Calculus

Line Integrals

Line Integrals of Scalar Functions

Page 1
Problems 1-2
Page 2
Problems 3-7

Definition

Suppose that we can describe a curve \(C\) by the vector function \(\mathbf{r} = \mathbf{r}\left( s \right),\) \(0 \le s \le S,\) where the variable \(s\) is the arc length of the curve (Figure \(1\text{).}\)

If a scalar function \(F\) is defined over the curve \(C,\) then the integral \(\int\limits_0^S {F\left( {\mathbf{r}\left( s \right)} \right)ds} \) is called a line integral of scalar function \(F\) along the curve \(C\) and denoted as
\[{\int\limits_C {F\left( {x,y,z} \right)ds} \;\;\;}\kern-0.3pt{\text{or}\;\;\int\limits_C {Fds} .}\]

A curve C defined by the vector function r(s)

Figure 1.

The line integral \(\int\limits_C {Fds}\) exists if the function \(F\) is continuous on the curve \(C.\)

Union of the two curves C1 and C2

Figure 2.

Properties of Line Integrals of Scalar Functions

The line integral of a scalar function has the following properties:

  1. The line integral of a scalar function over the smooth curve \(C\) does not depend on the orientation of the curve;
  1. If \({C_1}\) is a curve that begins at \(A\) and ends at \(B,\) and if \({C_2}\) is a curve that begins at \(B\) and ends at \(D\) (Figure \(2\)), then their union is defined to be the curve \({C_1} \cup {C_2}\) that progresses along the curve \({C_1}\) from \(A\) to \(B,\) and then along \({C_2}\) from \(B\) to \(D,\) so that\[\int\limits_{{C_1} \cup {C_2}} {Fds} = \int\limits_{{C_1}} {Fds} + \int\limits_{{C_2}} {Fds} ;\]
  2. If the smooth curve \(C\) is parameterized by \(\mathbf{r} = \mathbf{r}\left( t \right),\) \(\alpha \le t \le \beta \) and the scalar function \(F\) is continuous on the curve \(C,\) then\[
    {\int\limits_C {F\left( {x,y,z} \right)ds} }
    = {\int\limits_\alpha ^\beta {F\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right) \cdot}}\kern0pt{{ \sqrt {{{\left( {x’\left( t \right)} \right)}^2} + {{\left( {y’\left( t \right)} \right)}^2} + {{\left( {z’\left( t \right)} \right)}^2}} dt} }
    \]
  3. If \(C\) is a smooth curve in the \(xy\)-plane given by the equation \(y = f\left( x \right),\) \(a \le x \le b,\) then\[
    {\int\limits_C {F\left( {x,y} \right)ds} }
    = {\int\limits_a^b {F\left( {x,f\left( x \right)} \right) \cdot}}\kern0pt{{ \sqrt {1 + {{\left( {f’\left( x \right)} \right)}^2}} dx} ;}
    \]
  4. Similarly, if a smooth curve \(C\) in the \(xy\)-plane is defined by the equation \(x = \varphi \left( y \right),\) \(c \le y \le d,\) then\[
    {\int\limits_C {F\left( {x,y} \right)ds} }
    = {\int\limits_c^d {F\left( {\varphi \left( y \right),y} \right) \cdot}}\kern0pt{{ \sqrt {1 + {{\left( {\varphi’\left( y \right)} \right)}^2}} dy} ;}
    \]
  5. In polar coordinates the line integral \(\int\limits_C {F\left( {x,y} \right)ds} \) becomes\[
    {\int\limits_C {F\left( {x,y} \right)ds} }
    = {\int\limits_\alpha ^\beta {F\left( {r\cos \theta ,r\sin \theta } \right) \cdot}}\kern0pt{{ \sqrt {{r^2} + {{\left( {\frac{{dr}}{{d\theta }}} \right)}^2}} d\theta } ,}
    \]
    where the curve \(C\) is defined by the polar function \(r\left( \theta \right).\)

Solved Problems

Click on problem description to see solution.

 Example 1

Evaluate the line integral \(\int\limits_C {{x^2}yds} \) along the segment of the line \(y = x\) from the origin up to the point \(\left( {2,2} \right)\) (see Figure \(3\)).

