Line integrals of vector fields

1 Oriented curves

An oriented curve is a curve with a choice of direction. Each curve has two orientations: forward and backward. A parametrization of an oriented curve is a parametrization that travels the curve in the correct direction.

Note: the words “forward” and “backward” are subjective.

When a plane curve \(\gamma : [a,b] \to \mathbb {R}^2 \) is closed and simple (\(\gamma (a) = \gamma (b)\), and the curve doesn’t self intersect), then we say it is positively oriented if it is oriented with counterclockwise direction and negatively oriented if it is oriented with clockwise direction.

For example, the curve

\[ \gamma (t) = (\cos t , \sin t) \]

with \(0 \leq t \leq 2 \pi \) is positively oriented, while the curve

\[ \sigma (t) = ( \sin t, \cos t ) \]

with \(0 \leq t \leq 2 \pi \) is negatively oriented.

2 Line integrals of vector fields

Note: if you use a parametrization in the opposite direction, the integral changes sign.

If we let

\[ T(t) : = \frac {\gamma ' (t) }{ \vert \gamma ' (t) \vert } \]

be the unit tangent vector, then the line integral above becomes

\begin{align*} \int _C F \cdot ds & = \int _a ^b F(\gamma (t)) \cdot \gamma ' (t) \, dt \\ & = \int _a ^b F(\gamma (t)) \cdot \frac { \gamma ' (t)}{\vert \gamma ' (t) \vert } \, \vert \gamma ' (t) \vert \, dt \\ & = \int _C \left [ F \cdot T \right ] \, ds \end{align*}

The dot product \(F \cdot T\) represents “how much is \(F\) pointing in the direction in which \(\gamma \) is moving”. Therefore the integral can be interpreted as “how much did the force field \(F\) help \(\gamma \) perform its trajectory”.

The integral \(\int _C F \cdot ds\) is also called the work of \(F\) along the trajectory.

3 Exercises on line integrals

Let \(F (x,y,z) = \langle - 2y , z^2 + 3x , x - 1 \rangle \) and \(C\) the curve with parametrization \(\gamma : [0,2 ] \to \mathbb {R}^3\) given by
\[ \gamma (t) = ( t^2 - 3, 2 t , t^3 ). \]
Find
\[ \int _C F \cdot ds \]

Using the change of variables

\begin{align*} x(t) & = t^2 - 3\\ y(t) & = 2t \\ z(t) & = t^3 \end{align*}

we get

\[ F (\gamma (t) ) = \langle -4t , t^6 + 3t^2 - 9 , t^2 - 4 \rangle \]

We also need

\[ \gamma ' (t) = \langle 2t , 2, 3t^2 \rangle \]

Then

\begin{align*} \int _C F \cdot ds & = \int _0 ^2 \langle -4t , t^6 + 3t^2 - 9 , t^2 - 4 \rangle \cdot \langle 2t , 2, 3t^2 \rangle \, dt \\ & = \int _0 ^2 \left [ - 8 t^2 + 2t^6 + 6 t^2 - 18 + 3 t^4 - 12 t^2 \right ] \, dt \\ &= - \frac {64}{3} + \frac {256}{7} + \frac {48}{3} - 36 + \frac {96}{5} - 32 \end{align*}
Let \(F (x,y,z) = \langle z + 1 , x , y \rangle \) and \(C\) the curve with parametrization \(\gamma : [0, 3] \to \mathbb {R}^3\) given by
\[ \gamma (t) = ( e^{t} , - t^2 , t ). \]
Find
\[ \int _C F \cdot ds \]

Using the change of variables

\begin{align*} x(t) & = e^t \\ y(t) & = - t^2 \\ z(t) & = t \end{align*}

we get

\[ F (\gamma (t) ) = \langle t + 1 , e^t , - t^2 \rangle \]

We also need

\[ \gamma ' (t) = \langle e^t , - 2t , 1 \rangle \]

Then

\begin{align*} \int _C F \cdot ds & = \int _0 ^3 \langle t + 1 , e^t , - t^2 \rangle \cdot \langle e^t , - 2t , 1 \rangle \, dt \\ & = \int _ 0 ^3 \left [ te^t + e^t - 2 t e^t - t^2 \right ] \, dt \\ & = \left [ -t e^t + 2e^t - t^3 / 3 \right ] \vert _{t =0} ^3 \\ & = -3 e^3 + 2 e ^3 - 9 - 2 \\ & = - e ^3 - 11 \end{align*}

