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Section 3.3 Line integrals

We are finally ready to define the integral of a one-form over a curve in \(\mathbb{R}^n\text{.}\) Our strategy is to start with a parametrization for the curve, and then use the parametrization to pullback the integrand to the interval \([a,b]\text{,}\) over which we know how to integrate. In other words, we reduce a complicated problem to something that we already know how to solve!

However, ultimately we would like our integral to be independent of our choice of parametrization. We show that the integral is indeed invariant under orientation-preserving reparametrizations, and thus can be understood as an object defined solely in terms of the geometry of a curve and a choice of orientation. We also show that the integral changes sign under orientation-reversing reparametrizations, as expected.

Subsection 3.3.1 The pullback of a one-form along a parametric curve

Consider a one-form \(\omega\) on an open subset \(U\) of \(\mathbb{R}^n\text{.}\) Suppose that \(C\) is a curve which is contained in \(U\text{.}\) Our goal is to define an integral of the form “\(\int_C \omega\)” for the integral of the one-form \(\omega\) along the curve \(C\) (with a choice of orientation on \(C\)). However, this is not so obvious, as it is not clear what it means to “integrate along a curve”. To make sense of this, we use a parametrization for \(C\text{,}\) which is a map \(\alpha: [a,b] \to \mathbb{R}^n\text{.}\) The idea is to use the powerful concept of pullback, which we studied in Section 2.4, to pull back the one-form from \(U\) to the interval \([a,b]\text{,}\) and then we can integrate it, as we know how to integrate a one-form over an interval: Definition 3.1.1. Neat!

We can apply the general expression for the pullback of a one-form obtained in Lemma 2.4.5 to the case where \(\phi\) is replaced by a parametric curve \(\alpha: [a,b] \to \mathbb{R}^n\text{.}\) Let us write explicit expressions for the cases of \(\mathbb{R}^2\) and \(\mathbb{R}^3\text{.}\)

Let \(\alpha: [a,b] \to \mathbb{R}^2\) be a parametric curve as in Definition 3.2.1, with \(\alpha(t) = (x(t), y(t))\text{,}\) and let \(\omega = f(x,y)\ dx + g(x,y)\ dy\) be a one-form on an open subset \(U \subseteq \mathbb{R}^2\) containing \(C = \alpha([a,b])\text{.}\) Then \(\alpha^* \omega\) is a one-form on an open subset \(V \subseteq \mathbb{R}\) containing \([a,b]\) defined by

\begin{equation*} \alpha^* \omega = \left(f(\alpha(t)) \frac{dx}{dt} + g(\alpha(t)) \frac{dy}{dt} \right)\ dt. \end{equation*}

Similarly, given a parametric curve \(\alpha: [a,b] \to \mathbb{R}^3\text{,}\) with \(\alpha(t) = (x(t), y(t), z(t))\text{,}\) and a one-form \(\omega = f(x,y,z)\ dx + g(x,y,z)\ dy + h(x,y,z)\ dz \) on \(U \subseteq \mathbb{R}^3\text{,}\) the pullback \(\alpha^* \omega\) is given by:

\begin{equation*} \alpha^* \omega = \left( f(\alpha(t)) \frac{dx}{dt} + g(\alpha(t)) \frac{dy}{dt} + h(\alpha(t)) \frac{dz}{dt} \right) \ dt. \end{equation*}

Consider the counterclockwise parametrization of the unit circle introduced in Example 3.2.3, given by the function \(\alpha:[0,2\pi] \to \mathbb{R}^2\) with \(\alpha(\theta) = (\cos \theta, \sin \theta).\) Let \(\omega = x^2 y\ dx + e^y\ dy\) be a one-form on \(\mathbb{R}^2\text{.}\) We can pull it back along the parametrized unit circle to get a new one-form on \(\mathbb{R}\text{:}\)

\begin{align*} \alpha^* \omega =\amp \left( (\cos \theta)^2 \sin \theta \frac{d}{d \theta} \cos \theta + e^{\sin \theta} \frac{d}{d \theta} \sin \theta \right) \ d\theta \\ =\amp \left( - \cos^2 \theta \sin^2 \theta + \cos \theta e^{\sin \theta} \right)\ d \theta. \end{align*}

As is becoming customary, we can translate between the language of differential forms and the language of vector fields. If \(\mathbf{F}\) is the vector field associated to \(\omega\text{,}\) and \(\alpha\) is a parametric curve in \(\mathbb{R}^n\text{,}\) then the pullback one-form \(\alpha^* \omega\) can be written as

\begin{equation*} \alpha^* \omega = \left( \mathbf{F}(\alpha(t) ) \cdot \mathbf{T}(t) \right) \ dt, \end{equation*}

where \(\mathbf{T}(t)\) is the tangent vector to the parametric curve, and \(\cdot\) denotes the dot product of vectors.

