# Line Integrals of Vector Fields

• ### Definition

Suppose that a curve $$C$$ is defined by the vector function $$\mathbf{r} = \mathbf{r}\left( s \right),$$ $$0 \le s \le S,$$ where $$s$$ is the arc length of the curve. Then the derivative of the vector function

${\frac{{d\mathbf{r}}}{{dt}} = \boldsymbol{\tau} } = {\left( {\cos \alpha ,\cos \beta ,\cos \gamma } \right)}$

is the unit vector of the tangent line to this curve (Figure $$1$$).

Here $$\alpha, \beta$$ and $$\gamma$$ are the angles between the tangent line and the positive axis $$Ox, Oy$$ and $$Oz,$$ respectively.

We introduce the vector function $$\mathbf{F}\left( {P,Q,R} \right)$$ defined over the curve $$C$$ so that for the scalar function

${\mathbf{F} \cdot \boldsymbol{\tau} \text{ = }}\kern0pt{ P\cos \alpha + Q\cos \beta + R\cos \gamma }$

the line integral $$\int\limits_C {\left( {\mathbf{F} \cdot \boldsymbol{\tau}} \right)ds}$$ exists. Such an integral $$\int\limits_C {\left( {\mathbf{F} \cdot \boldsymbol{\tau}} \right)ds}$$ is called the line integral of the vector field $$\mathbf{F}$$ along the curve $$C$$ and is denoted as

$\int\limits_C {Pdx + Qdy + Rdz} .$

Thus, by definition,

${\int\limits_C {Pdx + Qdy + Rdz} } = {\int\limits_0^S {\left( {P\cos \alpha }\right.}}+{{\left.{ Q\cos \beta + R\cos \gamma } \right)ds},}$

where $$\boldsymbol{\tau} \left( {\cos \alpha ,\cos \beta ,\cos \gamma } \right)$$ is the unit vector of the tangent line to the curve $$C.$$

The latter formula can be written in the vector form:

${\int\limits_C {\left( {\mathbf{F} \cdot d\mathbf{r}} \right)} } = {\int\limits_0^S {\left( {\mathbf{F}\left( {\mathbf{r}\left( s \right)} \right) \cdot \boldsymbol{\tau} } \right)ds} ,}$

where $$d\mathbf{r} = \left( {dx,dy,dz} \right).$$

If a curve $$C$$ lies in the $$xy$$-plane and $$R = 0$$, we can write:

${\int\limits_C {Pdx + Qdy} } = {\int\limits_0^S {\left( {P\cos \alpha + Q\cos \beta } \right)ds} .}$

### Properties of Line Integrals of Vector Fields

The line integral of vector function has the following properties:

1. Let $$C$$ denote the curve $$AB$$ which is traversed from $$A$$ to $$B,$$ and let $$-C$$ denote the curve $$BA$$ with the opposite orientation − from $$B$$ to $$A.$$ Then
${\int\limits_{ – C} {\left( {\mathbf{F} \cdot d\mathbf{r}} \right)} }={ – \int\limits_C {\left( {\mathbf{F} \cdot d\mathbf{r}} \right)} ;}$
2. If $$C$$ is the union of the curves $${C_1}$$ and $${C_2}$$ (Figure $$2$$), then
${\int\limits_C {\left( {\mathbf{F} \cdot d\mathbf{r}} \right)} } = {\int\limits_{{C_1} \cup {C_2}} {\left( {\mathbf{F} \cdot d\mathbf{r}} \right)} } = {\int\limits_{{C_1}} {\left( {\mathbf{F} \cdot d\mathbf{r}} \right)} }+{ \int\limits_{{C_2}} {\left( {\mathbf{F} \cdot d\mathbf{r}} \right)} ;}$
3. If the curve $$C$$ is parameterized by $$\mathbf{r}\left( t \right) = \left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right),$$ $$\alpha \le t \le \beta ,$$ then
${\int\limits_C {Pdx + Qdy + Rdz} } = {\int\limits_\alpha ^\beta {\left[ {P\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right)\frac{{dx}}{{dt}} }\right.}+{\left.{ Q\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right)\frac{{dy}}{{dt}} }\right.}+{\left.{ R\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right)\frac{{dz}}{{dt}}} \right]dt} .}$
4. If $$C$$ lies in the $$xy$$-plane and is given by the equation $$y = f\left( x \right)$$ (in this case $$R = 0$$ and $$t = x$$), then the latter formula can be written as
${\int\limits_C {Pdx + Qdy} } = {\int\limits_a^b {\Big[ {P\left( {x,f\left( x \right)} \right) }}}+{{{ Q\left( {x,f\left( x \right)} \right)\frac{{df}}{{dx}}} \Big]dx} .}$

• ## Solved Problems

Click a problem to see the solution.

