# 6.2 Line integrals  (Page 7/20)

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${\int }_{{a}_{1}}^{{a}_{n}}f\left(x\right)dx={\int }_{{a}_{1}}^{{a}_{2}}f\left(x\right)dx+{\int }_{{a}_{1}}^{{a}_{3}}f\left(x\right)dx+\cdots +{\int }_{{a}_{n-1}}^{{a}_{n}}f\left(x\right)dx,$

which is analogous to property iv.

## Using properties to compute a vector line integral

Find the value of integral ${\int }_{C}\text{F}·\text{T}ds,$ where C is the rectangle (oriented counterclockwise) in a plane with vertices $\left(0,0\right),\left(2,0\right),\left(2,1\right),\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\left(0,1\right),$ and where $\text{F}=⟨x-2y,y-x⟩$ ( [link] ).

Note that curve C is the union of its four sides, and each side is smooth. Therefore C is piecewise smooth. Let ${C}_{1}$ represent the side from $\left(0,0\right)$ to $\left(2,0\right),$ let ${C}_{2}$ represent the side from $\left(2,0\right)$ to $\left(2,1\right),$ let ${C}_{3}$ represent the side from $\left(2,1\right)$ to $\left(0,1\right),$ and let ${C}_{4}$ represent the side from $\left(0,1\right)$ to $\left(0,0\right)$ ( [link] ). Then,

${\int }_{C}\text{F}·\text{T}d\text{r}={\int }_{{C}_{1}}\text{F}·\text{T}d\text{r}+{\int }_{{C}_{2}}\text{F}·\text{T}d\text{r}+{\int }_{{C}_{3}}\text{F}·\text{T}d\text{r}+{\int }_{{C}_{4}}\text{F}·\text{T}d\text{r}.$

We want to compute each of the four integrals on the right-hand side using [link] . Before doing this, we need a parameterization of each side of the rectangle. Here are four parameterizations (note that they traverse C counterclockwise):

$\begin{array}{c}{C}_{1}:⟨t,0⟩,0\le t\le 2\hfill \\ {C}_{2}:⟨2,t⟩,0\le t\le 1\hfill \\ {C}_{3}:⟨2-t,1⟩,0\le t\le 2\hfill \\ {C}_{4}:⟨0,1-t⟩,0\le t\le 1.\hfill \end{array}$

Therefore,

$\begin{array}{cc}\hfill {\int }_{{C}_{1}}\text{F}·\text{T}d\text{r}& ={\int }_{0}^{2}\text{F}\left(\text{r}\left(t\right)\right)·{r}^{\prime }\left(t\right)dt\hfill \\ & ={\int }_{0}^{2}⟨t-2\left(0\right),0-t⟩·⟨1,0⟩dt={\int }_{0}^{1}tdt\hfill \\ & ={\left[\frac{{t}^{2}}{2}\right]}_{0}^{2}=2.\hfill \end{array}$

Notice that the value of this integral is positive, which should not be surprising. As we move along curve C 1 from left to right, our movement flows in the general direction of the vector field itself. At any point along C 1 , the tangent vector to the curve and the corresponding vector in the field form an angle that is less than 90°. Therefore, the tangent vector and the force vector have a positive dot product all along C 1 , and the line integral will have positive value.

The calculations for the three other line integrals are done similarly:

$\begin{array}{cc}\hfill {\int }_{{C}_{2}}\text{F}·d\text{r}& ={\int }_{0}^{1}⟨2-2t,t-2⟩·⟨0,1⟩dt\hfill \\ & ={\int }_{0}^{1}\left(t-2\right)dt\hfill \\ & ={\left[\frac{{t}^{2}}{2}-2t\right]}_{0}^{1}=-\frac{3}{2},\hfill \end{array}$
$\begin{array}{cc}\hfill {\int }_{{C}_{3}}\text{F}·\text{T}ds& ={\int }_{0}^{2}⟨\left(2-t\right)-2,1-\left(2-t\right)⟩·⟨-1,0⟩dt\hfill \\ & ={\int }_{0}^{2}tdt=2,\hfill \end{array}$

and

$\begin{array}{cc}\hfill {\int }_{{C}_{4}}\text{F}·d\text{r}& ={\int }_{0}^{1}⟨-2\left(1-t\right),1-t⟩·⟨0,-1⟩dt\hfill \\ & ={\int }_{0}^{1}\left(t-1\right)dt\hfill \\ & ={\left[\frac{{t}^{2}}{2}-t\right]}_{0}^{1}=-\frac{1}{2}.\hfill \end{array}$

Thus, we have ${\int }_{C}\text{F}·d\text{r}=2.$

Calculate line integral ${\int }_{C}\text{F}·d\text{r},$ where F is vector field $⟨{y}^{2},2xy+1⟩$ and C is a triangle with vertices $\left(0,0\right),$ $\left(4,0\right),$ and $\left(0,5\right),$ oriented counterclockwise.

0

## 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 C in space. The mass per unit length (the linear density) of the wire is a continuous function $\rho \left(x,y,z\right).$ We can calculate the total mass of the wire using the scalar line integral ${\int }_{C}\rho \left(x,y,z\right)ds.$ 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 $\rho \left(x*,y*,z*\right)\text{Δ}s$ for some point $\left(x*,y*,z*\right)$ in the piece. Letting the length of the pieces shrink to zero with a limit yields the line integral ${\int }_{C}\rho \left(x,y,z\right)ds.$

## Calculating the mass of a wire

Calculate the mass of a spring in the shape of a curve parameterized by $⟨t,2\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}t,2\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}t⟩,$ $0\le t\le \frac{\pi }{2},$ with a density function given by $\rho \left(x,y,z\right)={e}^{x}+yz$ kg/m ( [link] ).

To calculate the mass of the spring, we must find the value of the scalar line integral ${\int }_{C}\left({e}^{x}+yz\right)ds,$ where C is the given helix. To calculate this integral, we write it in terms of t using [link] :

$\begin{array}{cc}\hfill {\int }_{C}{e}^{x}+yzds& ={\int }_{0}^{\pi \text{/}2}\left(\left({e}^{t}+4\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}t\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}t\right)\sqrt{1+{\left(-2\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}t\right)}^{2}+{\left(2\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}t\right)}^{2}}\right)dt\hfill \\ & ={\int }_{0}^{\pi \text{/}2}\left(\left({e}^{t}+4\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}t\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}t\right)\sqrt{5}\right)dt\hfill \\ & =\sqrt{5}{\left[{e}^{t}+2\phantom{\rule{0.2em}{0ex}}{\text{sin}}^{2}t\right]}_{t=0}^{t=\pi \text{/}2}\hfill \\ & =\sqrt{5\left({e}^{\pi \text{/}2}+1\right)}.\hfill \end{array}$

Therefore, the mass is $\sqrt{5}\left({e}^{\pi \text{/}2}+1\right)$ kg.

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