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2.4: Line Integrals

  • Page ID
    91901
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    We have already seen, in §1.6, one type of integral along curves. We are now going to see a second, that turns out to have significant connections to conservative vector fields. It arose from the concept of “work” in classical mechanics.

    Suppose that we wish to find the work done by a force \(\vecs{F} (\vecs{r} )\) moving a particle along a path \(\vecs{r} (t)\text{.}\) During the “infinitesimal time interval” 1 from \(t\) to \(t+\text{d}t\) the particle moves from \(\vecs{r} (t)\) to \(\vecs{r} (t)+\text{d}\vecs{r} \) with \(\text{d}\vecs{r} =\dfrac{d\vecs{r} }{dt}(t)\,\text{d}t\text{.}\) By definition, the work done during that infinitesimal time interval is

    \[ \vecs{F} \big(\vecs{r} (t)\big)\cdot\text{d}\vecs{r} = \vecs{F} \big(\vecs{r} (t)\big)\cdot \dfrac{d\vecs{r} }{dt}(t)\,\text{d}t \nonumber \]

    The total work done during the time interval from \(t_0\) to \(t_1\) is then

    \[ \text{Work} = \int_{t_0}^{t_1}\vecs{F} \big(\vecs{r} (t)\big)\cdot\dfrac{d\vecs{r} }{dt}(t)\,\text{d}t \nonumber \]

    There are some useful shorthand notations for this work.

    Definition 2.4.1

    Denote by \(\mathcal{C}\) the parametrized path \(\vecs{r} (t)\) with \(t_0\le t\le t_1\text{.}\) Then

    \[ \int_\mathcal{C}\vecs{F} \cdot\text{d}\vecs{r} =\int_\mathcal{C}\big(\vecs{F} _1\text{d}x+\vecs{F} _2\text{d}y+\vecs{F} _3\text{d}z\big) =\int_{t_0}^{t_1}\vecs{F} \big(\vecs{r} (t)\big)\cdot\dfrac{d\vecs{r} }{dt}(t)\,\text{d}t \nonumber \]

    If \(\mathcal{C}\) is a closed path, the notation \(\oint_\mathcal{C}\vecs{F} \cdot\text{d}\vecs{r} \) is also used.

    In the event that \(\vecs{F} \) is conservative, and we know the potential \(\varphi\text{,}\) the following theorem provides a really easy way to compute “work integrals”. The theorem is a generalization of the fundamental theorem of calculus, and indeed some people call it the fundamental theorem of line integrals.

    Theorem 2.4.2

    Let \(\vecs{F} =\nabla\varphi\) be a conservative vector field. Then if \(\mathcal{C}\) is any curve that starts at \(P_0\) and ends at \(P_1\text{,}\) we have 2

    \[ \int_\mathcal{C}\vecs{F} \cdot\text{d}\vecs{r} =\varphi(P_1)-\varphi(P_0) \nonumber \]

    Proof

    Let \(\vecs{r} (t)=\big(x(t),y(t),z(t)\big)\text{,}\) \(t_0\le t\le t_1\text{,}\) be any parametrization of \(\mathcal{C}\) with \(\vecs{r} (t_0)=P_0\) and \(\vecs{r} (t_1)=P_1\text{.}\) Then, by definition,

    \[\begin{align*} \int_\mathcal{C}\vecs{F} \cdot\text{d}\vecs{r} &=\int_{t_0}^{t_1}\vecs{F} \big(\vecs{r} (t)\big)\cdot\dfrac{d\vecs{r} }{dt}(t)\,\text{d}t =\int_{t_0}^{t_1}\vecs{ \nabla} \varphi\big(\vecs{r} (t)\big)\cdot\dfrac{d\vecs{r} }{dt}(t)\,\text{d}t\\ &=\int_{t_0}^{t_1}\Big[ \frac{\partial\varphi}{\partial x}\big(x(t),y(t),z(t)\big) \dfrac{dx}{dt}(t) +\frac{\partial\varphi}{\partial y}\big(x(t),y(t),z(t)\big) \dfrac{dy}{dt}(t)\\ &\hskip2.7in +\frac{\partial\varphi}{\partial z}\big(x(t),y(t),z(t)\big) \dfrac{dz}{dt}(t) \Big]\text{d}t\\ &=\int_{t_0}^{t_1}\dfrac{d\ }{dt}\Big[\varphi\big(x(t),y(t),z(t)\big) \Big]\text{d}t \qquad\text{by the chain rule in reverse}\\ &=\varphi\big(\vecs{r} (t_1)\big) - \varphi\big(\vecs{r} (t_0)\big) =\varphi(P_1) - \varphi(P_0) \end{align*}\]

    by the fundamental theorem of calculus.

    Observe that, in Theorem 2.4.2, the value, \(\varphi(P_1)-\varphi(P_0)\text{,}\) of the integral \(\int_\mathcal{C}\vecs{F} \cdot\text{d}\vecs{r} \) depended only on the endpoints \(P_0\) and \(P_1\) of the curve, not on the path that the curve followed to get to \(P_0\) from \(P_1\text{.}\) We shall see, in Theorem 2.4.7, below, that this happens only for conservative vector fields. Here are several examples of line integrals of vector fields that are not conservative.

    Example 2.4.3

    Set \(P_0=(0,0)\text{,}\) \(P_1=(1,1)\) and 3

    \[ \vecs{F} (x,y) = xy\,\hat{\pmb{\imath}} + (y^2+1)\,\hat{\pmb{\jmath}} \nonumber \]

    We shall consider three curves, all starting at \(P_0\) and ending at \(P_1\text{.}\)

    1. Let \(\mathcal{C}_1\) be the straight line from \(P_0\) to \(P_1\text{.}\)
    2. Let \(\mathcal{C}_2\) be the path, made from two straight lines, which follows the \(x\)-axis from \(P_0\) to \((1,0)\) and then follows the line \(x=1\) from \((1,0)\) to \(P_1\text{.}\)
    3. Let \(\mathcal{C}_3\) be the part of the parabola \(x=y^2\) from \(P_0\) to \(P_1\text{.}\)

    workIntegralA.svg

    We shall calculate the work \(\int_{\mathcal{C}_i}\vecs{F} \cdot\text{d}\vecs{r} \) for each of the curves.

    1. We parametrize \(\mathcal{C}_1\) by \(\vecs{r} (t) = t\,\hat{\pmb{\imath}}+t\,\hat{\pmb{\jmath}}\) with \(t\) running from \(0\) to \(1\text{.}\) Then \(x(t)=t\) and \(y(t)=t\) so that

      \[ \vecs{F} \big(\vecs{r} (t)\big) = t^2\,\hat{\pmb{\imath}} + (t^2+1)\,\hat{\pmb{\jmath}}\qquad\text{and}\qquad \dfrac{d\vecs{r} }{dt}(t) = \hat{\pmb{\imath}} + \hat{\pmb{\jmath}} \nonumber \]

      so that

      \[\begin{align*} \int_{\mathcal{C}_1}\vecs{F} \cdot\text{d}\vecs{r} &=\int_{0}^{1}\vecs{F} \big(\vecs{r} (t)\big)\cdot\dfrac{d\vecs{r} }{dt}(t)\,\text{d}t =\int_{0}^{1}\big[t^2\,\hat{\pmb{\imath}} + (t^2+1)\,\hat{\pmb{\jmath}}\big]\cdot[\hat{\pmb{\imath}} + \hat{\pmb{\jmath}}]\,\text{d}t\\ &=\int_{0}^{1}\big[2t^2+1\big]\,\text{d}t\\ &=\frac{5}{3} \end{align*}\]

