We started in Calculus 1 with integrals over an interval, then earlier in this course, we studied integrals over regions in the xy plane and over volumes. Now we'll see what we call **line integrals**, which are integrals along a path in three-dimensional space (later, we'll see integrals over surfaces).

Here's the idea: consider a surface in 3-D space, defined by \(z=f(x,y)\). Suppose there's a path along this surface, as shown below (pretend that there's a surface extending on both sides of the path). The curve \(C\) is the projection of this path onto the xy plane.

The line integral \[\int_C f(x,y) \ ds\] gives the area of this "curtain" under the curve.

To find \(ds\), zoom in a segment of \(C\) in the xy plane:

Since \(ds=\sqrt{dx^2+dy^2}\), \[\dfrac{ds}{dt} = \sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2} \longrightarrow ds = \sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\ dt.\] If \(\vec{r}(t) = \langle x(t),y(t) \rangle\), we can write this as \(ds = |\vec{r}\ '(t)| \ dt\), so the line integral becomes

\[\int_C f(x,y)\ ds = \int_C f\big(x(t),y(t)\big) \ |\vec{r}\ '(t)| \ dt.\]Evaluate the line integral \(\int_C 2+x^2y\ ds\), where \(C\) is the upper half of the unit circle \(x^2+y^2 = 1\).

Since we evaluate a line integral in terms of \(t\), we need to parametrize \(C\). We can do it as follows:

\[\begin{align} x &= \cos t\\ y &= \sqrt{1-x^2} = \sqrt{1-\cos^2 t} = \sin t \end{align}\]Since this is the upper half of the circle, \(0 \leq t \leq \pi\).

Then \(\vec{r}\ '(t) = \langle -\sin t, \cos t \rangle\), so \(|\vec{r}\ '(t)| = \sqrt{\sin^2 t + \cos^2 t} = 1\). Therefore, the line integral becomes

\[\begin{align} \int_0^\pi &2+\cos^2 t\ \sin t \ dt\\ &= 2t-\dfrac{1}{3}\cos^3 t\bigg|_0^\pi \ \ \ \ \textrm{ by u-substitution}\\ &= \ans{2\pi + \dfrac{2}{3}} \end{align}\]What if we parameterized the curve differently (recall that parameterization isn't unique)? It would be simpler to parameterize it as follows, but we'll find that the integral is tricker:

\[\begin{align} x &= t\\ y &= \sqrt{1-t^2} \end{align}\]In this case, \(-1 \leq t \leq 1\):

Then \(\vec{r}\ '(t) = \langle 1,\dfrac{-t}{\sqrt{1-t^2}} \rangle\), so \[|\vec{r}\ '(t)| = \sqrt{1+\left(\dfrac{-t}{\sqrt{1-t^2}}\right)^2} = \dfrac{1}{\sqrt{1-t^2}}.\] Therefore, the line integral becomes

\[\begin{align} \int_C f(x,y)\ ds &= \int_{-1}^1 (2+t^2\sqrt{1-t^2}) \cdot \dfrac{1}{\sqrt{1-t^2}}\ dt\\ &= \int_{-1}^1 \dfrac{2}{\sqrt{1-t^2}}+t^2 \ dt \end{align}\]To integrate this, we need to use trigonometric substitution: if we define \(t=\sin \theta\), then \(dt = \cos \theta \ d\theta\).

\[\begin{align} \int \dfrac{2}{\sqrt{1-t^2}}\ dt &= \int \dfrac{2}{\cos \theta} \cdot \cos \theta\ d\theta\\ &= 2\theta = 2\sin^{-1} t \end{align}\]Going back to the line integral:

\[\begin{align} \int_{-1}^1 &\dfrac{2}{\sqrt{1-t^2}} + t^2 \ dt\\ &= 2\sin^{-1} t+\dfrac{1}{3}t^3\bigg|_{-1}^1\\ &= \ans{2\pi + \dfrac{2}{3}} \end{align}\]Notice, of course, that we get the same answer in both cases; one had a simpler parameterization and an easier integral, and vice versa.

- Evaluate the integral \(\displaystyle\int_C x^2+y^2\ ds\), where \(C\) is the circle of radius 4 centered at the origin.
- Evaluate the integral \(\displaystyle\int_C \dfrac{x}{x^2+y^2}\ ds\), where \(C\) is the line segment from \((1,1)\) to \((10,10)\).

\(\displaystyle\int_0^{2\pi} 16 \cdot 4\ dt = 128\pi\)

\(\displaystyle\int_1^{10} \dfrac{1}{2t} \cdot \sqrt{2}\ dt = \dfrac{\sqrt{2}}{2}\ln 10\)

A line integral in three dimensions is similar: \[\int_C f\big(x(t),y(t),z(t)\big) |\vec{r}\ '(t)|\ dt,\] where \(\vec{r}(t) = \langle x(t),y(t),z(t) \rangle\).

