [latex]{\displaystyle\int_{C}}{\rho}{({x}, {y}, {z})}{ds}[/latex].

Example: calculating the mass of a wire

Calculate the mass of a spring in the shape of a helix parameterized by [latex]\langle{t},2\cos{t},2\sin{t}\rangle[/latex], [latex]0\leq{t}\leq\frac{\pi}2[/latex], with a density function given by [latex]\rho(x,y,z)=e^x+yz[/latex] kg/m.

Figure 1. The wire from Example “Calculating the Mass of a Wire”.

Show Solution

To calculate the mass of the spring, we must find the value of the scalar line integral [latex]\displaystyle\int_c(e^x+yz)ds[/latex], where [latex]C[/latex] is the given helix. To calculate this integral, we write it in terms of [latex]t[/latex] using The Scalar Line Integral Calculation Theorem Equation:

[latex]\begin{aligned} \displaystyle\int_ce^x+yzds&=\displaystyle\int_0^{\pi/2}\left((e^t+4\cos{t}\sin{t})\sqrt{1+(-2\cos{t}^2+(2\sin{t})^2}\right)dt \\ &=\displaystyle\int_0^{\pi/2}((e^t+4\cos{t}\sin{t})\sqrt{5})dt \\ &=\sqrt{5}[e^t+2\sin^2t]_{t=0}^{t=\pi/2} \\ &=\sqrt{5}(e^{\pi/2}+1) \end{aligned}[/latex].

Therefore, the mass is [latex]\sqrt{5}(e^{\pi/2}+1)[/latex] kg.

try it

Calculate the mass of a spring in the shape of a helix parameterized by [latex]{\bf{r}}(t)=\langle\cos{t},\sin{t},t\rangle[/latex], [latex]0\leq{t}\leq6\pi[/latex], with a density function given by [latex]\rho(x,y,z)=x+y+z[/latex] kg/m.

Show Solution

[latex]18\sqrt2\pi^2[/latex] kg.

Watch the following video to see the worked solution to the above Try It

You can view the transcript for “CP 6.20” here (opens in new window).When we first defined vector line integrals, we used the concept of work to motivate the definition. Therefore, it is not surprising that calculating the work done by a vector field representing a force is a standard use of vector line integrals. Recall that if an object moves along curve [latex]C[/latex] in force field [latex]{\bf{F}}[/latex], then the work required to move the object is given by [latex]\displaystyle\int_c{\bf{F}}\cdot \ d\bf{r}[/latex].

Example: calculating work

How much work is required to move an object in vector force field [latex]{\bf{F}}=\langle{y}x,xy,xz\rangle[/latex] along path [latex]{\bf{r}}(t)=\langle{t}^2,t,t^4\rangle[/latex], [latex]0\leq{t}\leq1[/latex]?  See Figure 2.

Show Solution

Let [latex]C[/latex] denote the given path. We need to find the value of [latex]\displaystyle\int_c{\bf{F}}\cdot \ d\bf{r}[/latex]. To do this, we use the equation to compute vector line integrals:

[latex]\begin{aligned} \displaystyle\int_c{\bf{F}}\cdot \ d\bf{r}&=\displaystyle\int_0^1(\langle{t}^5,t^3,t^6\rangle\cdot\langle2t,1,4t^3\rangle)dt\\ &=\displaystyle\int_0^1(2t^6+t^3+4t^9) \ dt \\ &=\left[\frac{2t^7}7+\frac{t^4}4+\frac{2t^{10}}5\right]_{t=0}^{t=1}=\frac{131}{140}. \end{aligned}[/latex]

Figure 2. The curve and vector field for Example “Calculating Work”.

Flux and Circulation

We close this section by discussing two key concepts related to line integrals: flux across a plane curve and circulation along a plane curve. Flux is used in applications to calculate fluid flow across a curve, and the concept of circulation is important for characterizing conservative gradient fields in terms of line integrals. Both these concepts are used heavily throughout the rest of this chapter. The idea of flux is especially important for Green’s theorem, and in higher dimensions for Stokes’ theorem and the divergence theorem.

Let [latex]C[/latex] be a plane curve and let [latex]{\bf{F}}[/latex] be a vector field in the plane. Imagine [latex]C[/latex] is a membrane across which fluid flows, but [latex]C[/latex] does not impede the flow of the fluid. In other words, [latex]C[/latex] is an idealized membrane invisible to the fluid. Suppose [latex]{\bf{F}}[/latex] represents the velocity field of the fluid. How could we quantify the rate at which the fluid is crossing [latex]C[/latex]?