Subsection 3.3.2 The definition of line integrals

We are now ready to define the integral of a one-form along a parametric curve: we pull back the one-form to the interval \([a,b]\) and integrate.

Definition 3.3.2. (Oriented) line integrals.

Let \(\omega\) be a one-form on an open subset \(U\) of \(\mathbb{R}^n\text{,}\) and let \(\alpha: [a,b] \to \mathbb{R}^n\) be a parametric curve whose image \(C \subset U\text{.}\) We define the line integral of \(\omega\) along \(\alpha\) as follows:

\begin{equation*} \int_\alpha \omega = \int_{[a,b]} \alpha^* \omega, \end{equation*}

where the integral on the right-hand-side is defined in Definition 3.1.1. 1 

In particular, the integral on the right-hand-side uses the canonical orientation on the interval \([a,b]\text{,}\) which is consistent with the orientation on the image curve \(C\) induced by the parametrization.

Explicitly, focusing on \(\mathbb{R}^2\) for simplicity, if \(\omega = f(x,y)\ dx + g(x,y)\ dy\text{,}\) and \(\alpha(t) = (x(t), y(t))\text{,}\) the integral reads:

\begin{equation*} \int_\alpha \omega = \int_a^b \left( f(\alpha(t) )\frac{dx}{dt} + g(\alpha(t)) \frac{dy}{dt} \right)\ dt. \end{equation*}

A similar expression of course holds in \(\mathbb{R}^3\) as well.

Such integrals are also called work integrals because of physical applications, as we will see in Section 3.5.

What is neat is that on the right-hand-side, we end up with a one-variable definite integral, which we certainly know how to integrate from Calculus I and II. So we have reduced the problem of computing integrals of one-forms along curves to standard definite integrals!

Consider the one-form \(\omega = y\ dx + \cos(x) \ dy\text{,}\) and the parametric curve \(\alpha:[0,1] \to \mathbb{R}^2\) with \(\alpha(t) = (t, t^2)\text{.}\)

First, for completeness we check that the parametric curve is well defined, according to Definition 3.2.1. \(\alpha\) is a smooth function on \(\mathbb{R}\text{;}\) \(\alpha'(t) = (1, 2t)\) which is never zero on \(\mathbb{R}\text{,}\) so in particular never zero on \([0,1]\text{;}\) \(\alpha\) is injective over \([0,1]\text{.}\) Thus \(\alpha\) defines a smooth parametric curve, and its image \(C = \alpha([0,1])\) has two boundary points at \(\alpha(0) = (0,0) \in \mathbb{R}^2\) and \(\alpha(1) = (1,1) \in \mathbb{R}^2\text{.}\) In fact, it is not difficult to see that \(\alpha\) is a parametrization of the parabola \(y-x^2 = 0\) between \((0,0)\) and \((1,1)\text{.}\)

We can compute the integral of \(\omega\) over the parametric curve \(\alpha\text{.}\) We get:

\begin{align*} \int_\alpha \omega =\amp \int_{[0,1]} \alpha^* \omega\\ =\amp \int_0^1 \left(f(x(t), y(t)) \frac{dx}{dt} + g(x(t),y(t)) \frac{dy}{dt} \right) \ dt\\ =\amp \int_0^1 \left( (t^2) (1) + \cos(t) (2t) \right)\ dt \\ =\amp \left( \frac{t^3}{3} + 2 (t \sin(t) + \cos(t) ) \right) \Big|_0^1 \\ =\amp \frac{1}{3} + 2 \sin(1) + 2 \cos(1) - 2\\ =\amp - \frac{5}{3} + 2 \sin(1) + 2 \cos(1). \end{align*}

Remark 3.3.4. Line integrals over piecewise parametric curves.

We note that we can easily generalize the definition of line integrals to piecewise parametric curves, as in Subsection 3.2.5. If the parametric curve is defined as a union of parametric curves, then to integrate along the curve we simply add up the integrals over the pieces.

