Line Integrals of Vector Fields

Solved Problems

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 \(3\)).

Solution.

The curve y=x^2 from O(0,0) to A(1,1) — Figure 3.

We use the formula

\[\int\limits_C {Pdx + Qdy} = \int\limits_a^b {\left[ {P\left( {x,f\left( x \right)} \right) + Q\left( {x,f\left( x \right)} \right)\frac{{df}}{{dx}}} \right]dx} .\]

Substituting \(y = {x^2}\) and \(dy = 2xdx\) in the integrand, we find the answer:

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

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 \(3\)).

Solution.

If \(y = f\left( x \right) = {x^2},\) then by the formula

\[\int\limits_C {Pdx + Qdy} = \int\limits_a^b {\left[ {P\left( {x,f\left( x \right)} \right) + Q\left( {x,f\left( x \right)} \right)\frac{{df}}{{dx}}} \right]dx} .\]

we obtain

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

Example 5.

Calculate the line integral \[\int\limits_C {{\frac{y}{x}} dx + dy} \] over the curve \(y = \ln x\) in the interval \(1 \le x \le e\) (Figure \(4\)).

Solution.

The segment of logarithmic curve y = ln(x) from x=1 to x=e. — Figure 4.

As \(y = \ln x,\) then the differential is \(dy = {\frac{{dx}}{x}}.\) According to the formula

\[\int\limits_C {Pdx + Qdy} = \int\limits_a^b {\left[ {P\left( {x,f\left( x \right)} \right) + Q\left( {x,f\left( x \right)} \right)\frac{{df}}{{dx}}} \right]dx},\]

we find the solution:

\[\int\limits_C {\frac{y}{x}dx + dy} = \int\limits_1^e {\left( {\frac{{\ln x}}{x} + \frac{1}{x}} \right)dx} = \int\limits_1^e {\frac{{\ln x}}{x}dx} + \int\limits_1^e {\frac{{dx}}{x}} = \int\limits_1^e {\ln xd\left( {\ln x} \right)} + \int\limits_1^e {\frac{{dx}}{x}} = \left. {\left[ {\frac{{{{\left( {\ln x} \right)}^2}}}{2} + \ln x} \right]} \right|_1^e = \left( {\frac{{{{\left( {\ln e} \right)}^2}}}{2} + \ln e} \right) - \left( {\frac{{{{\left( {\ln 1} \right)}^2}}}{2} + \ln 1} \right) = \frac{1}{2} + 1 - 0 = \frac{3}{2}.\]

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 \(5\)).

Solution.

A part of the circle lying in the first quadrant — Figure 5.

Obviously, the arc of the circle is described by the function \(y = \sqrt {{a^2} - {x^2}},\) where \(a\) is the radius of the circle. (We take the positive value of the root because \(y \gt 0\) in the first quadrant.) Then the differential is

\[dy = \frac{{dy}}{{dx}}dx = \frac{{d\sqrt {{a^2} - {x^2}} }}{{dx}}dx = - \frac{{xdx}}{{\sqrt {{a^2} - {x^2}} }}.\]

Since we integrate in the counterclockwise direction, the lower and upper limits of integration will be \(a\) and \(0,\) respectively. Hence,

\[\int\limits_C {{x^2}dx - xydy} = \int\limits_a^0 {{x^2}dx - x\sqrt {{a^2} - {x^2}} \left( { - \frac{{xdx}}{{\sqrt {{a^2} - {x^2}} }}} \right)dx} = \int\limits_a^0 {2{x^2}dx} = 2\left[ {\left. {\left( {\frac{{{x^3}}}{3}} \right)} \right|_a^0} \right] = - \frac{{2{a^3}}}{3}.\]

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

\[\mathbf{r}\left( t \right) = \left( {a\cos t,b\sin t} \right), 0 \le t \le {\frac{\pi }{2}}.\]

Solution.

A part of the ellipse lying in the first quadrant — Figure 6.

