### 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:

- 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
\[{\int\limits_{ – C} {\left( {\mathbf{F} \cdot d\mathbf{r}} \right)} }={ – \int\limits_C {\left( {\mathbf{F} \cdot d\mathbf{r}} \right)} ;}\]
- If \(C\) is the union of the curves \({C_1}\) and \({C_2}\) (Figure \(2\)), then
\[ {\int\limits_C {\left( {\mathbf{F} \cdot d\mathbf{r}} \right)} } = {\int\limits_{{C_1} \cup {C_2}} {\left( {\mathbf{F} \cdot d\mathbf{r}} \right)} } = {\int\limits_{{C_1}} {\left( {\mathbf{F} \cdot d\mathbf{r}} \right)} }+{ \int\limits_{{C_2}} {\left( {\mathbf{F} \cdot d\mathbf{r}} \right)} ;} \]
- 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
\[ {\int\limits_C {Pdx + Qdy + Rdz} } = {\int\limits_\alpha ^\beta {\left[ {P\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right)\frac{{dx}}{{dt}} }\right.}+{\left.{ Q\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right)\frac{{dy}}{{dt}} }\right.}+{\left.{ R\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right)\frac{{dz}}{{dt}}} \right]dt} .} \]
- 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 \(\mathbf{r}\left( t \right) =\) \( \left( {\cos t,\sin t} \right),\) \(0 \le t \le {\large\frac{\pi }{2}\normalsize}.\)### 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} ,}

\]

we find the answer:

\[

{\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.}

\]