Subsection 3.3.3 Reparametrization-invariance and orientability of line integrals

We defined line integrals in terms of a parametric curve \(\alpha: [a,b] \to \mathbb{R}^n\text{,}\) but in the end we would like the integral to be defined intrinsically in terms of the geometry of the image curve and a choice of orientation. To show that this is the case, we now show that our definition is invariant under orientation-preserving reparametrizations. We also show that it changes sign under orientation-reversing reparametrizations, which shows that line integrals are in fact oriented, as we want.

To prove this statement, let us first rewrite the integrals as integrals over intervals, using Definition 3.3.2. The integral on the left-hand-side is:

\begin{equation*} \int_\alpha \omega = \int_{[a,b]} \alpha^* \omega. \end{equation*}

As for the second integral, we are integrating over the parametric curve \(\alpha \circ \phi: [c,d] \to [a,b] \to \mathbb{R}^n\text{.}\) So we can write

\begin{equation*} \int_{\phi^* \alpha} \omega = \int_{[c,d]} (\alpha \circ \phi)^* \omega. \end{equation*}

However, from Exercise 2.4.3.6 we know that pulling back through the chain of maps \([c,d] \overset{\phi}{\to} [a,b] \overset{\alpha}{\to} \mathbb{R}^n\) is the same thing as doing it in two steps: first pulling back via \(\alpha\text{,}\) and then via \(\phi\text{.}\) In other words, \((\alpha \circ \phi)^* \omega = \phi^* ( \alpha^* \omega)\text{,}\) and we can write

\begin{equation*} \int_{\phi^* \alpha} \omega = \int_{[c,d]} \phi^* (\alpha^* \omega). \end{equation*}

The statement then is about the relation between \(\int_{[a,b]} \alpha^* \omega\) and \(\int_{[c,d]} \phi^* (\alpha^* \omega)\text{.}\) But we already studied such questions before in Lemma 3.1.5 and Lemma 3.1.6! Indeed, now that we wrote everything in terms of integrals of one-forms over intervals, we are back in the realm of Section 3.1. From Lemma 3.1.5 we know that the two integrals will be equal if \(\phi(c) = a\) and \(\phi(d) = b\text{,}\) i.e. \(\phi\) preserves the order of the endpoints of the interval (recall that invariance in this case is simply the statement of the substitution formula for definite integrals). If instead \(\phi\) exchanges the order of the endpoints (i.e. \(\phi(c) = b\) and \(\phi(d)=a\)), then by Lemma 3.1.6 the integrals differ by a sign.

But if \(\phi^* \alpha\) is orientation-preserving, then \(\phi'(u) > 0\) by Lemma 3.2.11, and so \(\phi\) is a strictly increasing function, which means that it must map \(c \to a\) and \(d \to b\text{,}\) as \(c \leq d\) and \(a \leq b\text{.}\) Therefore the integrals are equal. While if \(\phi^* \alpha\) is orientation-reversing, then \(\phi'(u) \lt 0\text{,}\) and so \(\phi \) is a strictly decreasing function, which means that it must map \(c \to b\) and \(a \to d\text{.}\) Therefore the integrals differ by a sign. This completes the proof of the lemma.

Great, our two principles of orientability and reparametrization-invariance are fulfilled for line integrals!

Let us consider the integral from Example 3.3.3 again. We consider the one-form \(\omega = y\ dx + \cos(x) \ dy\text{,}\) and the parametric curve \(\alpha:[0,1] \to \mathbb{R}^2\) with \(\alpha(t) = (t, t^2)\text{.}\)

Let us define two new parametrizations for the same curve. First, we define \(\beta: [0,\ln(2)] \to \mathbb{R}^2\) with \(\beta(t) = (e^t-1, (e^t-1)^2)\text{,}\) and \(\gamma: [-1,0] \to \mathbb{R}^2\) with \(\gamma(t) = (-t, t^2)\text{.}\) It is easy that \(\alpha, \beta\) and \(\gamma\) are all parametrizations of the parabola \(y - x^2 = 0\) between \((0,0)\) and \((1,1)\text{.}\)

Looking at the tangent vectors, we get \(\alpha'(t) = (1, 2t)\text{,}\) \(\beta'(t) = (e^t, 2 (e^t-1) e^t) \text{,}\) and \(\gamma'(t) =(-1, 2t)\text{.}\) Are those all inducing the same orientation on the curve? Let us look at the direction of the tangent vector at \((0,0)\text{,}\) which corresponds to \(t=0\) for all three parametrizations. We have \(\alpha'(0) = (1,0)\text{,}\) \(\beta'(0) = (1,0)\text{,}\) and \(\gamma'(0) = (-1,0)\text{.}\) So \(\alpha\) and \(\beta\) induce the same orientation on the curve (from \((0,0)\) to \((1,1)\)), while \(\gamma\) induces the opposite orientation. Therefore, we expect the integral of \(\omega\) over \(\alpha\) and \(\beta\) to both give the same answer, while the integral over \(\gamma\) should pick a sign.