    2. We split \(\mathcal{C}_2\) into two parts, \(\mathcal{C}_{2,x}\) running from \(P_0\) to \((1,0)\) along the \(x\)-axis and then \(\mathcal{C}_{2,y}\) running from \((1,0)\) to \(P_1\) along the line \(x=1\text{.}\) We parametrize \(\mathcal{C}_{2,x}\) by \(\vecs{r} (x) = x\,\hat{\pmb{\imath}}\) with \(x\) running from \(0\) to \(1\) and \(\mathcal{C}_{2,y}\) by \(\vecs{r} (y) = \hat{\pmb{\imath}}+y\,\hat{\pmb{\jmath}}\) with \(y\) running from \(0\) to \(1\text{.}\) Then 4 

      \[\begin{align*} &\int_{\mathcal{C}_2}\vecs{F} \cdot\text{d}\vecs{r} =\int_{\mathcal{C}_{2,x}}\vecs{F} \cdot\text{d}\vecs{r} + \int_{\mathcal{C}_{2,y}}\vecs{F} \cdot\text{d}\vecs{r} \\ &\hskip0.25in=\int_{0}^{1}\big[(x) (0)\,\hat{\pmb{\imath}} + (0^2+1)\,\hat{\pmb{\jmath}}\big]\cdot \overbrace{\dfrac{d\ }{dx}\big(x\,\hat{\pmb{\imath}}\big)}^{\hat{\pmb{\imath}}}\,\text{d}x\\ &\hskip1.25in + \int_{0}^{1}\big[(1) (y)\,\hat{\pmb{\imath}} + (y^2+1)\,\hat{\pmb{\jmath}}\big]\cdot \overbrace{\dfrac{d\ }{dy}\big(\hat{\pmb{\imath}}+y\,\hat{\pmb{\jmath}}\big)}^{\hat{\pmb{\jmath}}}\,\text{d}y\\ &\hskip0.25in=\int_{0}^{1}0\,\text{d}x + \int_{0}^{1}\big(y^2+1\big)\,\text{d}y\\ &\hskip0.25in=\frac{4}{3} \end{align*}\]

    3. We parametrize \(\mathcal{C}_3\) by \(\vecs{r} (t) = t^2\,\hat{\pmb{\imath}}+t\,\hat{\pmb{\jmath}}\) with \(t\) running from \(0\) to \(1\text{.}\) Then \(x(t)=t^2\) and \(y(t)=t\) so that

      \[ \vecs{F} \big(\vecs{r} (t)\big) = t^3\,\hat{\pmb{\imath}} + (t^2+1)\,\hat{\pmb{\jmath}}\qquad\text{and}\qquad \dfrac{d\vecs{r} }{dt}(t) = 2t\,\hat{\pmb{\imath}} + \hat{\pmb{\jmath}} \nonumber \]

      so that

      \[\begin{align*} \int_{\mathcal{C}_3}\vecs{F} \cdot\text{d}\vecs{r} & =\int_{0}^{1}\big[t^3\,\hat{\pmb{\imath}} + (t^2+1)\,\hat{\pmb{\jmath}}\big]\cdot[2t\,\hat{\pmb{\imath}} + \hat{\pmb{\jmath}}]\,\text{d}t\\ &=\int_{0}^{1}\big[2t^4+t^2+1\big]\,\text{d}t\\ &=\frac{2}{5}+\frac{1}{3}+1 = \frac{26}{15} \end{align*}\]

    Note that, despite the fact that \(\mathcal{C}_1\text{,}\) \(\mathcal{C}_2\) and \(\mathcal{C}_3\) all start at \(P_0\) and all end at \(P_1\text{,}\) the three integrals \(\int_{\mathcal{C}_1}\vecs{F} \cdot\text{d}\vecs{r} \text{,}\) \(\int_{\mathcal{C}_2}\vecs{F} \cdot\text{d}\vecs{r} \) and \(\int_{\mathcal{C}_3}\vecs{F} \cdot\text{d}\vecs{r} \) all have different values.

    Example 2.4.4

    Set 5 

    \[ \vecs{F} (x,y) = 2y\,\hat{\pmb{\imath}} + 3x\,\hat{\pmb{\jmath}} \nonumber \]

    This time we consider two curves.

    1. Let \(\mathcal{C}_1\) be circle \(x^2+y^2=1\text{,}\) traversed once counterclockwise, starting at \((1,0)\text{.}\)
    2. Let \(\mathcal{C}_2\) be (trivial) curve which just consists of the single point \((1,0)\text{.}\)

    We shall calculate the work \(\int_{\mathcal{C}_i}\vecs{F} \cdot\text{d}\vecs{r} \) for each curve.

    1. We parametrize \(\mathcal{C}_1\) by \(\vecs{r} (t) = \cos t\,\hat{\pmb{\imath}}+\sin t\,\hat{\pmb{\jmath}}\) with \(t\) running from \(0\) to \(2\pi\text{,}\) just as we did in Example 1.0.1. Then

      \[\begin{align*} \oint_{\mathcal{C}_1}\vecs{F} \cdot\text{d}\vecs{r} &=\int_{0}^{2\pi}\big[2\sin t\,\hat{\pmb{\imath}} + 3\cos t\,\hat{\pmb{\jmath}}\big] \cdot[-\sin t\,\hat{\pmb{\imath}} + \cos t\,\hat{\pmb{\jmath}}]\,\text{d}t\\ &=\int_{0}^{2\pi}\big[-2\sin^2 t+3\cos^2 t\big]\,\text{d}t \end{align*}\]

      You could evaluate these integrals using double angle trig identities like you did in first year calculus. But there is a sneaky, much easier, way. Because \(\sin^2 t\) and \(\cos^2 t\) are translates of each other, and both are periodic of period \(\pi\text{,}\) the two integrals \(\int_0^{2\pi}\sin ^2 t\,\text{d}t\) and \(\int_0^{2\pi}\cos ^2 t\,\text{d}t\) represent the same area and so are equal. See the figure below.

      sin2Graph.svg
      cos2Graph.svg

      Thus

      \[\begin{align*} \int_{0}^{2\pi} \sin^2 t\,\text{d}t &=\int_{0}^{2\pi} \cos^2 t\,\text{d}t =\int_{0}^{2\pi} \frac{1}{2}\big[\sin^2 t+\cos^2t\big]\,\text{d}t\\ &=\frac{1}{2}\int_{0}^{2\pi} \text{d}t =\pi \end{align*}\]

      and

      \[\begin{gather*} \oint_{\mathcal{C}_1}\vecs{F} \cdot\text{d}\vecs{r} =-2\int_{0}^{2\pi} \sin^2 t\,\text{d}t +3\int_{0}^{2\pi} \cos^2 t\,\text{d}t =\pi \end{gather*}\]

    2. We parametrize \(\mathcal{C}_2\) by \(\vecs{r} (t) = \hat{\pmb{\imath}}\) for all \(t\text{.}\) Then \(\dfrac{d\vecs{r} }{dt}(t) = \vecs{0}\) and \(\int_{\mathcal{C}_2}\vecs{F} \cdot\text{d}\vecs{r} =0\text{.}\)

    Again, despite the fact that \(\mathcal{C}_1\) and \(\mathcal{C}_2\) both start at \((1,0)\) and end at \((1,0)\text{,}\) the two integrals \(\int_{\mathcal{C}_1}\vecs{F} \cdot\text{d}\vecs{r} \) and \(\int_{\mathcal{C}_2}\vecs{F} \cdot\text{d}\vecs{r} \) are different.

    Example 2.4.5. Example 2.3.14, again

    In Example 2.3.14, we saw that the vector field

    \[\begin{align*} &\vecs{F} (x,y) = -\frac{y}{x^2+y^2}\hat{\pmb{\imath}} + \frac{x}{x^2+y^2}\hat{\pmb{\jmath}}\\ &\qquad \text{defined for all $(x,y)$ in $\mathbb{R}^2$ except $(x,y)=(0,0)$} \end{align*}\]

    passed the screening test of Theorem 2.3.9.a, and yet was not conservative. In this example, we will see that this \(\vecs{F} \) violates the conclusion of Theorem 2.4.2, thereby providing a second proof that \(\vecs{F} (x,y)\) is not conservative on \(\mathbb{R}^2\) with \((0,0)\) removed. For the curve \(\mathcal{C}\text{,}\) of Theorem 2.4.2, we use the circle parametrized by \(x=a\cos\theta,\ y=a\sin\theta\text{,}\) \(0\le\theta\le 2\pi\text{.}\) Then \(\text{d}x=-a\sin\theta\,\text{d}\theta\) and \(\text{d}y=a\cos\theta\,\text{d}\theta\) so that

    \[\begin{align*} \frac{1}{2\pi}\int_\mathcal{C} \frac{x\,\text{d}y-y\,\text{d}x}{x^2+y^2} &=\frac{1}{2\pi}\int_0^{2\pi} \frac{a^2\cos^2\theta\,\text{d}\theta+a^2\sin^2\theta\,\text{d}\theta} {a^2\cos^2\theta+a^2\sin^2\theta} =\frac{1}{2\pi}\int_0^{2\pi}\text{d}\theta\\ &=1 \end{align*}\]

    The curve \(\mathcal{C}\) has initial point

    \begin{align*} P_0&=(a\cos\theta,\ a\sin\theta)\big|_{\theta=0} = (a,0)\\ \end{align*}

    and final point

    \begin{align*} P_1&=(a\cos\theta,\ a\sin\theta)\big|_{\theta=2\pi} = (a,0) =P_0 \end{align*}

    So, if \(\vecs{F} \) were conservative with potential \(\varphi\text{,}\) Theorem 2.4.2 would give that

    \[ \frac{1}{2\pi}\int_\mathcal{C} \frac{x\,\text{d}y-y\,\text{d}x}{x^2+y^2} =\varphi(P_1) - \varphi(P_0)=0 \nonumber \]

    Consequently, \(\vecs{F} \) can't be conservative.

