Line Integrals of Scalar Functions

Solved Problems

Example 3.

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

Solution.

We use the formula

\[\int\limits_C {F\left( {x,y} \right)ds} = \int\limits_a^b {F\left( {x,f\left( x \right)} \right) \sqrt {1 + {{\left( {f'\left( x \right)} \right)}^2}} dx}.\]

Here

\[\sqrt {1 + {{\left( {f'\left( x \right)} \right)}^2}} = \sqrt {1 + {{\left[ {{{\left( {\ln x} \right)}^\prime }} \right]}^2}} = \sqrt {1 + \frac{1}{{{x^2}}}} = \frac{{\sqrt {1 + {x^2}} }}{x}.\]

Hence,

\[\int\limits_C {{x^2}ds} = \int\limits_1^e {{x^2}\frac{{\sqrt {1 + {x^2}} }}{x}dx} = \int\limits_1^e {\sqrt {1 + {x^2}} xdx} = \frac{1}{2}\int\limits_1^e {\sqrt {1 + {x^2}} d\left( {1 + {x^2}} \right)} = \frac{1}{2}\left[ {\left. {\left( {\frac{{{{\left( {1 + {x^2}} \right)}^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right)} \right|_1^e} \right] = \frac{1}{3}\left[ {{{\left( {1 + {e^2}} \right)}^{\frac{3}{2}}} - {2^{\frac{3}{2}}}} \right] = \frac{{\sqrt {{{\left( {1 + {e^2}} \right)}^3}} - \sqrt 8 }}{3} \approx 7,16.\]

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 \(A\left( {1,2} \right)\) (Figure \(4\)).

Solution.

The line segment from O(0,0) to A(1,2) — Figure 4.

First we find the equation of the line \(OA:\)

\[\frac{{y - {y_O}}}{{{y_A} - {y_O}}} = \frac{{x - {x_O}}}{{{x_A} - {x_O}}},\;\; \Rightarrow \frac{{y - 0}}{{2 - 0}} = \frac{{x - 0}}{{1 - 0}},\;\; \Rightarrow \frac{y}{2} = \frac{x}{1}\;\;\text{or}\;\;y = 2x.\]

Using the formula

\[\int\limits_C {F\left( {x,y} \right)ds} = \int\limits_a^b {F\left( {x,f\left( x \right)} \right) \sqrt {1 + {{\left( {f'\left( x \right)} \right)}^2}} dx},\]

we get the line integral:

\[\int\limits_C {ds} = \int\limits_0^1 {\sqrt {1 + {2^2}} dx} = \sqrt 5 \int\limits_0^1 {dx} = \sqrt 5 \cdot \left[ {\left. x \right|_0^1} \right] = \sqrt 5 .\]

Example 5.

Calculate the integral \[\int\limits_C {\left( {{x^2} + {y^2}} \right)zds},\] where the curve \(C\) is parameterized by

\[\mathbf{r}\left( t \right) = \left( {\sin 3t,\cos 3t,4t} \right), 0 \le t \le \pi.\]

Solution.

Using the formula

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

we can write

\[\int\limits_C {\left( {{x^2} + {y^2}} \right)zds} = \int\limits_0^\pi {\left( {{{\sin }^2}3t + {{\cos }^2}3t} \right) \cdot 4t \cdot \sqrt {{{\left( {3\cos 3t} \right)}^2} + {{\left( { - 3\sin 3t} \right)}^2} + 16} \,dt} = 4\int\limits_0^\pi {t\sqrt {9\left( {{{\cos }^2}3t + {{\sin }^2}3t} \right) + 16} \,dt} = 20\int\limits_0^\pi {tdt} = 20\left[ {\left. {\left( {\frac{{{t^2}}}{2}} \right)} \right|_0^\pi } \right] = 10{\pi ^2}.\]

Example 6.

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

Solution.

The line segment from A(0,-2) to B(4,0) — Figure 5.