Evaluate the line integral \(\int_C xy+2z\ ds\), where \(C\) is the line segment from \(P(1,0,0)\) to \(Q(0,1,1)\).

Start by parameterizing the line segment \(C\): \[r(t) = \langle 1-t,t,t \rangle, \ \ \ \ 0 \leq t \leq 1\] Then \(\vec{r}\ '(t) = \langle -1,1,1 \rangle\), so \(|\vec{r}\ '(t)| = \sqrt{3}\).

Rewriting the line integral in terms of t:

\[\begin{align} \int_C xy+2z\ ds &= \int_0^1 \big((1-t)t+2t\big)\sqrt{3}\ dt\\ &= \sqrt{3} \int_0^1 3t-t^2\ dt\\ &= \sqrt{3} \bigg[\dfrac{3}{2}t^2-\dfrac{1}{3}t^3 \bigg]_0^1\\ &= \ans{\dfrac{7\sqrt{3}}{6}} \end{align}\]Evaluate the integral \(\displaystyle\int_C xyz\ ds\), where \(C\) is the line segment from \((0,0,0)\) to \((1,2,3)\).

\(\dfrac{3\sqrt{14}}{2}\)

We can use line integrals to calculate arc length, when the function is \(f(x,y) = 1\) or \(f(x,y,z)=1\):

\[\textrm{Arc Length } = \int_C\ ds = \int_C |\vec{r}\ '(t)|\ dt\]The flight of an eagle is given by \[\vec{r}(t) = \langle 2400 \cos \dfrac{t}{2}, 2400 \sin \dfrac{t}{2}, 500t \rangle\] in ft. How far does the eagle travel over \(0 \leq t \leq 10\)?

Note that the magnitude of the velocity vector is 1300, meaning that the eagle traveled at 1300 ft/min for 10 minutes, for a total of 13,000 ft.

Find the length of the curve \(\vec{r}(t) = \left\langle 20 \sin \dfrac{t}{4}, 20\cos \dfrac{t}{4}, \dfrac{t}{2} \right\rangle\) for \(0 \leq t \leq 2\).

\(\sqrt{101}\)

A line integral arises when we describe the work done on an object moving along a path \(C\) in a force field.

At every point in this force field, the object feels a force, but only the tangential component of the force does any work: \[|\vec{F}_T| = |\vec{F}| \cos \theta,\] where \(\theta\) is the angle between \(\vec{F}\) and \(\vec{F}_T\).

\[\begin{align} \vec{F}_T = \vec{F} \cdot \vec{T} \ \ \ \ &: \textrm{ force (where } \vec{T} \textrm{ is the unit tangent vector)}\\ ds = |\vec{r}\ '(t)| \ dt \ \ \ &: \textrm{ distance traveled along the curve} \end{align}\]Recall that work is the product of force and distance, so

\[\begin{align} W &= \int_C \vec{F} \cdot \vec{T}\ ds\\ &= \int_a^b \vec{F} \cdot \dfrac{\vec{r}\ '(t)}{|\vec{r}\ '(t)|} |\vec{r}\ '(t)|\ dt \end{align}\]Simplifying:

\[\ans{W = \int_a^b \vec{F} \cdot \vec{r}\ '(t)\ dt}\]Note that work is a *scalar*.

An electric force field is described by \[\vec{F} = \dfrac{k \langle x,y,z \rangle}{(x^2+y^2+z^2)^{3/2}},\] where \(k\) is a constant. Suppose that \(k=1\) and find the work done by this force field in moving a particle from \((1,1,1)\) to \((3,3,3)\).

Start by parameterizing the line segment \(C\): \[r(t) = \langle 1+t,1+t,1+t \rangle, \ \ \ \ 0 \leq t \leq 2\] Then \(\vec{r}\ '(t) = \langle 1,1,1 \rangle\) and the force is written in parametric form as \[\vec{F} = \dfrac{1}{\big(3(1+t^2)\big)^{3/2}} \langle 1+t,1+t,1+t \rangle,\] so \[\vec{F} \cdot \vec{r}\ '(t) = \dfrac{3(1+t)}{\sqrt{27}(1+t)^3} = \dfrac{1}{\sqrt{3}(1+t)^2}.\]

The work, therefore, is

\[\begin{align} W &= \int_a^b \vec{F} \cdot \vec{r}\ '(t)\ dt\\ &= \int_0^2 \dfrac{1}{\sqrt{3}(1+t)^2}\ dt\\ &= -\dfrac{1}{\sqrt{3}(1+t)} \bigg|_0^2 = \ans{\dfrac{2}{3\sqrt{3}} = \dfrac{2\sqrt{3}}{9}} \end{align}\]Find the work done by moving an object from \((0,0)\) to \((2,8)\) along the parabola \(y=2x^2\) if \(\vec{F} = \langle y,x \rangle\).

\(W = \displaystyle\int_0^2 6t^2\ dt = 16\)