Recall that the line integral of [latex]{\bf{F}}[/latex] along [latex]C[/latex] is [latex]\displaystyle\int_c{\bf{F}}\cdot{\bf{T}} \ ds[/latex]—in other words, the line integral is the dot product of the vector field with the unit tangential vector with respect to arc length. If we replace the unit tangential vector with unit normal vector [latex]{\bf{N}}(t)[/latex] and instead compute integral [latex]\displaystyle\int_c{\bf{F}}\cdot{\bf{N}} \ ds[/latex], we determine the flux across [latex]C[/latex]. To be precise, the definition of integral [latex]\displaystyle\int_c{\bf{F}}\cdot{\bf{N}} \ ds[/latex] is the same as integral [latex]\displaystyle\int_c{\bf{F}}\cdot{\bf{T}} \ ds[/latex], except the [latex]{\bf{T}}[/latex] in the Riemann sum is replaced with [latex]{\bf{N}}[/latex]. Therefore, the flux across [latex]C[/latex] is defined as

[latex]\large{\displaystyle\int_c{\bf{F}}\cdot{\bf{N}} \ ds=\displaystyle\lim_{n\to\infty}\displaystyle\sum_{i=1}^n{\bf{F}}(P_i^*)\cdot{\bf{N}}(P_i^*)\Delta{s_i}}[/latex],

where [latex]P_i^*[/latex] and [latex]\Delta{s}[/latex] are defined as they were for integral [latex]\displaystyle\int_c{\bf{F}}\cdot{\bf{T}} \ ds[/latex]. Therefore, a flux integral is an integral that is perpendicular to a vector line integral, because [latex]{\bf{N}}[/latex] and [latex]{\bf{T}}[/latex] are perpendicular vectors.

If [latex]{\bf{F}}[/latex] is a velocity field of a fluid and [latex]C[/latex] is a curve that represents a membrane, then the flux of [latex]{\bf{F}}[/latex] across [latex]C[/latex] is the quantity of fluid flowing across [latex]C[/latex] per unit time, or the rate of flow.

More formally, let [latex]C[/latex] be a plane curve parameterized by [latex]{\bf{r}}(t)=\langle{x}(t),y(t)\rangle[/latex], [latex]a\leq{t}\leq{b}[/latex]. Let [latex]{\bf{n}}(t)=\langle{y}^\prime(t),-x^\prime(t)\rangle[/latex] be the vector that is normal to [latex]C[/latex] at the endpoint of [latex]{\bf{r}}(t)[/latex] and points to the right as we traverse [latex]C[/latex] in the positive direction (Figure 3). Then, [latex]{\bf{N}}(t)=\frac{{\bf{n}}(t)}{||{\bf{n}}(t)||}[/latex] is the unit normal vector to [latex]C[/latex] at the endpoint of [latex]{\bf{r}}(t)[/latex] that points to the right as we traverse [latex]C[/latex].

Definition

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The flux of [latex]\bf{F}[/latex] across [latex]C[/latex] is line integral [latex]\displaystyle\int_C{\bf{F}}\cdot\frac{{\bf{n}}(t)}{||{\bf{n}}(t)||}[ds[/latex].

Figure 3. The flux of vector field [latex]\bf{F}[/latex] across curve [latex]C[/latex] is computed by an integral similar to a vector line integral.

We now give a formula for calculating the flux across a curve. This formula is analogous to the formula used to calculate a vector line integral (see the equation to compute vector line integrals).

Theorem: calculating flux across a curve

Let [latex]{\bf{F}}[/latex] be a vector field and let [latex]C[/latex] be a smooth curve with parameterization [latex]{\bf{r}}(t)=\langle{x}(t),y(t)\rangle[/latex], [latex]a\leq{t}\leq{b}[/latex]. Let [latex]{\bf{n}}(t)=\langle{y}^\prime(t),-x^\prime(t)\rangle[/latex]. The flux of [latex]{\bf{F}}[/latex] across [latex]C[/latex] is

[latex]\displaystyle\int_C{\bf{F}}\cdot{\bf{N}}ds=\displaystyle\int_a^b{\bf{F}}({\bf{r}}(t))\cdot{\bf{n}}(t)dt[/latex].

Proof

The proof of The Calculating Flux Across A Curve Theorem Equation is similar to the proof of The Scalar Line Integration Calculation Theorem Equation. Before deriving the formula, note that [latex]||{\bf{n}}(t)||=||\langle y'(t),-x'(t)\rangle||=\sqrt{(y^\prime(t))^2+(x^\prime(t))^2}=||{\bf{r}}^\prime(t)||[/latex]. Therefore,

[latex]\begin{aligned} \displaystyle\int_C{\bf{F}}\cdot{\bf{N}}ds&=\displaystyle\int_C{\bf{F}}\cdot\frac{{\bf{n}}}{||{\bf{n}}(t)||}ds \\ &=\displaystyle\int_a^b{\bf{F}}\frac{{\bf{n}}}{||{\bf{n}}(t)||}||{\bf{r}}^\prime(t)||dt \\ &=\displaystyle\int_a^b{\bf{F}}[({\bf{r}}(t))\cdot{\bf{n}}(t)dt \end{aligned}[[/latex].