Let us do the calculation for fun. The integral over \(\alpha\) was already performed in Example 3.3.3. For \(\beta\text{,}\) we get:

\begin{align*} \int_\beta \omega =\amp \int_{[0,\ln(2)]} \beta^* \omega\\ =\amp \int_0^{\ln(2)} \left(f(x(t), y(t)) \frac{dx}{dt} + g(x(t),y(t)) \frac{dy}{dt} \right) \ dt\\ =\amp \int_0^{\ln(2)} \left( (e^t-1)^2 e^t + 2 e^t (e^t-1) \cos(e^t-1) \right)\ dt \\ =\amp \int_0^1 \left( u^2 + 2 u \cos(u) \right)\ du, \end{align*}

where we did the substitution \(u = e^t-1\text{.}\) But this is the same integral as in Example 3.3.3, so we get the same result indeed.

As for \(\gamma\text{,}\) we get:

\begin{align*} \int_\gamma \omega =\amp \int_{[-1,0]} \gamma^* \omega\\ =\amp \int_{-1}^{0} \left(f(x(t), y(t)) \frac{dx}{dt} + g(x(t),y(t)) \frac{dy}{dt} \right) \ dt\\ =\amp \int_{-1}^{0} \left( - t^2 + 2 t \cos(-t) \right)\ dt \\ =\amp \int_1^0 \left( u^2 + 2 u \cos(u) \right)\ du, \end{align*}

where we did the substitution \(u=-t\text{.}\) By exchanging the limits of integration, we see that this is minus the integral in Example 3.3.3, and thus we get a minus sign as expected.

Subsection 3.3.4 Line integrals in terms of vector fields

Now that we know how to integrate one-forms along curves, we can translate the definition in terms of the associated vector fields. This is straightforward, since we saw in Subsection 3.3.1 how to rephrase the pullback of a one-form along a curve in terms of the associated vector field.

This is how line integrals, or “work integrals”, are generally defined in standard vector calculus textbooks. 2  The integrand is justified in the context of the calculation of work in physics (as we will see in Section 3.5, such line integrals can be used to calculate work), but it is not clear why taking the dot product between the vector field and the tangent vector to the parametric curve is the right thing to do in general. In our context, the integrand arises naturally by pulling back the one-form in order to be able to integrate over an interval. So it gives a natural geometric interpretation to these work integrals.

Note that the symbol \(\mathbf{T}(t)\) is sometimes used to denote the normalized or unit tangent vector (i.e. our tangent vector divided by its norm \(|\mathbf{T}(t)|\)), in which case \(dt\) should be replaced by \(ds = |\mathbf{T}(t)|\ dt\text{.}\)

Remark 3.3.8.

In standard vector calculus textbooks, such as CLP4, the following notation is often used. Instead of writing \(\alpha: [a,b] \to \mathbb{R}^n\) for the parametric curve, the symbol \(\mathbf{r} = \mathbf{r}(t)\) is often used, with \(a \leq t \leq b\text{,}\) as if it was the position function of an object moving along the image curve \(C\text{.}\) Then the tangent (or velocity) vector is written as

\begin{equation*} \mathbf{T} = \frac{d \mathbf{r}}{d t}\text{,} \end{equation*}

standing for the velocity of the object moving along the curve, and the notation

\begin{equation*} d \mathbf{r} = \frac{d \mathbf{r}}{d t}\ dt \end{equation*}

is used. With this notation, one can rewrite the line integral of a vector field along the curve as

\begin{equation*} \int_C \mathbf{F} \cdot d \mathbf{r}. \end{equation*}

The notation makes sense, as we know that the integral is invariant under orientation-preserving reparametrizations, so we can rewrite it in terms of the image curve \(C\) itself, with the orientation specified by the direction of travel along the curve. With this notation however one needs to keep in mind that to evaluate the integral we need to compose the vector field with the parametrization to rewrite \(\mathbf{F}\) as a function of the parameter \(t\) before integrating in \(t\) over the interval \([a,b]\text{.}\)

Exercises 3.3.5 Exercises

1.

