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 r = r (s), 0 ≤ s ≤ S, where the variable s is the arc length of the curve (Figure 1).

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

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

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.
  3. 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
    \[\int\limits_C {F\left( {x,y,z} \right)ds} = \int\limits_\alpha ^\beta {F\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right) \sqrt {{{\left( {x'\left( t \right)} \right)}^2} + {{\left( {y'\left( t \right)} \right)}^2} + {{\left( {z'\left( t \right)} \right)}^2}} dt}. \]
  4. 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
    \[\int\limits_C {F\left( {x,y} \right)ds} = \int\limits_a^b {F\left( {x,f\left( x \right)} \right) \sqrt {1 + {{\left( {f'\left( x \right)} \right)}^2}} dx} .\]
  5. 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
    \[\int\limits_C {F\left( {x,y} \right)ds} = \int\limits_c^d {F\left( {\varphi \left( y \right),y} \right) \sqrt {1 + {{\left( {\varphi'\left( y \right)} \right)}^2}} dy} ;\]
  6. In polar coordinates the line integral \(\int\limits_C {F\left( {x,y} \right)ds} \) becomes
    \[\int\limits_C {F\left( {x,y} \right)ds} = \int\limits_\alpha ^\beta {F\left( {r\cos \theta ,r\sin \theta } \right) \sqrt {{r^2} + {{\left( {\frac{{dr}}{{d\theta }}} \right)}^2}} d\theta } ,\]
    where the curve \(C\) is defined by the polar function \(r\left( \theta \right).\)

Solved Problems

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

\[x = a\cos t, y = a\sin t, 0 \le t \le {\frac{\pi }{2}}.\]

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

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

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

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

See more problems on Page 2.

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