[latex]_\blacksquare[/latex]

Example: flux across a curve

Calculate the flux of [latex]{\bf{F}}=\langle2x,2y\rangle[/latex] across a unit circle oriented counterclockwise (Figure 4).

Figure 4. A unit circle in vector field [latex]{\bf{F}}=\langle2x,2y\rangle[/latex].

Show Solution

To compute the flux, we first need a parameterization of the unit circle. We can use the standard parameterization [latex]{\bf{r}}(t)\langle\cos{t},\sin{t}\rangle[/latex], [latex]0\leq{t}\leq2\pi[/latex]. The normal vector to a unit circle is [latex]\langle\cos{t},\sin{t}\rangle[/latex]. Therefore, the flux is

[latex]\begin{aligned} \displaystyle\int_C{\bf{F}}\cdot{\bf{N}}ds&=\displaystyle\int_0^{2\pi}\langle2\cos{t},2\sin{t}\rangle\cdot\langle\cos{t},\sin{t}\rangle \ dt \\ &=\displaystyle\int_0^{2\pi}(2\cos^2{t},2\sin^2{t}) \ dt = 2\displaystyle\int_0^{2\pi}(\cos^2{t},\sin^2{t}) \ dt \\ &=2\displaystyle\int_0^{2\pi} \ dt = 4\pi \end{aligned}[/latex].

try it

Calculate the flux of [latex]{\bf{F}}=\langle{x}+y,2y\rangle[/latex] across the line segment from [latex](0, 0)[/latex] to [latex](2, 3)[/latex] where the curve is oriented from left to right.

Show Solution

[latex]3/2[/latex]

Let [latex]{\bf{F}}(x,y)=\langle{P}(x,y),Q(x,y)\rangle[/latex] be a two-dimensional vector field. Recall that integral [latex]\displaystyle\int_C{\bf{F}}\cdot{\bf{T}}ds[/latex] is sometimes written as [latex]\displaystyle\int_CPdx+Qdy[/latex]. Analogously, flux [latex]\displaystyle\int_C{\bf{F}}\cdot{\bf{N}}ds[/latex] is sometimes written in the notation [latex]\displaystyle\int_C-Qdx+Pdy[/latex], because the unit normal vector [latex]\bf{N}[/latex] is perpendicular to the unit tangent [latex]\bf{T}[/latex]. Rotating the vector [latex]d{\bf{r}}=\langle{d}x,dy\rangle[/latex] by 90° results in vector [latex]\langle{d}y,-dx\rangle[/latex]. Therefore, the line integral in Example “Using Properties to Compute a Vector Line Integral” can be written as [latex]\displaystyle\int_C-2ydx+2xdy[/latex].

Now that we have defined flux, we can turn our attention to circulation. The line integral of vector field [latex]\bf{F}[/latex] along an oriented closed curve is called the circulation of [latex]\bf{F}[/latex] along [latex]C[/latex]. Circulation line integrals have their own notation: [latex]\displaystyle\oint_C{\bf{F}}\cdot{\bf{T}}ds[/latex]. The circle on the integral symbol denotes that [latex]C[/latex] is “circular” in that it has no endpoints. Example “Evaluating a Vector Line Integral” shows a calculation of circulation.

To see where the term circulation comes from and what it measures, let [latex]\bf{v}[/latex] represent the velocity field of a fluid and let [latex]C[/latex] be an oriented closed curve. At a particular point [latex]P[/latex], the closer the direction of [latex]{\bf{v}}(P)[/latex] is to the direction of [latex]{\bf{T}}(P)[/latex], the larger the value of the dot product [latex]{\bf{v}}(P)\cdot{\bf{T}}(P)[/latex]. The maximum value of [latex]{\bf{v}}(P)\cdot{\bf{T}}(P)[/latex] occurs when the two vectors are pointing in the exact same direction; the minimum value of [latex]{\bf{v}}(P)\cdot{\bf{T}}(P)[/latex] occurs when the two vectors are pointing in opposite directions. Thus, the value of the circulation [latex]\displaystyle\oint_C{\bf{v}}\cdot{\bf{T}}ds[/latex] measures the tendency of the fluid to move in the direction of [latex]C[/latex].