Consider the one-form \(\omega = x y\ dx + z^2\ dy + z\ dz\) on \(\mathbb{R}^3\text{.}\) Find its pullback along the helix centered around the \(z\)-axis and with tangent vector \(\mathbf{T}(t) = (-3 \sin t, 3 \cos t, 4)\text{,}\) \(0 \leq t \leq 2 \pi\text{,}\) and initial position \(\mathbf{r}(0) = (3,0,1)\text{.}\)

Solution.

First, we find a parametrization for the helix. Integrating the tangent vector, we know that the parametrization must be given by \(\alpha:[0,2 \pi] \to \mathbb{R}^3\) with \(\alpha(t) = (3 \cos t + A, 3 \sin t + B, 4 t + C)\) for some constants \(A,B,C\text{.}\) Using the fact that \(\alpha(0) = (3,0,1)\text{,}\) we conclude that \(A=0\text{,}\) \(B=0\text{,}\) and \(C = 1\text{.}\) Thus the parametrization is \(\alpha(t) = (3 \cos t, 3 \sin t, 4 t + 1)\text{.}\)

We can then calculate the pullback \(\alpha^* \omega\text{.}\) We get:

\begin{align*} \alpha^* \omega =\amp \left( (3 \cos t)(3 \sin t) (-3 \sin t) + (4t+1)^2 (3 \cos t) + (4 t + 1) (4) \right) \ dt\\ =\amp \left( - 27 \cos t \sin^2 t + 3 (4t+1)^2 \cos t + 4 (4t+1) \right)\ dt. \end{align*}

2.

Consider the one-form \(\omega = \frac{x}{x^2+y^2}\ dx + \frac{y}{x^2+y^2}\ dy + z\ dz\text{.}\) Explain why you cannot pull it back along the parametric curve \(\alpha:[0,2] \to \mathbb{R}^3\) with \(\alpha(t) = (t-1, t-1, t^2)\text{.}\)

Solution.

The key here is to be careful about the domain of definition of the one-form \(\omega\text{.}\) The largest open subset \(U \subseteq \mathbb{R}^3\) over which \(\omega\) is defined is all points in \(\mathbb{R}^3\) such that \(x^2+y^2 \neq 0\text{,}\) so that the denominators in the component functions do not vanish. But \(x^2 + y^2 = 0\) if and only if \((x,y) = (0,0)\text{,}\) so the domain of definition of \(\omega\) is \(U = \{ (x,y,z) \in \mathbb{R}^3~|~ (x,y) \neq (0,0) \}\text{.}\) In other words, it consists of all points in \(\mathbb{R}^3\) except the points on the \(z\)-axis.

To be able to pull back our one-form along the parametric curve \(\alpha\text{,}\) we must make sure that its image \(C = \alpha([0,2])\) lies within \(U\text{.}\) Unfortunately, we see that at \(t=1\text{,}\) \(\alpha(1) = (0,0,1)\text{,}\) so the parametric curve intersects the \(z\)-axis, which is not in \(U\text{!}\) Thus we can't pull back along \(\alpha\text{.}\)

Just for fun, let's see what would happen if we had naively tried to pull back along \(\alpha\text{.}\) What we would get is:

\begin{equation*} \alpha^* \omega = \left( \frac{t-1}{2(t-1)^2} + \frac{t-1}{2(t-1)^2} + t^2 (2t) \right) \ dt = \left(\frac{1}{t-1} + 2 t^3 \right)\ dt. \end{equation*}

The problem is that this is not defined at \(t=1\text{,}\) which is part of the interval \([0,2]\) for the parametric curve. In other words, the result of the pullback is not actually a well defined one-form on an open subset containing \([0,2]\text{,}\) since it is not defined at \(t=1\text{.}\)

3.

Find the integral of the one-form \(\omega = x \ dx + x \ dy + y \ dz\) in \(\mathbb{R}^3\) along one turn clockwise around the circle of radius two in the \(xy\)-plane and centered at the origin.

Solution.

