Calculus

Line Integrals

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Line Integrals of Scalar Functions

  • 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 \(1\text{).}\)

    A curve C defined by the vector function r(s)
    Figure 1.

    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

    \[{\int\limits_C {F\left( {x,y,z} \right)ds} \;\;\;}\kern-0.3pt{\text{or}\;\;\int\limits_C {Fds} .}\]

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

    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;
    2. 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\[\int\limits_{{C_1} \cup {C_2}} {Fds} = \int\limits_{{C_1}} {Fds} + \int\limits_{{C_2}} {Fds} ;\]
    3. Union of the two curves C1 and C2
      Figure 2.
    4. 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
      \[ {\int\limits_C {F\left( {x,y,z} \right)ds} } = {\int\limits_\alpha ^\beta {F\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right) \cdot}}\kern0pt{{ \sqrt {{{\left( {x’\left( t \right)} \right)}^2} + {{\left( {y’\left( t \right)} \right)}^2} + {{\left( {z’\left( t \right)} \right)}^2}} dt} } \]
    5. 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
      \[ {\int\limits_C {F\left( {x,y} \right)ds} } = {\int\limits_a^b {F\left( {x,f\left( x \right)} \right) \cdot}}\kern0pt{{ \sqrt {1 + {{\left( {f’\left( x \right)} \right)}^2}} dx} ;} \]
    6. 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
      \[ {\int\limits_C {F\left( {x,y} \right)ds} } = {\int\limits_c^d {F\left( {\varphi \left( y \right),y} \right) \cdot}}\kern0pt{{ \sqrt {1 + {{\left( {\varphi’\left( y \right)} \right)}^2}} dy} ;} \]
    7. 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 a problem to see the 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 line 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.

    Segment of the line y = x from the origin up to the point (2,2)
    Figure 3.

    \[
    {\int\limits_C {{x^2}yds} }={ \int\limits_0^2 {{x^2} \cdot x\sqrt {1 + {1^2}} dx} }
    = {\sqrt 2 \int\limits_0^2 {{x^3}dx} }
    = {\sqrt 2 \left[ {\left. {\left( {\frac{{{x^4}}}{4}} \right)} \right|_0^2} \right] }
    = {4\sqrt 2 .}
    \]

    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