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Section 3.5 Applications of line integrals

We mentioned in Subsection 3.3.2 that line integrals are sometimes called “work integrals”. In this section we explain why, and work through an example where line integrals can be used to calculate the work done. In this section we use vector field notation instead of one-forms, as this is what is most commonly encountered in such applications.

Subsection 3.5.1 Work

If you pick something off the ground, you expend energy. In physics, this is called “work”, because you move an object that is under the influence of a force field. If you move an object along a given path in a force field, how can you find the work done?

The idea is to use the well known “slicing” principle that turns a problem into an integration question. Suppose that there is a force field \(\mathbf{F}(x,y,z)\) in \(\mathbb{R}^3\text{,}\) and that we move an object along a path specified by a position vector \(\mathbf{r}(t)\) with \(a \leq t \leq b\text{.}\) If the force field was constant and did not depend on the position \((x,y,z)\) of the object, from general physics principles the work done on the object moving along the path would be \(\mathbf{F} \cdot (\mathbf{r}(b) - \mathbf{r}(a) )\text{,}\) i.e. the dot product of the force and the displacement (in other words, it is the magnitude of the force times the displacement in the direction of the force). However, as the force field depends on the position of the object, we cannot easily calculate the work directly. But we can slice the problem and sum over slices to rewrite the calculation as an integral.

We slice our time interval \([a,b]\) into small slices of width \(\Delta t\text{.}\) Over a small time interval \(\Delta t\text{,}\) the object moves from position \(\mathbf{r}(t)\) to position \(\mathbf{r}(t) + \Delta \mathbf{r}\text{,}\) where \(\Delta \mathbf{r} = ( \Delta x, \Delta y, \Delta z)\text{.}\) If the time interval is small, we can assume that the force is constant, and the work done during this time interval can be calculated, to first order, by \(\mathbf{F}(\mathbf{r}(t)) \cdot \Delta \mathbf{r}\text{.}\) Now we sum over slices, and take the limit where we have an infinite number of infinitesimal time intervals; this turns the calculation into a definite integral:

\begin{equation*} W = \int_a^b \mathbf{F} \cdot d \mathbf{r} = \int_a^b \mathbf{F} \cdot \frac{d \mathbf{r}}{d t}\ dt. \end{equation*}

We of course recognize the line integral of a vector field as defined in Lemma 3.3.7. This is why line integrals are called work integrals: if the vector field is a force field, the line integral over a parametrized curve calculates the work done when the objects moves along this curve.

Consider the force field in \(\mathbb{R}^2\) given by \(\mathbf{F}(x,y) = (y, 5 x)\text{.}\) We will calculate the work done when moving an object along two closed curves:

  1. going once around the unit circle counterclockwise, starting and ending at \((1,0)\text{;}\)

  2. going once around a square counterclockwise, with vertices \((1,1), (1,-1), (-1,-1), (-1,1)\text{,}\) and starting and ending at \((1,1)\text{.}\)

Let us start with the circle (call it \(C\)). We parametrize the circle by \(\mathbf{r}(t) = (\cos t, \sin t)\text{,}\) with \(0 \leq t \leq 2 \pi\text{.}\) The velocity vector is then \(\frac{d \mathbf{r}}{d t} = (- \sin t, \cos t)\text{.}\) To calculate the work done, we evaluate the line integral along the circle:

\begin{align*} W = \int_C \mathbf{F} \cdot \frac{d \mathbf{r}}{d t}\ dt =\amp \int_0^{2 \pi} \left( (\sin t) (- \sin t) + (5 \cos t)(\cos t) \right)\ dt \\ =\amp \int_0^{2 \pi} \left( - \sin^2 t + 5 \cos^2 t \right)\ dt\\ = \amp \int_0^{2 \pi} \left( 6 \cos^2 t - 1 \right)\ dt\\ = \amp \int_0^{2 \pi} \left( 3 (1 + \cos(2 t)) - 1 \right)\ dt\\ =\amp \int_0^{2 \pi} \left( 2 + 3 \cos(2 t) \right)\ dt \\ =\amp 2(2 \pi) + \frac{3}{2} \sin(4 \pi) - \frac{3}{2} \sin(0)\\ = \amp 4 \pi. \end{align*}

We see that even if the curve is closed (i.e. it starts and ends at the same point), the work done is non-zero: this is because the force field is not conservative. If it was conservative, by Corollary 3.4.3 the work would have been zero.