 Example 2

Calculate the line integral \(\int\limits_C {{y^2}ds},\) where \(C\) is a part of the circle \(x = a\cos t,\) \(y = a\sin t,\) \(0 \le t \le {\large\frac{\pi }{2}\normalsize}.\)

 Example 3

Evaluate the line integral \(\int\limits_C {{x^2}ds},\) where \(C\) is a the curve given by the equation \(y = f\left( x \right) = \ln x,\) \(1 \le x \le e.\)

 Example 4

Evaluate the integral \(\int\limits_C {ds}\) over the plane curve, where \(C\) is the line segment from \(O\left( {0,0} \right)\) to \(A\left( {1,2} \right)\) (see Figure \(4\)).

 Example 5

Calculate the integral \(\int\limits_C {\left( {{x^2} + {y^2}} \right)zds},\) where the curve \(C\) is parameterized by \(\mathbf{r}\left( t \right) = \left( {\sin 3t,\cos 3t,4t} \right)\), \(0 \le t \le \pi .\)

 Example 6

Calculate the ine integral \(\int\limits_C {\large\frac{{ds}}{{y – x}}\normalsize},\) where the curve \(C\) is the line segment from \(\left( {0, – 2} \right)\) to \(\left( {4, 0} \right)\) (Figure \(5\)).

 Example 7

Calculate the line integral \(\int\limits_C {xyds},\) where the curve \(C\) is the part of ellipse \({\large\frac{{{x^2}}}{{{a^2}}}\normalsize} + {\large\frac{{{y^2}}}{{{b^2}}}\normalsize} = 1\) lying in the first quadrant (Figure \(6\)).

Example 1.

Evaluate the line integral \(\int\limits_C {{x^2}yds} \) along the segment of the line \(y = x\) from the origin up to the point \(\left( {2,2} \right)\) (see Figure \(3\)).

Solution.

\[
{\int\limits_C {{x^2}yds} }={ \int\limits_0^2 {{x^2} \cdot x\sqrt {1 + {1^2}} dx} }
= {\sqrt 2 \int\limits_0^2 {{x^3}dx} }
= {\sqrt 2 \left[ {\left. {\left( {\frac{{{x^4}}}{4}} \right)} \right|_0^2} \right] }
= {4\sqrt 2 .}
\]
Segment of the line y = x from the origin up to the point (2,2)

Figure 3.

Example 2.

Calculate the line integral \(\int\limits_C {{y^2}ds},\) where \(C\) is a part of the circle \(x = a\cos t,\) \(y = a\sin t,\) \(0 \le t \le {\large\frac{\pi }{2}\normalsize}.\)

Solution.

The arc length differential is

\[
{ds }={ \sqrt {{{\left( {x’\left( t \right)} \right)}^2} + {{\left( {y’\left( t \right)} \right)}^2}} dt }
= {\sqrt {{a^2}{{\sin }^2}t + {a^2}{{\cos }^2}t} dt }={ adt.}
\]

Then applying the formula

\[
{\int\limits_C {F\left( {x,y,z} \right)ds} }
= {\int\limits_\alpha ^\beta {F\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right) \cdot}}\kern0pt{{ \sqrt {{{\left( {x’\left( t \right)} \right)}^2} + {{\left( {y’\left( t \right)} \right)}^2} + {{\left( {z’\left( t \right)} \right)}^2}} dt} }
\]

in the \(xy\)-plane, we obtain

\[
{\int\limits_C {{y^2}ds} }={ \int\limits_0^{\large\frac{\pi }{2}\normalsize} {{a^2}{{\sin }^2}t \cdot adt} }
= {{a^3}\int\limits_0^{\large\frac{\pi }{2}\normalsize} {{{\sin }^2}tdt} }
= {\frac{{{a^3}}}{2}\int\limits_0^{\large\frac{\pi }{2}\normalsize} {\left( {1 – \cos 2t} \right)dt} }
= {\frac{{{a^3}}}{2}\left[ {\left. {\left( {t – \frac{{\sin 2t}}{2}} \right)} \right|_0^{\large\frac{\pi }{2}\normalsize}} \right] }
= {\frac{{{a^3}}}{2} \cdot \frac{\pi }{2} }
= {\frac{{{a^3}\pi }}{4}.}
\]
Page 1
Problems 1-2
Page 2
Problems 3-7