# Calculus

## Line Integrals # Line Integrals of Vector Fields

## Definition

Suppose that a curve C is defined by the vector function r = r (s), 0 ≤ sS, 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

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

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,

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

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
$\int\limits_C {\left( {\mathbf{F} \cdot d\mathbf{r}} \right)} = \int\limits_{{C_1} \cup {C_2}} {\left( {\mathbf{F} \cdot d\mathbf{r}} \right)} = \int\limits_{{C_1}} {\left( {\mathbf{F} \cdot d\mathbf{r}} \right)} + \int\limits_{{C_2}} {\left( {\mathbf{F} \cdot d\mathbf{r}} \right)} ;$
4. 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
$\int\limits_C {Pdx + Qdy + Rdz} = \int\limits_\alpha ^\beta {\left[ {P\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right)\frac{{dx}}{{dt}} + Q\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right)\frac{{dy}}{{dt}} + R\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right)\frac{{dz}}{{dt}}} \right]dt} .$
5. 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 or tap a problem to see the solution.

### Example 1

Evaluate the integral $\int\limits_C {ydx - xdy}$ over the curve $$C$$ parameterized by

$\mathbf{r}\left( t \right) = \left( {\cos t,\sin t} \right), 0 \le t \le {\frac{\pi }{2}}.$

### 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 $$\left( {2,8} \right).$$

### Example 1.

Evaluate the integral $\int\limits_C {ydx - xdy}$ over the curve $$C$$ parameterized by

$\mathbf{r}\left( t \right) = \left( {\cos t,\sin t} \right), 0 \le t \le {\frac{\pi }{2}}.$

Solution.

Using the formula

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

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

### 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

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

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

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

See more problems on Page 2.