    Path Independence

    This brings us to the following question. Let \(\vecs{F} \) be any fixed vector field. When is it true that, given any two fixed points \(P_0\) and \(P_1\text{,}\) the integrals

    \[ \int_{\mathcal{C}}\vecs{F} \cdot\text{d}\vecs{r} = \int_{\mathcal{C}'}\vecs{F} \cdot\text{d}\vecs{r} \nonumber \]

    for all curves \(\mathcal{C}\text{,}\) \(\mathcal{C}'\) that start at \(P_0\) and end at \(P_1\text{?}\) When can we ignore the path taken? If this is the case we say that “\(\int_{\mathcal{C}}\vecs{F} \cdot\text{d}\vecs{r} \) is independent of the path chosen” and we write

    \[ \int_{P_0}^{P_1}\vecs{F} \cdot\text{d}\vecs{r} =\int_{\mathcal{C}}\vecs{F} \cdot\text{d}\vecs{r} \nonumber \]

    for any path \(\mathcal{C}\) from \(P_0\) to \(P_1\text{.}\) The point of this section is that there is an intimate relation between path independence and conservativeness of vector fields, that we will get to in Theorem 2.4.7.

    For simplicity, we will consider only vector fields that are defined and continuous on all of \(\mathbb{R}^2\) (i.e. the \(xy\)-plane) or \(\mathbb{R}^3\) (i.e. the usual three dimensional world). Some discussion about what happens for vector fields that are defined only on part of \(\mathbb{R}^2\) or \(\mathbb{R}^3\) is given in the optional §4.5.

    First we show that if there is one pair of (not necessarily distinct) points \(P_0\text{,}\) \(P_1\) such that

    \[ \int_{\mathcal{C}_1}\vecs{F} \cdot\text{d}\vecs{r} = \int_{\mathcal{C}_2}\vecs{F} \cdot\text{d}\vecs{r} \nonumber \]

    for all curves \(\mathcal{C}_1\text{,}\) \(\mathcal{C}_2\) that start at \(P_0\) and end at \(P_1\text{,}\) then it is also true that, for any other pair of points \(P_0'\text{,}\) \(P_1'\)

    \[ \int_{\mathcal{C}'_1}\vecs{F} \cdot\text{d}\vecs{r} = \int_{\mathcal{C}'_2}\vecs{F} \cdot\text{d}\vecs{r} \nonumber \]

    for all curves \(\mathcal{C}'_1\text{,}\) \(\mathcal{C}'_2\) that start at \(P'_0\) and end at \(P'_1\text{.}\) This might seem unlikely at first, but the idea of the proof is really intuitive.

    Theorem 2.4.6

    Let \(\vecs{F} \) be a vector field that is defined and continuous on all of \(\mathbb{R}^2\) (or \(\mathbb{R}^3\)). Let \(P_0\text{,}\) \(P_1\text{,}\) \(P'_0\text{,}\) \(P'_1\) be any four points in \(\mathbb{R}^2\) (or \(\mathbb{R}^3\)). Assume that

    \[ \int_{\mathcal{C}_1}\vecs{F} \cdot\text{d}\vecs{r} = \int_{\mathcal{C}_2}\vecs{F} \cdot\text{d}\vecs{r} \nonumber \]

    for all curves \(\mathcal{C}_1\text{,}\) \(\mathcal{C}_2\) that start at \(P_0\) and end at \(P_1\text{.}\) Then

    \[ \int_{\mathcal{C}'_1}\vecs{F} \cdot\text{d}\vecs{r} = \int_{\mathcal{C}'_2}\vecs{F} \cdot\text{d}\vecs{r} \nonumber \]

    for all curves \(\mathcal{C}'_1\text{,}\) \(\mathcal{C}'_2\) that start at \(P'_0\) and end at \(P'_1\text{.}\)

    Proof

    Let \(\mathcal{C}'_1\) and \(\mathcal{C}'_2\) be any two curves that start at \(P'_0\) and end at \(P'_1\text{.}\)

    pathIndep.svg

    We start by choosing any two (auxiliary) curves

    • \(\mathcal{C}_\ell\) that starts at \(P_0\) and ends at \(P'_0\) and
    • \(\mathcal{C}_r\) that starts at \(P'_1\) and ends at \(P_1\text{.}\)

    and then we define the curves

    • \(\mathcal{C}_1\) to be \(\mathcal{C}_\ell\text{,}\) followed by \(C'_1\text{,}\) followed by \(\mathcal{C}_r\) and
    • \(\mathcal{C}_2\) to be \(\mathcal{C}_\ell\text{,}\) followed by \(C'_2\text{,}\) followed by \(\mathcal{C}_r\text{.}\)

    Then both \(\mathcal{C}_1\) and \(\mathcal{C}_2\) start at \(P_0\) and end at \(P_1\text{,}\) so that, by hypothesis,

    \[ \int_{\mathcal{C}_1}\vecs{F} \cdot\text{d}\vecs{r} = \int_{\mathcal{C}_2}\vecs{F} \cdot\text{d}\vecs{r} \nonumber \]

    and, from the construction of \(\mathcal{C}_1\) and \(\mathcal{C}_2\text{,}\)

    \[\begin{align*} &\int_{\mathcal{C}_\ell}\vecs{F} \cdot\text{d}\vecs{r} +\int_{\mathcal{C}'_1}\vecs{F} \cdot\text{d}\vecs{r} +\int_{\mathcal{C}_r}\vecs{F} \cdot\text{d}\vecs{r} = \int_{\mathcal{C}_\ell}\vecs{F} \cdot\text{d}\vecs{r} +\int_{\mathcal{C}'_2}\vecs{F} \cdot\text{d}\vecs{r} +\int_{\mathcal{C}_r}\vecs{F} \cdot\text{d}\vecs{r} \\ \implies & \int_{\mathcal{C}'_1}\vecs{F} \cdot\text{d}\vecs{r} = \int_{\mathcal{C}'_2}\vecs{F} \cdot\text{d}\vecs{r} \end{align*}\]

    as desired.

    We are now ready for our main theorem on conservative fields.

    Theorem 2.4.7

    Let \(\vecs{F} \) be a vector field that is defined and continuous on all of \(\mathbb{R}^2\) (or \(\mathbb{R}^3\)). Then the following three statements are equivalent.

    1. \(\vecs{F} \) is conservative. That is, there exists a function \(\varphi\) such that \(\vecs{F} =\vecs{ \nabla} \varphi\text{.}\)
    2. The integral \(\oint_\mathcal{C}\vecs{F} \cdot\text{d}\vecs{r} =0\) for any closed curve \(\mathcal{C}\text{.}\)
    3. The integral \(\int\vecs{F} \cdot\text{d}\vecs{r} \) is path independent. That is, for any points \(P_0\text{,}\) \(P_1\) we have \(\int_{\mathcal{C}_1}\vecs{F} \cdot\text{d}\vecs{r} = \int_{\mathcal{C}_2}\vecs{F} \cdot\text{d}\vecs{r} \) for all curves \(\mathcal{C}_1\text{,}\) \(\mathcal{C}_2\) that start at \(P_0\) and end at \(P_1\text{.}\)

    That is, if any one of the three statements are true, then all three are true.

    Proof

    It suffices for us to prove 6  that

    • the truth of (a) implies the truth of (b) and
    • the truth of (b) implies the truth of (c) and
    • the truth of (c) implies the truth of (a).

    That's exactly what we will do.

