“Line” Integrals of Functions

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Integrals in one variable (Riemann Integrals)

In Calc 1 you learn the definition of the definite integral of a function \(f:\mathbb{R} \to \mathbb{R}\) over an interval from \(a\) to \(b\):

\[\int_a^b f(x)\; dx\]

is defined as the signed area between the graph of \(f\) and the \(x\)-axis, starting at \(x=a\) and going to \(x = b\).

We can visualize the same region embedded into \(\mathbb{R}^3\), in the \(xz\)-plane:

“Line” integrals of function of two variables

We are now very close to the idea of a “line” integral in two variables. The Riemann integral above was defined as the area of the vertical surface between the \(x\)-axis and the graph of the function \(f\), with the surface intersection the \(x\)-axis forming the interval \([a,b]\).

Now suppose we have a planar curve \(C\) inside the \(xy\)-plane, and a function \(f:\mathbb{R}^2 \to \mathbb{R}\) whose domain contains the curve \(C\). We can now form a vertical surface whose intersection with the \(xy\)-plane will be the curve \(C\), and which will extend between the \(xy\)-plane and the surface \(z=f(x,y)\). The signed area (the parts of the area under the \(xy\)-plane will be negative) of the surface will be the line integral of the function \(f\) over the curve \(C\):

\[\int_C f(x,y)\;ds\]

The \(ds\) indicates that this is a line integral with respect to the arc length. There are also two other types of line integrals: with respect to \(x\) and with respect to \(y\).

Line integral with respect to \(x\)

The basic idea of a line integral with respect to \(x\) is simple: we are still looking at the area of the same surface as before, but now we perceive the surface as if we saw it from a direction parallel to the \(y\)-axis: we ignore any changes in the “depth” of the surface, or the \(y\) coordinate, and only look at the changes in \(x\). It is almost like an area of the projection of the surface onto the \(xz\)-plane, with one complication – if you imagine a point traveling along the curve \(C\) from one end to the other, and just watch the \(x\)-coordinate, it may happen that for some parts of the curve, as the point travels, the \(x\)-coordinate will decrease.

The illustration below shows a curve where the parts of the curve where the \(x\)-coordinate increases are shown in blue, while the part where the \(x\)-coordinate decreases are shown in red.

(Of course it depends on the direction in which the curve is traveled. If we change the direction, the pieces of curve will get switched, the red ones will be blue and the blue ones will be red.)

The following illustration shows the projection of the line integral surface onto the \(xz\)-plane, with the areas with decreasing \(x\)-coordinate colored red, and areas with increasing \(x\)-coordinate colored blue.

If we add all the blue areas and subtract all the red areas, we get the line integral of \(f\) over the curve \(C\) with respect to \(x\):

\[\int_C f(x,y)\; dx\]

Of course these areas are again signed areas, which means that for areas under the \(xy\)-plane, the sign will be switched. Also, the integral depends on the direction in which the curve is traversed. If we take the same curve and travel through it in the opposite direction, we will get the opposite integral. A curve that is the same as \(C\) but traveled in the opposite direction is sometimes denoted as \(-C\). With this notation,

\[\int_{-C} f(x,y)\;dx = - \int_C f(x,y)\; dx\]

This is unlike the integral with respect to the arc length which does not depend on the direction:

\[\int_{-C} f(x,y)\;ds = \int_C f(x,y)\; ds\]

Line integral with respect to \(y\)

Instead of projecting the line integral surface onto the \(xz\)-plane, we can instead project it onto the \(yz\)-plane, and get the line integral with respect to \(y\). The idea is exactly the same as for the integrals with respect to \(x\), except for the direction of projection. The areas projected onto the \(yz\)-plane are added or subtracted depending on whether the \(y\)-coordinate is increasing or decreasing.

In the illustration below, the \(y\)-coordinate is always increasing, and the line integral is simply the area of the projection in the \(yz\)-plane.

Line integrals over parametric curves

Suppose the curve \(C\) is parametrized by a vector function \(\mathbf{r}(t) = \left\langle x(t), y(t)\right\rangle\), \(a \le t \le b\), where \(\mathbf{r}\) is a differentiable vector function such that \(\left\lvert\mathbf{r}'(t) \right\rvert \neq 0\) for any \(t \in (a,b)\). Then we can use the following substitution to convert the line integral into a single variable integral over the interval \([a,b]\):

\[\begin{aligned} x &= x(t)\\ y &= y(t)\\ dx &= x'(t)\;dt\\ dy &= y'(t)\;dt\\ ds &= \left\lvert\mathbf{r}'(t) \right\rvert dt = \sqrt{(x'(t))^2 + (y'(t))^2}\;dt \end{aligned}\]

Then we get

\[\begin{aligned} \int_C f(x,y)\; ds &= \int_a^b f\left(\mathbf{r}(t)\right)\left\lvert\mathbf{r}'(t) \right\rvert dt = \int_a^b f\left(x(t), y(t)\right)\sqrt{(x'(t))^2 + (y'(t))^2}\;dt\\ \int_C f(x,y)\;dx &= \int_a^b f\left(x(t), y(t)\right)x'(t)\;dt\\ \int_C f(x,y)\;dy &= \int_a^b f\left(x(t), y(t)\right)y'(t)\;dt\\ \end{aligned}\]

Examples

A line integral with respect to the arc length:

Calculate

\[\int_C x\;ds\]

where \(C\) is the curve parametrized by \(\mathbf{r}(t) = \left\langle t,t^2\right\rangle\) for \(0 \le t \le 1\).