Now consider the square (call it \(S\)). It is a piecewise parametric curve, so we need to parametrize the four line segments separately. We use the following parametrizations:

  • (\(L_1\)) From \((1,1)\) to \((-1,1)\) : \(\qquad \mathbf{r}_1(t) = (1-t, 1)\text{,}\) \(0 \leq t \leq 2\text{;}\)

  • (\(L_2\)) From \((-1,1)\) to \((-1,-1)\text{:}\) \(\qquad \mathbf{r}_2(t) = (-1, 1-t)\text{,}\) \(0 \leq t \leq 2\text{;}\)

  • (\(L_3\)) From \((-1,-1)\) to \((1,-1)\text{:}\) \(\qquad \mathbf{r}_3(t) = (-1+t, -1)\text{,}\) \(0 \leq t \leq 2\text{;}\)

  • (\(L_4\)) From \((1,-1)\) to \((1,1)\text{:}\) \(\qquad \mathbf{r}_4(t) = (1, -1+t)\text{,}\) \(0 \leq t \leq 2\text{.}\)

The work done is then calculated by summing the four line integrals:

\begin{align*} W =\amp \int_S \mathbf{F} \cdot \frac{d \mathbf{r}}{d t}\ dt \\ =\amp \int_{L_1} \mathbf{F} \cdot \frac{d \mathbf{r}_1}{d t}\ dt+\int_{L_2} \mathbf{F} \cdot \frac{d \mathbf{r}_2}{d t}\ dt+\int_{L_3} \mathbf{F} \cdot \frac{d \mathbf{r}_3}{d t}\ dt+\int_{L_4} \mathbf{F} \cdot \frac{d \mathbf{r}_4}{d t}\ dt\\ =\amp \int_0^2((1)(-1) +5(1-t)(0))\ dt + \int_0^2 ((1-t)(0) + 5(-1)(-1))\ dt\\ \amp + \int_0^2 ((-1)(1)+5(-1+t)(0))\ dt + \int_0^2((-1+t)(0) + 5 (1)(1))\ dt\\ =\amp -2 + 10 -2 + 10 \\ =\amp 16. \end{align*}

We see that the work is again non-zero, and in fact it is not the same as the work done when going around the unit circle. This is as expected: as the force is non-conservative, the work done should depend on the path chosen.

Subsection 3.5.2 Conservation of energy

A force field that can be written as the gradient of a potential is called “conservative” for a reason. The name comes from physics, as it is related to conservation of energy, as we now see.

Let \(\mathbf{F}\) be a force field in \(\mathbb{R}^3\text{,}\) and suppose that an object moves along a parametrized path \(\mathbf{r}(t)\) from \(t=a\) to \(t=b\) (let's call this parametric curve \(\alpha\)). By Newton's law, we know that

\begin{equation*} \mathbf{F} = m \frac{d^2 \mathbf{r}}{d t^2}, \end{equation*}

where \(m\) is the mass of the object. From the discussion above, we see that the work done by the force on the object is:

\begin{align*} W =\amp \int_\alpha \mathbf{F} \cdot d \mathbf{r}\\ =\amp m \int_a^b \frac{d^2 \mathbf{r} }{d t^2} \cdot \frac{d \mathbf{r}}{d t} \ dt\\ =\amp m \int_a^b \frac{d}{dt} \left( \frac{d \mathbf{r}}{d t} \cdot \frac{d \mathbf{r}}{d t} \right) \ dt - m \int_a^b \frac{d \mathbf{r}}{d t} \cdot \frac{d^2 \mathbf{r} }{d t^2} \ dt. \end{align*}

We notice that the second integral on the last line is the same as the integral on the previous line, so we end up with the statement that

\begin{align*} m \int_a^b \frac{d^2 \mathbf{r} }{d t^2} \cdot \frac{d \mathbf{r}}{d t} \ dt =\amp \frac{1}{2} m \int_a^b \frac{d}{dt} \left( \frac{d \mathbf{r}}{d t} \cdot \frac{d \mathbf{r}}{d t} \right) \ dt\\ =\amp \frac{m}{2} |\mathbf{r}'(b)|^2 - \frac{m}{2} |\mathbf{r}'(a)|^2. \end{align*}

Since \(\mathbf{r}'(t) = \mathbf{v}(t)\) is the velocity of the object, we recognize the expression \(\frac{m}{2} |\mathbf{r}'(t)|^2\) as being the kinetic energy of the object at the point parametrized by \(t\) on its path \(C\text{,}\) which we will denote by \(T(t)\text{.}\) Thus we conclude that the work is the difference in kinetic energy

\begin{equation*} W = T(b) - T(a) \end{equation*}

between the kinetic energy \(T(b)\) of the object at the ending point of the path and its kinetic energy \(T(a)\) at the starting point.