The circle has equation \(x^2+y^2 = 4\text{.}\) We parametrize one turn clockwise around the circle with \(\alpha:[0,2 \pi] \to \mathbb{R}^3\text{,}\) \(\alpha(t) = (2 \sin t, 2 \cos t, 0)\) (recall that we are in \(\mathbb{R}^3\)). The pullback of the one-form is

\begin{equation*} \alpha^* \omega = \left( (2 \sin t) (2 \cos t) + (2 \sin t) (- 2 \sin t) + (2 \cos t) (0) \right)\ dt = 2 \left(\sin(2 t) + \cos(2t) - 1\right) \ dt. \end{equation*}

The line integral becomes

\begin{align*} \int_{[0,2 \pi]} \alpha^* \omega =\amp 2 \int_0^{2 \pi} \left(\sin(2t)+\cos(2t)-1 \right) dt \\ =\amp (- \cos(4\pi) + \cos(0) ) +(\sin(4 \pi) - \sin(0) ) - 2 (2 \pi)\\ =\amp - 4 \pi. \end{align*}

4.

Consider the vector field \(\mathbf{F}(x,y) = (e^y, e^x)\) in \(\mathbb{R}^2\text{.}\) Find its line integral along the oriented curve obtained by first moving from the origin to the point \((0,1)\text{,}\) and then from \((0,1)\) to the point \((4,0)\) along straight lines.

Solution.

This is a piecewise parametric curve, so we need to split it into line segments. For the first segment from \((0,0)\) to \((0,1)\text{,}\) we can write a parametrization as \(\alpha_1:[0,1] \to \mathbb{R}^2\) with \(\alpha_1(t) = (0,t)\text{.}\) For the second line segment from \((0,1)\) to \((0,4)\text{,}\) we write a parametrization as \(\alpha_2:[0,1] \to \mathbb{R}^2\) with \(\alpha_2(t) = (4 t, 1-t)\text{.}\) The tangent vectors are:

\begin{equation*} \mathbf{T}_1(t) = (0,1), \qquad \mathbf{T}_2(t) = (4, -1). \end{equation*}

We thus get:

\begin{equation*} \mathbf{F}(x(t), y(t)) \cdot \mathbf{T}_1(t) = 1, \qquad \mathbf{F}(x(t),y(t)) \cdot \mathbf{T}_2(t) = 4 e^{1-t} - e^{4 t}. \end{equation*}

The line integral becomes:

\begin{align*} \int_{[0,1]}\amp \mathbf{F}(x(t), y(t)) \cdot \mathbf{T}_1(t) \ dt + \int_{[0,1]} \mathbf{F}(x(t),y(t)) \cdot \mathbf{T}_2(t) \ dt \\ =\amp \int_0^1 \ dt + \int_0^1 \left( 4 e^{1-t} - e^{4 t} \right)\ dt\\ =\amp 1 + \left( - 4 e^{1-t} - \frac{1}{4} e^{4 t} \right)_{0}^1\\ =\amp 1 - 4 - \frac{1}{4} e^4 + 4 e + \frac{1}{4}\\ =\amp 4 e - \frac{1}{4} e^4 - \frac{11}{4}. \end{align*}

5.

Let \(C\) be the curve from \((0,0,0)\) to \((1,1,1)\) along the intersection of the surfaces \(y = x^2\) and \(z = x^3\text{.}\) Find the integral of the vector field \(\mathbf{F}(x,y,z) = (x^2, xy, z^2)\) along this curve.

Solution.

We first need to parametrize the curve. A point in \(\mathbb{R}^3\) on the surface \(y=x^2\) has coordinates \((t,t^2,z)\) for \((z,t) \in \mathbb{R}^2\text{.}\) At the intersection with the surface \(z = x^3\text{,}\) we must also have \(z = t^3\text{.}\) It then follows that points on the intersection of the two surfaces have coordinates \((t,t^2,t^3)\) with \(t \in \mathbb{R}\text{.}\) Now we want our curve to start at \((0,0,0)\) and end at \((1,1,1)\text{.}\) So our parameter must go from \(t=0\) to \(t=1\text{.}\) We thus end up with the parametrization

\begin{equation*} \alpha:[0,1] \to \mathbb{R}^3, \qquad \alpha(t) = (t, t^2, t^3). \end{equation*}

The tangent vector is \(\mathbf{T}(t) = (1, 2t, 3 t^2)\text{.}\) The line integral becomes:

\begin{align*} \int_C \mathbf{F} \cdot \mathbf{T}\ dt =\amp \int_0^1 \left( (t^2, t^3, t^6) \cdot (1, 2t, 3 t^2) \right)\ dt \\ =\amp \int_0^1 \left( t^2 + 2 t^4 + 3 t^8 \right)\ dt \\ =\amp \frac{1}{3} + \frac{2}{5} + \frac{3}{9} \\ =\amp \frac{16}{15}. \end{align*}