    (a)\(\implies\)(b): Let \(\mathcal{C}\) be a closed curve that starts at \(P_0\) and then ends back at \(P_0\text{.}\) Then, by Theorem 2.4.2 with \(P_1=P_0\text{,}\)

    \[ \oint_\mathcal{C}\vecs{F} \cdot\text{d}\vecs{r} =\varphi(P_0) - \varphi(P_0)=0 \nonumber \]

    (b)\(\implies\)(c): Pick any point \(P_0\) and set \(P_1=P_0\text{.}\) Then we are assuming that \(\oint_\mathcal{C}\vecs{F} \cdot\text{d}\vecs{r} =0\) for all curves that start at \(P_0\) and end at \(P_1\text{.}\) In particular \(\int_\mathcal{C}\vecs{F} \cdot\text{d}\vecs{r} \) takes the same value for all curves that start at \(P_0\) and end at \(P_1\text{.}\) So Theorem 2.4.6 immediately yields property (c).

    (c)\(\implies\)(a): We are to show that \(\vecs{F} \) is conservative. We'll start by guessing \(\varphi\) and then we'll verify that, for our chosen \(\varphi\text{,}\) we really do have \(\vecs{F} =\vecs{ \nabla} \varphi\text{.}\) Our guess for \(\varphi\) is motivated by Theorem 2.4.2. If our \(\vecs{F} \) really is conservative, its potential is going to have to obey \(\int_\mathcal{C}\vecs{F} \cdot\text{d}\vecs{r} =\varphi(P_1) - \varphi(P_0)\) for any curve \(\mathcal{C}\) that starts at \(P_0\) and ends at \(P_1\text{.}\) Let's choose \(P_0=\vecs{0}\text{.}\) Remembering, from Definition 2.3.1.a, that adding a constant to a potential always yields another potential, we can always choose \(\varphi(\vecs{0})=0\text{.}\) Then \(\varphi(P_1)=\int_\mathcal{C}\vecs{F} \cdot\text{d}\vecs{r} \) for any curve \(\mathcal{C}\) that starts at \(\vecs{0}\) and ends at \(P_1\text{.}\) So define, for each point \(\textbf{x}\text{,}\) \(\varphi(\textbf{x})=\int_\mathcal{C}\vecs{F} \cdot\text{d}\vecs{r} \) for any curve \(\mathcal{C}\) that starts at \(\vecs{0}\) and ends at \(\textbf{x}\text{.}\) Note that, since we we are assuming that (c) is true, the integral \(\int_\mathcal{C}\vecs{F} \cdot\text{d}\vecs{r} \) takes the same value for all curves \(\mathcal{C}\) that start at \(\vecs{0}\) and end at \(\textbf{x}\text{.}\)

    We now verify that, for this chosen \(\varphi\text{,}\) we really do have \(\vecs{F} =\vecs{ \nabla} \varphi\text{.}\) Fix any point \(\textbf{x}\) and any curve \(\mathcal{C}_{\textbf{x}}\) that starts at the origin and ends at \(\textbf{x}\text{.}\) For any vector \(\textbf{u}\text{,}\) let \(\mathcal{D}_{\textbf{u}}\) be the curve with parametrization

    \[ \vecs{r} _{\textbf{u}}(t)=\textbf{x}+t\textbf{u}\qquad 0\le t\le 1 \nonumber \]

    This curve is a line segment that starts at \(\textbf{x}\) at \(t=0\) and ends at \(\textbf{x}+\textbf{u}\) at \(t=1\text{.}\) Observe that \({\vecs{r} \,}'_{\textbf{u}}(t)=\textbf{u}\text{.}\) Recall that, by assumption, \(\varphi(\textbf{x}+s\textbf{u})=\int_\mathcal{C}\vecs{F} \cdot\text{d}\vecs{r} \) for any curve \(\mathcal{C}\) that starts at \(\vecs{0}\) and ends at \(\textbf{x}+s\textbf{u}\text{.}\) So

    \[ \varphi(\textbf{x}+s\textbf{u}) =\int_{\mathcal{C}_{\textbf{x}}+\mathcal{D}_{s\textbf{u}}}\vecs{F} \cdot d\vecs{r} \nonumber \]

    where \(C_{\textbf{x}}+D_{s\textbf{u}}\) is the curve which first follows \(C_{\textbf{x}}\) from the origin to \(\textbf{x}\) and then follows \(D_{s\textbf{u}}\) from \(x\) to \(\textbf{x}+s\textbf{u}\text{.}\) We have

    \[\begin{align*} \int_{C_{\textbf{x}}+D_{s\textbf{u}}}\vecs{F} \cdot d\vecs{r} &=\int_{C_{\textbf{x}}}\vecs{F} \cdot d\vecs{r} +\int_{D_{s\textbf{u}}}\vecs{F} \cdot d\vecs{r} \\ &=\int_{C_{\textbf{x}}}\vecs{F} \cdot d\vecs{r} +\int_0^1 \vecs{F} (\textbf{x}+ts\textbf{u})\cdot (s\textbf{u})\,dt \end{align*}\]

    In the second integral, make the change of variables \(\tau=ts\text{,}\) \(\text{d}\tau=s\text{d}t\text{.}\) This gives

    \[ \varphi(\textbf{x}+s\textbf{u})=\int_{C_{\textbf{x}}}\vecs{F} \cdot d\vecs{r} +\int_0^s \vecs{F} (\textbf{x}+\tau\textbf{u})\cdot \textbf{u}\,d\tau \nonumber \]

    By the fundamental theorem of calculus, applied to the second integral,

    \[ \dfrac{d\ }{ds}\varphi(\textbf{x}+s\textbf{u})\Big|_{s=0} =\vecs{F} (\textbf{x}+s\textbf{u})\cdot \textbf{u}\Big|_{s=0}=\vecs{F} (\textbf{x})\cdot \textbf{u} \nonumber \]

    Applying this with \(\textbf{u}=\hat{\pmb{\imath}},\ \hat{\pmb{\jmath}},\ \hat{\mathbf{k}}\) gives us

    \[ \Big(\frac{\partial\varphi}{\partial x}(\textbf{x})\,,\, \frac{\partial\varphi}{\partial y}(\textbf{x})\,,\, \frac{\partial\varphi}{\partial z}(\textbf{x})\Big) =\big(\vecs{F} (\textbf{x})\cdot\hat{\pmb{\imath}}\,,\,\vecs{F} (\textbf{x})\cdot\hat{\pmb{\jmath}}\,,\,\vecs{F} (\textbf{x})\cdot\hat{\mathbf{k}}\big) \nonumber \]

    which is

    \[ \nabla\varphi(\textbf{x})=\vecs{F} (\textbf{x}) \nonumber \]

    as desired.

    Using this result, we can completely characterize conservative fields on \(\mathbb{R}^2\) and \(\mathbb{R}^3\text{.}\)

    Theorem 2.4.8

    Let \(\vecs{F} \) be a vector field that is defined and has continuous first order partial derivatives on all of \(\mathbb{R} ^2\) (or \(\mathbb{R}^3\)). Then \(\vecs{F} \) is conservative if and only if it passes the screening test \(\vecs{\nabla} \times\vecs{F} =\vecs{0}\text{,}\) i.e. is curl free.

    Warning 2.4.9

    Note that in Theorem 2.4.8 we are assuming that \(\vecs{F} \) passes the screening test on all of \(\mathbb{R}^2\) or \(\mathbb{R}^3\text{.}\) We have already seen, in Example 2.3.14, that if the screening test fails at even a single point, for example because the vector field is not defined at that point, then \(\vecs{F} \) need not be conservative. We'll explore what happens in such cases in the (optional) §4.5. We'll see that something can be salvaged.

    Proof of Theorem 2.4.8.

    We'll give the proof for the \(\mathbb{R}^2\) case. The proof for the \(\mathbb{R}^3\) case is very similar. We have already seen, in Theorem 2.3.9, that if \(\vecs{F} \) is conservative, then it passes the screening test and there is nothing more to do.