Now assume that \(\mathbf{F}\) is a conservative force field. It can then be written as the gradient of a potential. We now change our conventions (only for this section), to be consistent with the physics literature, and introduce a minus sign. We write: \(\mathbf{F} = -\boldsymbol{\nabla} V\) for some potential function \(V\text{.}\) By the Fundamental Theorem of line integrals, the work can then be evaluated as follows:

\begin{align*} W =\amp \int_\alpha \mathbf{F} \cdot d \mathbf{r}\\ =\amp - \int_a^b \boldsymbol{\nabla} V \cdot d \mathbf{r}\\ =\amp - (V(\mathbf{r}(b)) - V(\mathbf{r}(a)). \end{align*}

Comparing with our previous calculation of the work, we conclude that

\begin{equation*} T(b) - T(a) = - (V(\mathbf{r}(b)) - V(\mathbf{r}(a))), \end{equation*}

or, rearranging,

\begin{equation*} T(b) + V(\mathbf{r}(b) = T(a) + V(\mathbf{r}(a)). \end{equation*}

This statement is the well known law of energy conservation in physics! Indeed, the potential \(V\) is interpreted as the potential energy, and the equality says that the total sum of the object's kinetic and potential energies remains constant as the object moves along the path. This is why such forces are called “conservative”!

Exercises 3.5.3 Exercises

1.

The force exerted by an electric charge at the origin on a charged particle at a point \((x,y,z) \in \mathbb{R}^3\) is

\begin{equation*} \mathbf{F}(x,y,z) = \frac{K}{(x^2+y^2+z^2)^{3/2}} (x,y,z), \end{equation*}

where \(K\) is a constant. Find the work done as the particle moves along a straight line from \((1,0,0)\) to \((2,2,3)\text{.}\)

Solution.

We parametrize the line as \(\alpha:[0,1] \to \mathbb{R}^3\) with \(\alpha(t) = (1+t, 2t, 3t)\text{.}\) The tangent vector is \(\mathbf{T}(t) = (1,2,3)\text{.}\) The line integral thus becomes

\begin{equation*} \int_\alpha \mathbf{F} \cdot d \mathbf{r} = K \int_0^1 \frac{1}{( (1+t)^2 + (2t)^2 + (3t)^2 )^{3/2}} ((1+t) + 2(2t) + 3(3t) ) \ dt . \end{equation*}

We do the substitution \(u = (1+t)^2 + (2t)^2 + (3t)^2\text{,}\) with \(du = 2 \left((1+t) + 2(2t) + 3(3t) \right)\ dt\text{,}\) and \(u(0) = 1,~u(1)= 4+4+9 = 17\text{.}\) The integral becomes

\begin{align*} \int_\alpha \mathbf{F} \cdot d \mathbf{r} =\amp \frac{K}{2} \int_1^{17} u^{-3/2}\ du \\ =\amp K \left(1-\frac{1}{\sqrt{17}} \right). \end{align*}

2.

True or False. A force field \(\mathbf{F}(x,y,z) = k (x,y,z)\text{,}\) with \(k\) any constant, does no work on a particle that moves once around the unit circle in the \(xy\)-plane.

Solution.

This is true, since the force field is conservative, and the integral of a conservative vector field around a closed curve is always zero. To show that the force field is conservative, consider the potential \(f(x,y,z) = \frac{k}{2}(x^2+y^2+z^2).\) Then

\begin{equation*} \boldsymbol{\nabla}f = k (x,y,z) = \mathbf{F}. \end{equation*}

While the above is a sufficient solution, let us compute the line integral for fun, to see that we get zero indeed. We parametrize the circle as \(\alpha:[0,2\pi] \to \mathbb{R}^3\text{,}\) \(\alpha(t) = (\cos t, \sin t, 0)\text{,}\) with tangent vector \(\mathbf{T}(t) = (-\sin t, \cos t, 0)\text{.}\) The line integral becomes

\begin{align*} \int_\alpha \mathbf{F} \cdot d \mathbf{r} =\amp k \int_0^{2 \pi} \left( - \sin t \cos t + \sin t \cos t +0 \right)\ dt \\ =\amp 0, \end{align*}

as expected.

3.

Find the work done by the force field \(\mathbf{F}(x,y,z) = ( x^2+y, x + y, 0)\) when moving an object from \((1,1,1)\) to \((0,0,0)\text{.}\)

Solution.

The question does not specify the path taken between \((1,1,1)\) and \((0,0,0)\text{;}\) so we can only calculate the work if the force is conservative, in which case its line integral does not depend on the path chosen.

Fortunately, the force is conservative. Pick the potential \(f(x,y,z) = \frac{x^3}{3} + x y + \frac{y^2}{2}\text{.}\) Then

\begin{equation*} \boldsymbol{\nabla} f = \left( x^2 + y, x + y, 0 \right) = \mathbf{F}. \end{equation*}

Then, using the Fundamental Theorem for line integrals, we calculate the work:

\begin{align*} \int_C \mathbf{F} \cdot d \mathbf{r} =\amp \int_C \boldsymbol{\nabla} f \cdot d \mathbf{r}\\ =\amp f(0,0,0) - f(1,1,1) \\ =\amp -\frac{1}{3} - 1 - \frac{1}{2}\\ =\amp -\frac{11}{6}. \end{align*}