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].

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.

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.

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].

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]