Find the equation of the line \(AB:\)

\[\frac{{y - {y_A}}}{{{y_B} - {y_A}}} = \frac{{x - {x_A}}}{{{x_B} - {x_A}}},\;\; \Rightarrow \frac{{y + 2}}{{0 + 2}} = \frac{{x - 0}}{{4 - 0}},\;\; \Rightarrow \frac{y + 2}{2} = \frac{x}{4}\;\;\text{or}\;\;\; y = \frac {x}{2} - 2.\]

By the formula

\[\int\limits_C {F\left( {x,y} \right)ds} = \int\limits_a^b {F\left( {x,f\left( x \right)} \right) \sqrt {1 + {{\left( {f^\prime\left( x \right)} \right)}^2}} dx},\]

we find the integral:

\[\int\limits_C {\frac{{ds}}{{y - x}}} = \int\limits_0^4 {\frac{1}{{\frac{x}{2} - 2 - x}} \sqrt {1 + {{\left( {\frac{1}{2}} \right)}^2}} dx} = \int\limits_0^4 {\frac{1}{{ - 2 - \frac{x}{2}}}\sqrt {\frac{5}{4}} dx} = - \frac{{\sqrt 5 }}{2}\int\limits_0^4 {\frac{{dx}}{{x + 4}}} = - \sqrt 5 \left[ {\left. {\left( {\ln \left| {x + 4} \right|} \right)} \right|_0^4} \right] = - \sqrt 5 \left( {\ln 8 - \ln 4} \right) = - \sqrt 5 \ln 2 \approx - 1,55.\]

Example 7.

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

Solution.

Part of ellipse lying in the first quadrant — Figure 6.

The equation of the ellipse can be written in the parametric form:

\[ \left\{ \begin{array}{l} x = a\cos t\\ y = b\sin t \end{array} \right.,\;\; 0 \le t \le 2\pi .\]

For the arc of the ellipse lying in the first quadrant, we have \(0 \le t \le {\frac{\pi }{2}}.\)

Consequently, by the formula

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

the line integral becomes

\[I = \int\limits_C {xyds} = \int\limits_0^{\frac{\pi }{2}} {a\cos t \cdot b\sin t \sqrt {{{\left( { - a\sin t} \right)}^2} + {{\left( {b\cos t} \right)}^2}} dt} = \int\limits_0^{\frac{\pi }{2}} {a\cos t \cdot b\sin t \sqrt {{a^2}{{\sin }^2}t + {b^2}{\cos^2}t} \,dt} .\]

Make the substitution. Let \(a\sin t = u\) or \(\sin t = {\frac{u}{a}}.\) Then

\[a\cos tdt = du,\;\;\; {b^2}{\cos ^2}t = {b^2}\left( {1 - {{\sin }^2}t} \right) = {b^2}\left[ {1 - {{\left( {\frac{u}{a}} \right)}^2}} \right].\]

Refine the limits of integration. When \(t = 0,\) we have \(u = 0,\) and when \(t = {\frac{\pi }{2}}\), we get \(u = a.\) As a result, the integral can be written as

\[I = \int\limits_0^a {u \cdot b \cdot \frac{{du}}{a} \sqrt {{u^2} + \frac{{{b^2}}}{{{a^2}}}\left( {{a^2} - {u^2}} \right)} du} = \frac{b}{a}\int\limits_0^a {\sqrt {{u^2} + {b^2} - \frac{{{b^2}}}{{{a^2}}}{u^2}} \,udu} = \frac{b}{a}\int\limits_0^a {\sqrt {{b^2} + \left( {1 - \frac{{{b^2}}}{{{a^2}}}} \right){u^2}} \,udu} .\]

It is convenient to make one more substitution of variable:

\[p = {b^2} + \left( {1 - \frac{{{b^2}}}{{{a^2}}}} \right){u^2},\;\; \Rightarrow dp = \left( {1 - \frac{{{b^2}}}{{{a^2}}}} \right) \cdot 2udu,\;\; \Rightarrow udu = \frac{{{a^2}dp}}{{2\left( {{a^2} - {b^2}} \right)}}.\]

When \(u = 0,\) we have \(p = {b^2}\), and accordingly, if \(u = a,\) then \(p = {a^2}.\) Thus,

\[I = \frac{b}{a}\int\limits_{{b^2}}^{{a^2}} {\sqrt p \frac{{{a^2}}}{{2\left( {{a^2} - {b^2}} \right)}}dp} = \frac{{ab}}{{2\left( {{a^2} - {b^2}} \right)}}\int\limits_{{b^2}}^{{a^2}} {\sqrt p dp} = \frac{{ab}}{{2\left( {{a^2} - {b^2}} \right)}} \left[ {\left. {\left( {\frac{{{p^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right)} \right|_{{b^2}}^{{a^2}}} \right] = \frac{{ab}}{{2\left( {{a^2} - {b^2}} \right)}} \cdot \frac{2}{3}\left( {{a^3} - {b^3}} \right) = \frac{{ab}}{{3\left( {a - b} \right)}} \cdot \left( {a - b} \right) \left( {{a^2} + ab + {b^2}} \right) = \frac{{ab\left( {{a^2} + ab + {b^2}} \right)}}{{3\left( {a + b} \right)}}.\]