    So we now have to assume that \(\vecs{F} \) obeys \(\frac{\partial F_1}{\partial y}(x,y) = \frac{\partial F_2}{\partial x}(x,y)\) on all of \(\mathbb{R}^2\) and prove that it is conservative. We'll do so using the strategy of Example 2.3.13 to find a function \(\varphi(x,y)\text{,}\) that obeys

    \[ \begin{split} \frac{\partial \varphi}{\partial x}(x,y) &= F_1(x,y) \\ \frac{\partial \varphi}{\partial y}(x,y) &= F_2(x,y) \end{split} \nonumber \]

    The partial derivative \(\frac{\partial \ }{\partial x}\) treats \(y\) as a constant. So \(\varphi(x,y)\) obeys the first equation if and only if there is a function \(\psi(y)\) with

    \[ \varphi(x,y) =\int_0^x F_1(X,y)\,\text{d}X \ +\ \psi(y) \nonumber \]

    This \(\varphi(x,y)\) will also obey the second equation if and only if

    \[\begin{align*} F_2(x,y)&= \frac{\partial \varphi}{\partial y}(x,y)\\ &=\frac{\partial \ }{\partial y}\Big(\int_0^x F_1(X,y)\,\text{d}X\ +\ \psi(y)\Big)\\ &=\int_0^x \frac{\partial F_1}{\partial y}(X,y)\,\text{d}X\ +\ \psi'(y) \end{align*}\]

    So we have to find a \(\psi(y)\) that obeys

    \[ \psi'(y) = F_2(x,y) - \int_0^x \frac{\partial F_1}{\partial y}(X,y)\,\text{d}X \nonumber \]

    This looks bad — no matter what \(\psi(y)\) is, the left hand side is independent of \(x\text{,}\) while it looks like the right hand side depends on \(x\text{.}\) Fortunately our screening test hypothesis now rides in to the rescue 7. (We haven't used it yet, and it has to come in somewhere.)

    \[\begin{align*} F_2(x,y) - \int_0^x \frac{\partial F_1}{\partial y}(X,y)\,\text{d}X &=F_2(x,y) - \int_0^x \frac{\partial F_2}{\partial x}(X,y)\,\text{d}X\\ &= F_2(x,y) - F_2(X,y)\Big|_{X=0}^{X=x}\\ &=F_2(0,y) \end{align*}\]

    In going from the first line to the second line we used the fundamental theorem of calculus. So choosing

    \[ \psi(y) = \int_0^y F_2(0,Y)\,\text{d}Y +C \nonumber \]

    for any constant \(C\text{,}\) does the trick.

    Exercises

    Stage 1

    1

    Evaluate \(\int_\mathcal{C} x^2y^2\,\text{d}x+x^3y\,\text{d}y\) counterclockwise around the square with vertices \((0,0)\text{,}\) \((1,0)\text{,}\) \((1,1)\) and \((0,1)\text{.}\)

    2

    For each of the following fields, decide which of the following holds:

    1. The characterization of conservative vector fields, Theorem 2.4.8 (with Theorem 2.3.9), tells us \(\vecs{F} \) is conservative.
    2. The characterization of conservative vector fields, Theorem 2.4.8 (with Theorem 2.3.9), tells us \(\vecs{F} \) is not conservative.
    3. The characterization of conservative vector fields, Theorem 2.4.8 (with Theorem 2.3.9), does not tell us whether \(\vecs{F} \) is conservative or not.
    1. \(\displaystyle \vecs{F} =x\hat{\pmb{\imath}} + z\hat{\pmb{\jmath}} + y\hat{\mathbf{k}}\)
    2. \(\displaystyle \vecs{F} =y^2z\hat{\pmb{\imath}} + x^2z\hat{\pmb{\jmath}} + x^2y\hat{\mathbf{k}}\)
    3. \(\displaystyle \vecs{F} =(ye^{xy}+1)\hat{\pmb{\imath}} + (xe^{xy}+z)\hat{\pmb{\jmath}} + \left( \frac1z+y\right)\hat{\mathbf{k}}\)
    4. \(\displaystyle \vecs{F} =y\cos(xy)\hat{\pmb{\imath}} + x\sin(xy)\hat{\pmb{\jmath}} \)
    3

    Let \(\varphi(x,y,z)=e^{x^2+y^2}+\cos(z^2)\text{,}\) and define \(\vecs{F} = \vecs{ \nabla} \varphi\text{.}\) Evaluate \(\int_C\vecs{F} \cdot\text{d}\vecs{r} \) over the closed curve \(C\) that is an ellipse traversed clockwise, centred at \((1,2,3)\text{,}\) passing through the points \((\sqrt5-1,-2,\sqrt5-3)\text{,}\) \(((\sqrt5-2)/2,-1/2,(\sqrt5-6)/2)\text{,}\) and \((-2,\sqrt 3-2,\sqrt3-3)\text{.}\)

    4

    Let \(P_1\) and \(P_2\) be points in \(\mathbb R^2\text{.}\) Let \(A\) and \(B\) be paths from \(P_1\) to \(P_2\text{,}\) as shown below.

    image-191.svg

    Suppose \(\vecs{F} \) is a conservative vector field in \(\mathbb R^2\) with \(\int_A \vecs{F} \cdot\text{d}\vecs{r} =5\text{.}\) What is \(\int_B \vecs{F} \cdot\text{d}\vecs{r} ?\)

    Let \(\vecs{F} (x, y, z) = e^x \sin y\,\hat{\pmb{\imath}} + \big[ ae^x \cos y + bz\big]\,\hat{\pmb{\jmath}} + cx\, \hat{\mathbf{k}}\text{.}\) For which values of the constants \(a\text{,}\) \(b\text{,}\) \(c\) is \(\int_C\vecs{F} \cdot\text{d}\vecs{r} =0\) for all closed paths \(C\text{?}\)

    6

    Consider the four vector fields sketched below. Exactly one of those vector fields is conservative. Determine which three vector fields are not conservative and explain why.

    VFa.svg
    (a)
    VFb.svg
    (b)
    VFc.svg
    (c)
    VFd.svg
    (d)

    Consider the vector field

    \[ \vecs{F} (x, y, z) = \frac{x-2y}{x^2+y^2}\,\hat{\pmb{\imath}} +\frac{2x + y}{x^2+y^2}\,\hat{\pmb{\jmath}} + z\,\hat{\mathbf{k}} \nonumber \]

    1. Determine the domain of \(\vecs{F} \text{.}\)
    2. Compute \(\vecs{ \nabla} \times \vecs{F} \text{.}\) Simplify the result.
    3. Evaluate the line integral

      \[ \int_C \vecs{F} \cdot \text{d}\vecs{r} \nonumber \]

      where \(C\) is the circle of radius \(2\) in the plane \(z = 3\text{,}\) centered at \((0, 0, 3)\) and traversed counter-clockwise if viewed from the positive \(z\)-axis, i.e. viewed “from above”.
    4. Is \(\vecs{F} \) conservative?
    8

    Find the work, \(\int_\mathcal{C}\vecs{F} \cdot \text{d}\vecs{r} \text{,}\) done by the force field \(\vecs{F} =(x+y)\hat{\pmb{\imath}}+(x-z)\hat{\pmb{\jmath}}+(z-y)\hat{\mathbf{k}}\) in moving an object from \((1,0,-1)\) to \((0,-2,3)\text{.}\) Does the work done depend on the path used to get from \((1,0,-1)\) to \((0,-2,3)\text{?}\)

    Stage 2

    9

    Consider the vector field

    \[ \textbf{V}(x, y) = (e^x \cos y + x^2, x^2y + 3) \nonumber \]

    Evaluate the line integral \(\int_C\textbf{V}\cdot \text{d}\vecs{r} \) along the oriented curve \(C\) obtained by moving from \((0, 0)\) to \((1,0)\) to \((1, \pi)\) and finally to \((0, \pi)\) along straight line segments.

    10

    Evaluate \(\int_\mathcal{C}\vecs{F} \cdot \text{d}\vecs{r} \) for

    1. \(\vecs{F} (x,y)=xy\,\hat{\pmb{\imath}}-x^2\,\hat{\pmb{\jmath}}\) along \(y=x^2\) from \((0,0)\) to \((1,1)\text{.}\)
    2. \(\vecs{F} (x,y,z)=(x-z)\,\hat{\pmb{\imath}}+(y-z)\,\hat{\pmb{\jmath}}-(x+y)\,\hat{\mathbf{k}}\) along the polygonal path from \((0,0,0)\) to \((1,0,0)\) to \((1,1,0)\) to \((1,1,1)\text{.}\)
    11 

    Let \(\mathcal{C}\) be the part of the curve of intersection of \(xyz=8\) and \(x=2y\) which lies between the points \((2,1,4)\) and \((4,2,1)\text{.}\) Calculate

    \[ \int_\mathcal{C} \vecs{F} \cdot \text{d}\vecs{r} \nonumber \]

    where

    \[ \vecs{F} = x^2\,\hat{\pmb{\imath}}+(x-2y)\,\hat{\pmb{\jmath}}+x^2 y\,\hat{\mathbf{k}} \nonumber \]

    12 

    Let \(\ \vecs{F} = e^x\sin y\,\hat{\pmb{\imath}}+[ae^x\cos y+bz]\,\hat{\pmb{\jmath}}+cx\,\hat{\mathbf{k}}\text{.}\) For which values of the constants \(a,\ b,\ c\) is \(\ \int_C\vecs{F} \cdot \text{d}\vecs{r} =0\ \) for all closed paths \(C\text{?}\)