### Example 1

Evaluate the integral $$\int\limits_C {ydx – xdy}$$ over the curve $$C$$ parameterized by $$\mathbf{r}\left( t \right) =$$ $$\left( {\cos t,\sin t} \right),$$ $$0 \le t \le {\large\frac{\pi }{2}\normalsize}.$$

### Example 2

Evaluate the line integral $$\int\limits_C {xdy – ydx}$$ along the curve $$C$$ defined by the equation $$y = {x^3}$$ from the origin $$\left( {0,0} \right)$$ to $$\left( {2,8} \right).$$

### Example 3

Calculate $$\int\limits_C {\sqrt x dx + \sqrt y dy}$$ along the curve $$y = {x^2}$$ from $$O\left( {0,0} \right)$$ to $$A\left( {1,1} \right)$$ (Figure $$3$$).

### Example 4

Evaluate the line integral $$\int\limits_C {ydx + xdy}$$ along the curve $$y = {x^2}$$ from the point $$O\left( {0,0} \right)$$ to the point $$A\left( {1,1} \right)$$ (Figure $$3$$).

### Example 5

Calculate the line integral $$\int\limits_C {{\large\frac{y}{x}\normalsize} dx + dy}$$ over the curve $$y = \ln x$$ in the interval $$1 \le x \le e$$ (Figure $$4$$).

### Example 6

Evaluate the line integral $$\int\limits_C {{x^2}dx – xydy},$$ where $$C$$ is the part of the circle lying in the first quadrant and traversed in the counterclockwise direction (Figure $$5$$).

### Example 7

Calculate the line integral $$\int\limits_C {{y^2}dx + xydy},$$ where the curve $$C$$ is the part of the ellipse (Figure $$6$$) parameterized by $$\mathbf{r}\left( t \right) =$$ $$\left( {a\cos t,b\sin t} \right),$$ $$0 \le t \le {\large\frac{\pi }{2}\normalsize}.$$

### Example 8

Find the integral $$\int\limits_C {xdx + ydy }$$ $$+{ \left( {x + y – 1} \right)dz}$$ along the curve $$C,$$ where $$C$$ is the line segment $$AB$$ traversed in the direction from the point $$A\left( {1,1,1} \right)$$ to the point $$B\left( {2,3,4} \right)$$ (Figure $$7$$).

### Example 1.

Evaluate the integral $$\int\limits_C {ydx – xdy}$$ over the curve $$C$$ parameterized by $$\mathbf{r}\left( t \right) =$$ $$\left( {\cos t,\sin t} \right),$$ $$0 \le t \le {\large\frac{\pi }{2}\normalsize}.$$

Solution.

Using the formula

${\int\limits_C {Pdx + Qdy} } = {\int\limits_\alpha ^\beta {\left[ {P\left( {x\left( t \right),y\left( t \right)} \right)\frac{{dx}}{{dt}} }\right.}}+{{\left.{ Q\left( {x\left( t \right),y\left( t \right)} \right)\frac{{dy}}{{dt}}} \right]dt} ,}$

${\int\limits_C {ydx – xdy} } = {\int\limits_0^{\large\frac{\pi }{2}\normalsize} {\left( {y\frac{{dx}}{{dt}} – x\frac{{dy}}{{dt}}} \right)dt} } = {\int\limits_0^{\large\frac{\pi }{2}\normalsize} {\left( {\sin t\frac{{d\left( {\cos t} \right)}}{{dt}} }\right.}-{\left.{ \cos t\frac{{d\left( {\sin t} \right)}}{{dt}}} \right)dt} } = {\int\limits_0^{\large\frac{\pi }{2}\normalsize} {\left( {\sin t\left( { – \sin t} \right) }\right.}-{\left.{ \cos t\cos t} \right)dt} } = { – \int\limits_0^{\large\frac{\pi }{2}\normalsize} {\left( {{{\sin }^2}t + {{\cos }^2}t} \right)dt} } = { – \int\limits_0^{\large\frac{\pi }{2}\normalsize} {dt} } = { – \frac{\pi }{2}.}$

### Example 2.

Evaluate the line integral $$\int\limits_C {xdy – ydx}$$ along the curve $$C$$ defined by the equation $$y = {x^3}$$ from the origin $$\left( {0,0} \right)$$ to $$\left( {2,8} \right).$$

Solution.

To find the given integral, we use the formula

${\int\limits_C {Pdx + Qdy} } = {\int\limits_a^b {\left[ {P\left( {x,f\left( x \right)} \right) }\right.}}+{{\left.{ Q\left( {x,f\left( x \right)} \right)\frac{{df}}{{dx}}} \right]dx} .}$

Substituting $$y = {x^3}$$ and $$dy = 3{x^2}dx$$ in the integrand, we obtain

${\int\limits_C {xdy – ydx} } = {\int\limits_0^2 {x \cdot 3{x^2}dx – {x^3}dx} } = {\int\limits_0^2 {2{x^3}dx} } = {2\left[ {\left. {\left( {\frac{{{x^4}}}{4}} \right)} \right|_0^8} \right] }={ 8.}$

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Problems 1-2
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Problems 3-8