We will substitute \(x = t\), \(y = t^2\) and \[ds = \left\lvert\mathbf{r}'(t) \right\rvert dt = \sqrt{1 + 4t^2}\;dt\] which will give us

\[\begin{aligned} \int_C x\;ds &= \int_0^1 t\sqrt{1 + 4t^2}\;dt\\ \color{green}{\text{substitute: }} & \color{green}{u = 1 + 4t^2, \quad du = 8t\;dt}\\ &= \frac{1}{8}\int_1^4 \sqrt{u}\;du\\ &= \frac{1}{8}\left(\left.\frac{3}{2}\sqrt{u^3}\right\rvert_{1}^{4}\right)\\ &= \frac{3}{16}\left(8 - 1\right)\\ &= \frac{21}{16} \end{aligned}\]

A line integral with respect to \(x\)

Calculate

\[\int_C x^2 + y^2\;dx\]

where the curve \(C\) is parametrized by \(\mathbf{r}(t) = \left\langle \sin(t), 1-\cos(t)\right\rangle\) for \(0 \le t \le \pi\).

Here

\[\begin{aligned} x &= \sin(t)\\ y &= 1 - \cos(t)\\ dx &= \cos(t)\;dt \end{aligned}\]

and so

\[\begin{aligned} \int_C x^2 + y^2\;dx &= \int_0^{\pi} (\sin^2(t) + 1 - 2\cos(t) + \cos^2(t))\cos(t)\;dt\\ &= 2\int_0^\pi \cos(t) - \cos^2(t)\;dt\\ &= 2\int_0^\pi \cos(t)\;dt - 2\int_0^\pi \cos^2(t)\;dt\\ &= 0 - 2\int_0^\pi \frac{1 + \cos(2t)}{2}\;dt\\ &= 0 - \int_0^\pi \;dt - \int_0^\pi \cos(2t)\;dt\\ &= 0 - \pi - 0\\ &= -\pi \end{aligned}\]

Video on line integrals of functions

Try it yourself

Line integrals in three variables

Given a function of three variables \(f:\mathbb{R}^3 \to \mathbb{R}\) and a curve \(C \subset \mathbb{R}^3\), we can define line integrals of \(f\) over the curve \(C\) the same way as in \(\mathbb{R}^2\). We will end up with four different line integrals:

• Integral with respect to the arc length: \(\displaystyle\int_C f(x,y,z)\;ds\)

• Integral with respect to \(x\): \(\displaystyle\int_C f(x,y,z)\;dx\)

• Integral with respect to \(y\): \(\displaystyle\int_C f(x,y,z)\;dy\)

• Integral with respect to \(z\): \(\displaystyle\int_C f(x,y,z)\;dz\)

Suppose the curve \(C\) is parametrized by a vector function \(\mathbf{r}(t) = \left\langle x(t), y(t), z(t)\right\rangle\), \(a \le t \le b\), where \(\mathbf{r}\) is a differentiable vector function such that \(\left\lvert\mathbf{r}'(t) \right\rvert \neq 0\) for any \(t \in (a,b)\). Then we can use the following substitution to convert the line integral into a single variable integral over the interval \([a,b]\):

\[\begin{aligned} x &= x(t)\\ y &= y(t)\\ z &= z(t)\\ dx &= x'(t)\;dt\\ dy &= y'(t)\;dt\\ dz &= z'(t)\;dt\\ ds &= \left\lvert\mathbf{r}'(t) \right\rvert dt = \sqrt{(x'(t))^2 + (y'(t))^2 + (z'(t))^2}\;dt \end{aligned}\]

Then we get

\[\begin{aligned} \int_C f(x,y,z)\; ds &= \int_a^b f\left(\mathbf{r}(t)\right)\left\lvert\mathbf{r}'(t) \right\rvert dt = \int_a^b f\left(x(t), y(t), z(t)\right)\sqrt{(x'(t))^2 + (y'(t))^2 + (z'(t))^2}\;dt\\ \int_C f(x,y,z)\;dx &= \int_a^b f\left(x(t), y(t), z(t)\right)x'(t)\;dt\\ \int_C f(x,y,z)\;dy &= \int_a^b f\left(x(t), y(t), z(t)\right)y'(t)\;dt\\ \int_C f(x,y,z)\;dz &= \int_a^b f\left(x(t), y(t), z(t)\right)z'(t)\;dt\\ \end{aligned}\]