    13

    Let \(\vecs{F} = 6x^2yz^2\,\hat{\pmb{\imath}} + (2x^3z^2 + 2y - xz)\,\hat{\pmb{\jmath}} + 4x^3yz\,\hat{\mathbf{k}}\) and let \(\textbf{G} = yz\,\hat{\pmb{\imath}} + xy\,\hat{\mathbf{k}}\text{.}\)

    1. For what value of the constant \(\lambda\) is the vector field \(\textbf{H} = \vecs{F} + \lambda\textbf{G}\) conservative on 3-space?
    2. Find a scalar potential \(\phi(x,y,z)\) for the conservative field \(\textbf{H}\) referred to in part (a).
    3. Find \(\int_C\vecs{F} \cdot \text{d}\vecs{r} \) if \(C\) is the curve of intersection of the two surfaces \(z = x\) and \(y = e^{xz}\) from the point \((0, 1, 0)\) to the point \((1, e, 1)\text{.}\)
    14 

    Find the work done by the force field \(\vecs{F} (x,y,z) = (x - y^2\,,\, y - z^2\,,\, z - x^2)\) on a particle that moves along the line segment from \((0, 0, 1)\) to \((2, 1, 0)\text{.}\)

    15 

    Let \(\ \vecs{F} = \frac{x}{x^2+y^2}\,\hat{\pmb{\imath}}+\frac{y}{x^2+y^2}\,\hat{\pmb{\jmath}}+x^3\,\hat{\mathbf{k}}\text{.}\) Let \(P\) be the path which starts at \((1,0,0)\text{,}\) ends at \(\big(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},\frac{1}{2}\ln 2\big)\) and follows

    \[ x^2+y^2=1\qquad xe^z=1 \nonumber \]

    Find the work done in moving a particle along \(P\) in the field \(\vecs{F} \text{.}\)

    16 

    Let \(\vecs{F} = \big(yz\cos x\,,\,z\sin x+2yz\,,\,y\sin x+y^2-\sin z\big)\) and let \(C\) be the line segment \(\vecs{r} (t) = (t,t,t)\text{,}\) for \(0\le t\le\pi/2\text{.}\) Evaluate \(\displaystyle\int_C\vecs{F} \cdot\text{d}\vecs{r} \text{.}\)

    17 

    Let \(C\) be the upper half of the unit circle centred on \((1, 0)\) (i.e. that part of the circle which lies above the \(x\)-axis), oriented clockwise. Compute the line integral \(\int_C xy\,\text{d}y\text{.}\)

    18 

    Show that the following line integral is independent of path and evaluate the integral.

    \[\begin{gather*} \int_C (ye^x + \sin y)\,\text{d}x + (e^x + \sin y + x \cos y)\,\text{d}y \end{gather*}\]

    where \(C\) is any path from \((1, 0)\) to \((0, \pi/2)\text{.}\)

    19 

    Evaluate the integral

    \[\begin{gather*} \int_C xy \,\text{d}x + yz \,\text{d}y + zx \,\text{d}z \end{gather*}\]

    around the triangle with vertices \((1, 0, 0)\text{,}\) \((0, 1, 0)\text{,}\) and \((0, 0, 1)\text{,}\) oriented clockwise as seen from the point \((1, 1, 1)\text{.}\)

    20 

    Evaluate the line integral \(\int_C \vecs{F} \cdot \text{d}\vecs{r} \text{,}\) where \(\vecs{F} \) is the conservative vector field

    \[ \vecs{F} (x, y, z) = \big(y + ze^x , x + e^y \sin z, z + e^x + e^y \cos z\big) \nonumber \]

    and \(C\) is the curve given by the parametrization

    \[ \vecs{r} (t) = (t, e^t , \sin t),\qquad t \text{ from } 0 \text{ to } \pi. \nonumber \]

    21 
    1. For which values of the constants \(\alpha\text{,}\) \(\beta\) and \(\gamma\) is the vector field

      \[ \vecs{F} (x,y,z) = \alpha e^y\,\hat{\pmb{\imath}}+(xe^y+\beta\cos z)\,\hat{\pmb{\jmath}}-\gamma y\sin z\,\hat{\mathbf{k}} \nonumber \]

      conservative?
    2. For those values of \(\alpha\text{,}\) \(\beta\) and \(\gamma\) found in part (a), calculate \(\int_C \vecs{F} \cdot \text{d}\vecs{r} \text{,}\) where \(C\) is the curve parametrized by \(x=t^2\text{,}\) \(y=e^t\text{,}\) \(z=\pi t\text{,}\) \(0\le t\le 1\text{.}\)
    22 

    Consider the vector field \(\vecs{F} (x,y,z) = (\cos x, 2 + \sin y, e^z)\text{.}\)

    1. Compute the curl of \(\vecs{F} \text{.}\)
    2. Is there a function \(f\) such that \(\vecs{F} = \vecs{ \nabla} f\text{?}\) Justify your answer.
    3. Compute the integral \(\int_C \vecs{F} \cdot\text{d}\vecs{r} \) along the curve \(C\) parametrized by \(\vecs{r} (t) = (t, \cos t, \sin t)\) with \(0 \le t \le 3\pi\text{.}\)
    23 
    1. Consider the vector field

      \[ \vecs{F} (x, y, z) = \left(z + e^y , xe^y - e^z \sin y, 1 + x + e^z \cos y\right) \nonumber \]

      Find the curl of \(\vecs{F} \text{.}\) Is \(\vecs{F} \) conservative?
    2. Find the integral \(\int_C \vecs{F} \cdot \text{d}\vecs{r} \) of the field \(\vecs{F} \) from (a) where \(C\) is the curve with parametrization

      \[ \vecs{r} (t) = (t^2 , \sin t, \cos^2 t) \nonumber \]

      where \(t\) ranges from \(0\) to \(\pi\text{.}\)
    24 

    A physicist studies a vector field \(\vecs{F} \text{.}\) From experiments, it is known that \(\vecs{F} \) is of the form

    \[ \vecs{F} = (x - a)ye^x\,\hat{\pmb{\imath}} + (xe^x + z^3 )\,\hat{\pmb{\jmath}} + byz^2\,\hat{\mathbf{k}} \nonumber \]

    where \(a\) and \(b\) are some real numbers. From theoretical considerations, it is known that \(\vecs{F} \) is conservative.

    1. Determine \(a\) and \(b\text{.}\)
    2. Find a potential \(f(x,y,z)\) such that \(\vecs{ \nabla} f = \vecs{F} \text{.}\)
    3. Evaluate the line intgeral \(\int_C\vecs{F} \cdot\text{d}\vecs{r} \) where \(C\) is the curve defined by

      \[ \vecs{r} (t) = \big(t\,,\, \cos 2t\,,\, \cos t\big), \qquad 0 \le t \le \pi \nonumber \]

    4. Evaluate the line integral

      \[ I = \int_C (x + 1)ye^x \,\text{d}x + (xe^x + z^3 ) \,\text{d}y + 4yz^2 \,\text{d}z, \nonumber \]

      where \(C\) is the same curve as in part (c). [Note: the “4” in the last term is not a misprint!].

    Questions 2.4.2.25 and 2.4.2.26 ask you to evaluate line integrals of vector fields that are not conservative, but that can be expressed as a sum of a conservative vector field and another vector field that can be written concisely.