4 Fundamental Theorem of Calculus II

The classic Fundamental Theorem of Calculus says that the integral of the derivative of a function \(F(x)\) is the function \(F\) itself:

\[ \int _a^b F' (x) \, dx = F(b) - F(a) \]

Something similar happens when we take the line integral of a gradient. Consider a differentiable function \(f : D \to \mathbb {R}\) with \(D \subset \mathbb {R}^3\) and its gradient vector field \( \nabla f \). Also take an oriented curve \(C \subset \mathbb {R}^3\) and a parametrization \(\gamma : [a,b] \to \mathbb {R}^3\). Recall that by the chain rule, we have

\[ \frac {d}{dt} (f ( \gamma (t)) ) = \nabla f (\gamma (t) ) \cdot \gamma ' (t) . \]

Therefore

\begin{align*} \int _C \nabla f \cdot ds & = \int _a^b \nabla f (\gamma (t)) \cdot \gamma ' (t) \, dt \\ & = \int _a^b \frac {d}{dt} ( f ( \gamma (t) ) ) \, dt \\ & = f (\gamma (b)) - f (\gamma (a)) \end{align*}
Let \(F (x,y,z) = \langle x , \cos y , e^z \rangle \) and \(C\) the curve with parametrization \(\gamma : [0, \pi ] \to \mathbb {R}^3\) given by
\[ \gamma (t) = ( t^2 \sqrt { t^2 + 1 } , e^{t^2} , t^2 \cos t ). \]
Find
\[ \int _C F \cdot ds \]

Note that

\[ \text {curl}(F) = \langle 0,0,0 \rangle , \]

and the domain of \(F\) is \(\mathbb {R}^3\), so \(F\) is conservative. To find the potential function, we do partial integration,

\begin{align*} f(x,y,z) & = x^2 / 2 + g_1 (y,z) \\ f(x,y,z) & = \sin y + g_2 (x,z) \\ f(x,y,z) & = e^z + g_3(x,y) \end{align*}

Matching terms, we get

\[ f (x,y,z) = x^2 /2 + \sin y + e^z + C \]

On the other hand, the endpoints of \(C\) are

\begin{align*} \gamma (0) & = ( 0, 1, 0 ) \\ \gamma (\pi ) & = ( \pi ^2 \sqrt {\pi ^2 + 1 } , e ^{\pi ^2} , - \pi ^2 ) \end{align*}

Therefore,

\begin{align*} \int _C F \cdot ds & = f( \pi ^2 \sqrt {\pi ^2 + 1 } , e ^{\pi ^2} , - \pi ^2 ) - f (0,1,0) \\ & = \pi ^4 (\pi ^2 + 1) /2 + \sin (e ^{\pi ^2}) + e ^{- \pi ^2} - \sin (1) - 1 \end{align*}
Show that the vector field
\[ F(x,y) = \left \langle \frac {-y}{x^2 + y^2 } , \frac {x}{x^2 + y^2 } \right \rangle \]
is not conservative, even though it has zero curl.

Let \(C \subset \mathbb {R}^2 \) be the unit circle oriented counterclockwise. Take the parametrization \(\gamma (t) = (\cos t , \sin t)\) with \(0 \leq t \leq 2 \pi \). Then

\begin{align*} F(\gamma (t)) & = \left \langle \frac {- \sin t}{ \cos ^2 t + \sin ^2 t } , \frac {\cos t}{ \cos ^2 t + \sin ^2 t } \right \rangle \\ & = \langle - \sin t, \cos t \rangle \end{align*}

Also,

\[ \gamma ' (t) = \langle - \sin t , \cos t \rangle \]

Therefore,

\begin{align*} \int _{C} F \cdot ds & = \int _0 ^{2 \pi } F(\gamma (t)) \cdot \gamma ' (t) \, dt \\ & = \int _0 ^{2 \pi } (\sin ^2 t + \cos ^2 t ) \, dt \\ & = \int _0 ^{2 \pi } 1 \, dt \\ & = 2 \pi \neq 0 \end{align*}

If \(F\) was conservative, we would have \(F = \nabla f\) for some scalar function \(f (x,y)\). By the Fundamental Theorem of Calculus, we would have

\begin{align*} \int _C F \cdot ds & = f(\gamma (2 \pi )) - f (\gamma (0)) \\ &= f (1,0) - f (1,0) \\ & = 0 \end{align*}