We write all the expressions in terms of the parameter \(t:\)

\[{y^2} = {b^2}{\sin ^2}t,\;\; \frac{{dx}}{{dt}} = - a\sin t,\;\; xy = ab\cos t\sin t,\;\; \frac{{dy}}{{dt}} = b\cos t.\]

Then using the formula

\[\int\limits_C {Pdx + Qdy} = \int\limits_\alpha ^\beta {\left[ {P\left( {x\left( t \right),y\left( t \right)} \right)\frac{{dx}}{{dt}} + Q\left( {x\left( t \right),y\left( t \right)} \right)\frac{{dy}}{{dt}}} \right]dt} ,\]

we obtain

\[\int\limits_C {{y^2}dx + xydy} = \int\limits_0^{\frac{\pi }{2}} {\left[ {{b^2}{{\sin }^2}t \cdot \left( { - a\sin t} \right) + ab\cos t\sin t \cdot b\cos t} \right]dt} = \int\limits_0^{\frac{\pi }{2}} {\left[ { - a{b^2}{{\sin }^3}t + a{b^2}{{\cos }^2}t\sin t} \right]dt} = a{b^2}\int\limits_0^{\frac{\pi }{2}} {\left[ {\sin t\left( {{{\cos }^2}t - {{\sin }^2}t} \right)} \right]dt} = a{b^2}\int\limits_0^{\frac{\pi }{2}} {\sin t\cos 2tdt} = a{b^2}\int\limits_0^{\frac{\pi }{2}} {\left[ {\frac{1}{2}\left( {\sin 3t + \sin \left( { - t} \right)} \right)} \right]dt} = \frac{{a{b^2}}}{2}\int\limits_0^{\frac{\pi }{2}} {\left( {\sin 3t - \sin t} \right)dt} = \frac{{a{b^2}}}{2}\left[ {\left. {\left( { - \frac{{\cos 3t}}{3} + \cos t} \right)} \right|_0^{\frac{\pi }{2}}} \right] = \frac{{a{b^2}}}{2}\left[ {0 - \left( { - \frac{1}{3} + 1} \right)} \right] = - \frac{{a{b^2}}}{3}.\]

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 \(7\)).

Solution.

The line segment from A(1,1,1) to B(2,3,4) — Figure 7.

We first find the equation of the line \(AB:\)

\[\frac{{x - {x_A}}}{{{x_B} - {x_A}}} = \frac{{y - {y_A}}}{{{y_B} - {y_A}}} = \frac{{z - {z_A}}}{{{z_B} - {z_A}}},\;\; \Rightarrow \frac{{x - 1}}{{2 - 1}} = \frac{{y - 1}}{{3 - 1}} = \frac{{z - 1}}{{4 - 1}},\;\; \Rightarrow \frac{{x - 1}}{1} = \frac{{y - 1}}{2} = \frac{{z - 1}}{3}.\]

Introduce the parameter \(t:\)

\[\frac{{x - 1}}{1} = \frac{{y - 1}}{2} = \frac{{z - 1}}{3} = t,\]

and write the equation in parametric form:

\[ \left\{ \begin{array}{l} x - 1 = t\\ y - 1 = 2t\\ z - 1 = 3t \end{array} \right.,\;\; \Rightarrow \left\{ \begin{array}{l} x = t + 1\\ y = 2t + 1\\ z = 3t + 1 \end{array} \right..\]

Then we apply the formula

\[\int\limits_C {Pdx + Qdy + Rdz} = \int\limits_\alpha ^\beta {\left[ {P\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right)\frac{{dx}}{{dt}} + Q\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right)\frac{{dy}}{{dt}} + R\left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right)\frac{{dz}}{{dt}}} \right]dt}.\]

Obviously, the parameter \(t\) varies in the interval \(\left[ {0,1} \right].\) Then the line integral becomes

\[\int\limits_C {xdx + ydy + \left( {x + y - 1} \right)dz} = \int\limits_0^1 {\left[ {\left( {t + 1} \right)\frac{{d\left( {t + 1} \right)}}{{dt}} + \left( {2t + 1} \right)\frac{{d\left( {2t + 1} \right)}}{{dt}} + \left( {t + 1 + 2t + 1 - 1} \right)\frac{{d\left( {3t + 1} \right)}}{{dt}}} \right]dt} = \int\limits_0^1 {\left[ {t + 1 + \left( {2t + 1} \right) \cdot 2 + \left( {3t + 1} \right) \cdot 3} \right]dt} = \int\limits_0^1 {\left( {14t + 6} \right)dt} = \left. {\left( {7{t^2} + 6t} \right)} \right|_0^1 = 13.\]