    25 

    Let

    \[ \vecs{F} = \big(y^2 e^{3z} +Axy^3\big)\,\hat{\pmb{\imath}} +(2xye^{3z}+3x^2y^2\big)\,\hat{\pmb{\jmath}} +Bxy^2e^{3z}\,\hat{\mathbf{k}} \nonumber \]

    1. Find all values of \(A\) and \(B\) for which the vector field \(\vecs{F} \) is conservative.
    2. If \(A\) and \(B\) have values found in (a), find a potential function for \(\vecs{F} \text{.}\)
    3. Let \(C\) be the curve with parametrization \(\vecs{r} (t) = e^{2t}\,\hat{\pmb{\imath}} + e^{-t}\,\hat{\pmb{\jmath}} + \ln(1 + t) \,\hat{\mathbf{k}}\) from \((1, 1, 0)\) to \(\big(e^2\,,\,\frac{1}{e}\,,\,\ln 2\big)\text{.}\) Evaluate

      \[ \int_C (y^2 e^{3z} + xy^3)\,\text{d}x + (2xye^{3z} + 3x^2 y^2 )\,\text{d}y + 3xy^2 e^{3z}\,\text{d}z. \nonumber \]

    26 
    1. For which value(s) of the constants \(a,b\) is the vector field

      \[ \vecs{F} =\big(2x\sin(\pi y)-e^z\big)\hat{\pmb{\imath}} +\big(ax^2\cos(\pi y)-3e^z\big)\hat{\pmb{\jmath}} -\big(x+by\big)e^z\hat{\mathbf{k}} \nonumber \]

      conservative?
    2. Let \(\vecs{F} \) be a conservative field from part (a). Find all functions \(\varphi\) for which \(\vecs{F} =\vecs{ \nabla} \varphi\text{.}\)
    3. Let \(\vecs{F} \) be a conservative field from part (a). Evaluate \(\int_C \vecs{F} \cdot \text{d}\vecs{r} \) where \(C\) is the intersection of \(y=x\) and \(z=\ln(1+x)\) from \((0,0,0)\) to \((1,1,\ln 2)\text{.}\)
    4. Evaluate \(\int_C \textbf{G}\cdot \text{d}\vecs{r} \) where

      \[ \textbf{G}=\left(2x\sin(\pi y)-e^z\right)\,\hat{\pmb{\imath}} +\left(\pi x^2\cos(\pi y)-3e^z\right)\,\hat{\pmb{\jmath}} -xe^z\,\hat{\mathbf{k}} \nonumber \]

      and \(C\) is the intersection of \(y=x\) and \(z=\ln(1+x)\) from \((0,0,0)\) to \((1,1,\ln 2)\text{.}\)
    27 

    Consider the vector field

    \[ \vecs{F} (x, y, z) = -2y \cos x \sin x\,\hat{\pmb{\imath}} + (\cos^2 x + (1 + yz) e^{yz} )\,\hat{\pmb{\jmath}} + y^2 e^{yz}\,\hat{\mathbf{k}} \nonumber \]

    1. Find a real valued function \(f (x, y, z)\) such that \(\vecs{F} = \vecs{ \nabla} f\text{.}\)
    2. Evaluate the line integral

      \[ \int_C \vecs{F} \cdot \text{d}\vecs{r} \nonumber \]

      where \(C\) is the arc of the curve \(\vecs{r} (t) = \big(t, e^t , t^2 - \pi^2\big)\), \(0 \le t \le \pi\text{,}\) traversed from \((0, 1, -\pi^2 )\) to \((\pi, e^\pi , 0)\text{.}\)
    28 

    Consider the vector field \(\vecs{F} (x, y, z) = 2x\,\hat{\pmb{\imath}} + 2y\,\hat{\pmb{\jmath}} + 2z\,\hat{\mathbf{k}}\text{.}\)

    1. Compute \(\vecs{ \nabla} \times\vecs{F} \text{.}\)
    2. If \(C\) is any path from \((0, 0, 0)\) to \((a_1 , a_2 , a_3)\) and \(\textbf{a} = a_1\,\hat{\pmb{\imath}} + a_2\,\hat{\pmb{\jmath}} + a_3\,\hat{\mathbf{k}}\text{,}\) show that \(\int_C \vecs{F} \cdot \text{d}\vecs{r} = \textbf{a}\cdot\textbf{a}\text{.}\)
    29 

    Let \(C\) be the parameterized curve given by

    \[ \vecs{r} (t) = \big(\cos t, \sin t, t\big),\qquad 0 \le t \le \frac{\pi}{2} \nonumber \]

    and let

    \[ \vecs{F} = \big(e^{yz}\,,\, xze^{yz} + ze^y\,,\, xye^{yz} + e^y\big) \nonumber \]

    1. Compute and simplify \(\vecs{ \nabla} \times\vecs{F} \text{.}\)
    2. Compute the work integral \(\int_C\vecs{F} \cdot\text{d}\vecs{r} \text{.}\)
    30 
    1. Show that the planar vector field

      \[ \vecs{F} (x, y) = \big(2xy \cos(x^2)\,,\, \sin(x^2) - \sin(y)\big) \nonumber \]

      is conservative.
    2. Find a potential function for \(\vecs{F} \text{.}\)
    3. For the vector field \(\vecs{F} \) from above compute \(\int_C\vecs{F} \cdot\text{d}\vecs{r} \text{,}\) where \(C\) is the part of the graph \(x = \sin(y)\) from \(y = \pi/2\) to \(y = \pi\text{.}\)
    31 

    Consider the following force field, in which \(m,n,p,q\) are constants:

    \[ \vecs{F} = (mxyz + z^2 - ny^2)\,\hat{\pmb{\imath}} + (x^2 z - 4xy)\,\hat{\pmb{\jmath}} + (x^2y + pxz + qz^3)\,\hat{\mathbf{k}} \nonumber \]

    1. Find all values of \(m,n,p,q\) such that \(\oint_\mathcal{C} \vecs{F} \cdot \text{d}\vecs{r} = 0\) for all piecewise smooth closed curves \(\mathcal{C}\) in \(\mathbb{R}^3\text{.}\)
    2. For every possible choice of \(m,n,p,q\) in (a), find the work done by \(\vecs{F} \) in moving a particle from the bottom to the top of the sphere \(x^2 + y^2 + z^2 = 2z\text{.}\) (The direction of \(\hat{\mathbf{k}}\) defines “up”.)

    Stage 3

    32

    Let \(C\) be the curve from \((0,0,0)\) to \((1,1,1)\) along the intersection of the surfaces \(y=x^2\) and \(z=x^3\text{.}\)

    1. Find \(\int_C \rho\, \text{d}s\) if \(s\) is arc length along \(C\) and \(\rho=8x+36z\text{.}\)
    2. Find \(\int_C \vecs{F} \cdot \text{d}\vecs{r} \) if \(\vecs{F} =\sin y\,\hat{\pmb{\imath}} + (x\cos y+z)\,\hat{\pmb{\jmath}}+ (y+z)\,\hat{\mathbf{k}}\text{.}\)
    33 

    The curve \(C\) is the helix that winds around the cylinder \(x^2 + y^2 = 1\) (counterclockwise, as viewed from the positive \(z\)-axis, looking down on the \(xy\)-plane). It starts at the point \((1, 0, 0)\text{,}\) winds around the cylinder once, and ends at the point \((1, 0, 1)\text{.}\) Compute the line integral of the vector field

    \[ \vecs{F} (x, y, z) = (-y, x, z^2) \nonumber \]

    along \(C\text{.}\)

    34 

    Evaluate the line integrals below. (Use any method you like.)

    1. \(\int_C (x^2 + y)\,\text{d}x + x\,\text{d}y\text{,}\) where \(C\) is the arc of the parabola \(y = 9 - x^2\) from \((-3, 0)\) to \((3, 0)\text{.}\)
    2. \(\int_C \vecs{F} \cdot\hat{\textbf{n}}\, \text{d}s\text{,}\) where \(\vecs{F} (x, y) = 2x^2\hat{\pmb{\imath}} + ye^x\hat{\pmb{\jmath}}\text{,}\) \(C\) is the boundary of the square \(0 \le x \le 1\text{,}\) \(0 \le y \le 1\text{.}\) Here \(\hat{\textbf{n}}\) is the unit normal vector pointing outward from the square, and \(s\) is arc length.
    35 

    A particle of mass \(m = 1\) has position \(\vecs{r} _0 = \hat{\pmb{\jmath}}\) and velocity \(\vecs{v} _0 = \hat{\pmb{\imath}} + \hat{\mathbf{k}}\) at time \(t = 0\text{.}\) The particle moves under a force \(\vecs{F} (t) = \hat{\pmb{\jmath}} - \sin t\, \hat{\mathbf{k}}\text{,}\) where \(t\) denotes time.

    1. Find the position \(\vecs{r} (t)\) of the particle as a function of \(t\text{.}\)
    2. Find the position \(\vecs{r} _1\) of the particle when it crosses the plane \(x = \pi/2\) for the first time after time \(t = 0\text{.}\)
    3. Determine the work done by \(\vecs{F} \) in moving the particle from \(\vecs{r} _0\) to \(\vecs{r} _1\text{.}\)

    Questions 2.4.2.36 and 2.4.2.37 ask you to find a path that leads to a particular value of a line integral. Many such paths are possible — you only need to find one.

    36 
    1. Consider the vector field \(\vecs{F} \big(x,y\big)= (3y, x-1)\) in \(\mathbb{R}^2\). Compute the line integral

      \[ \int_L \vecs{F} \cdot\text{d}\vecs{r} \nonumber \]

      where \(L\) is the line segment from \((2, 2)\) to \((1, 1)\text{.}\)
    2. Find an oriented path \(C\) from \((2, 2)\) to \((1, 1)\) such that

      \[ \int_C \vecs{F} \cdot\text{d}\vecs{r} =4 \nonumber \]

      where \(\vecs{F} \) is the vector field from (a).
    37 

    Let \(\vecs{F} = (2y + 2)\,\hat{\pmb{\imath}}\) be a vector field on \(\mathbb{R}^2\text{.}\) Find an oriented curve \(C\) from \((0, 0)\) to \((2, 0)\) such that \(\int_C \vecs{F} \cdot \text{d}\vecs{r} = 8\text{.}\)

    38 

    Let

    \[ \vecs{F} (x, y) = \big(1, y g(y)\big) \nonumber \]

    and suppose that \(g(y)\) is a function defined everywhere with everywhere continuous partials. Show that for any curve \(C\) whose endpoints \(P\) and \(Q\) lie on the \(x\)-axis,

    \[ \text{distance between } P \text{ and } Q = \left|\int_C\vecs{F} \cdot\text{d}\vecs{r} \right| \nonumber \]

    39 

    Let \(S\) be the surface \(z = 2 + x^2 - 3 y^2\) and let \(\vecs{F} (x , y , z) = ( xz + axy^2 )\hat{\pmb{\imath}} + yz\hat{\pmb{\jmath}} + z^2\hat{\mathbf{k}}\text{.}\) Consider the points \(P_1 = ( 1 , 1 , 0 )\) and \(P_2 = ( 0 , 0 , 2 )\) on the surface \(S\text{.}\)

    Find a value of the constant \(a\) so that \(\int_{C_1} \vecs{F} \cdot \text{d}\vecs{r} = \int_{C_2} \vecs{F} \cdot \text{d}\vecs{r} \) for any two curves \(C_1\) and \(C_2\) on the surface \(S\) from \(P_1\) to \(P_2\text{.}\)

    40 

    Consider the vector field \(\vecs{F} \) defined as

    \[ \vecs{F} (x,y,z) = \Big( (1 + ax^2 )ye^{3x^2} - bxz \cos(x^2 z)\,,\, xe^{3x^2} \,,\, x^2 \cos(x^2 z) \Big) \nonumber \]

    where \(a\) and \(b\) are real valued constants.

    1. Compute \(\vecs{ \nabla} \times\vecs{F} \text{.}\)
    2. Determine for which values \(a\) and \(b\) the vector field \(\vecs{F} \) is conservative.
    3. For the values of \(a\) and \(b\) obtained in part (b), find a potential function \(f\) such that \(\vecs{ \nabla} f = \vecs{F} \text{.}\)
    4. Evaluate the line integral

      \[ \int_C \Big(ye^{3x^2} + 2xz\cos(x^2 z)\Big) \text{d}x + xe^{3x^2} \text{d}y + x^2 \cos(x^2 z) \text{d}z \nonumber \]

      where \(C\) is the arc of the curve \((t, t, t^3)\) starting at the point \((0, 0, 0)\) and ending at the point \((1, 1, 1)\text{.}\)
    41 

    Let \(C\) be the curve from \((0,0,0)\) to \((1,1,1)\) along the intersection of the surfaces \(y=x^2\) and \(z=x^3\text{.}\)

    1. Find \(\int_C \vecs{F} \cdot \text{d}\vecs{r} \) if \(\vecs{F} =(xz-y)\,\hat{\pmb{\imath}} + (z+x)\,\hat{\pmb{\jmath}}+ y\,\hat{\mathbf{k}}\text{.}\)
    2. Find \(\int_C \rho\, \text{d}s\) if \(s\) is arc length along \(C\) and \(\rho(x,y,z)=8x+36z\text{.}\)
    3. Find \(\int_C \vecs{F} \cdot \text{d}\vecs{r} \) if \(\vecs{F} =\sin y\,\hat{\pmb{\imath}} + (x\cos y+z)\,\hat{\pmb{\jmath}}+ (y+z)\,\hat{\mathbf{k}}\text{.}\)
    42 

    The vector field \(\vecs{F} (x,y,z)= Ax^3y^2z\,\hat{\pmb{\imath}}+\big(z^3+Bx^4yz\big)\,\hat{\pmb{\jmath}} +\big(3yz^2-x^4y^2\big)\,\hat{\mathbf{k}}\) is conservative on \(\mathbb{R}^3\text{.}\)

    1. Find the values of the constants \(A\) and \(B\text{.}\)
    2. Find a potential \(\varphi\) such that \(\vecs{F} =\vecs{ \nabla} \varphi\) on \(\mathbb{R}^3\text{.}\)
    3. If \(\mathcal{C}\) is the curve \(y=-x,\ z=x^2\) from \((0,0,0)\) to \((1,-1,1)\text{,}\) evaluate \(I=\int_\mathcal{C} \vecs{F} \cdot \text{d}\vecs{r} \text{.}\)
    4. Evaluate \(J=\int_\mathcal{C} (z-4x^3y^2z)\text{d}x+(z^3-x^4yz)\text{d}y+(3yz^2-x^4y^2)\text{d}z\text{,}\) where \(\mathcal{C}\) is the curve of part (c).
    5. Let \(\mathcal{T}\) be the closed triangular path with vertices \((1,0,0)\text{,}\) \((0,1,0)\) and \((0,0,1)\text{,}\) oriented counterclockwise as seen from the point \((1,1,1)\text{.}\) Evaluate \(\int_\mathcal{T}(z\hat{\pmb{\imath}}+\vecs{F} )\cdot \text{d}\vecs{r} \text{.}\)
    43 

    A particle of mass

    \[ m=2 \nonumber \]

    is acted on by a force

    \[ \vecs{F} = \big(4t\,,\,6t^2\,,\,-4t\big) \nonumber \]

    At \(t=0\text{,}\) the particle has velocity zero and is located at the point \((1,2,3)\text{.}\)

    1. Find the velocity vector \(\vecs{v} (t)\) for \(t\ge 0\text{.}\)
    2. Find the position vector \(\vecs{r} (t)\) for \(t\ge 0\text{.}\)
    3. Find \(\kappa(t)\) the curvature of the path traversed by the particle for \(t\ge 0\text{.}\)
    4. Find the work done by the force on the particle from \(t=0\) to \(t=T\text{.}\)
    44 

    The position of an airplane at time \(t\) is given by \(x=y=\frac{4\sqrt{2}}{3}t^{3/2},\ z=t(2-t)\) from take-off at \(t=0\) to landing at \(t=2\text{.}\)

    1. What is the total distance the plane travels on this flight?
    2. Find the radius of curvature \(\kappa\) at the apex of the flight, which occurs at \(t=1\text{.}\)
    3. Two external forces are applied to the plane during the flight: the force of gravity \(\textbf{G} =-M g\,\hat{\mathbf{k}}\text{,}\) where \(M\) is the mass of the plane and \(g\) is a constant; and a friction force \(\vecs{F} =-|\vecs{v} |^2\vecs{v} \text{,}\) where \(\vecs{v} \) is the velocity of the plane. Find the work done by each of these forces during the flight.
    4. One half-hour later, a bird follows the exact same flight — path as the plane, travelling at a constant speed \(v=3\text{.}\) One can show that at the apex of the path, i.e. when the bird is at \(\big(\frac{4\sqrt{2}}{3},\frac{4\sqrt{2}}{3},1\big)\text{,}\) the principal unit normal \(\hat{\textbf{N}}\) to the path points in the \(-\hat{\mathbf{k}}\) direction. Find the bird's (vector) acceleration at that moment.
    1. Yes, yes. We should first consider short time intervals \(\Delta t \gt 0\) and then take the limit \(\Delta t\rightarrow 0\) at the end. But you have undoubtedly used this type of argument so many times before that you would be thoroughly bored by it.
    2. So \(\varphi\) acts a bit like the antiderivative of first year calculus.
    3. The reader should check that this vector field is not conservative.
    4. You might like to think about why we can split up the integral like this.
    5. Again, the reader should verify that this vector field is not conservative.
    6. This is a pretty efficient, and standard, way to structure the proof of the equivalence of three statements.
    7. or bails us out, or saves our bacon, or \(\ldots\)

    This page titled 2.4: Line Integrals is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform.