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17.2 (part 1) Line Integrals of Functions

Line integral: evaluation / accumulation of a function along a curve.

\[ \int_{C} f(x,y) \, ds \]

Where:

  • \( C \): the curve
  • \( ds \): length of a small segment of curve

\( ds \): we get from length:

A sketch of a curve segment from t=a to t=b, with a small portion marked as ds.
\[ \int_{a}^{b} |\vec{r}'(t)| \, dt \]

length of small segment: \( ds = |\vec{r}'(t)| \, dt \)

In fact, length of curve is a special case of line integral with \( f(x,y) = 1 \):

\[ \int_{C} ds = \int_{a}^{b} |\vec{r}'(t)| \, dt \]

In 3D, \[ \int_{C} f(x,y,z) \, ds \]

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The potentially complicated part is the parametrization of curve \( C \).

\( \hookrightarrow \) write an equation for it

Example

\[ \int_{C} xe^y \, ds \]
A 2D Cartesian coordinate system showing a vector/line segment starting at the origin (0,0) and ending at (4,-1).

\( C \): line segment from \( (0,0) \) to \( (4,-1) \)

Parametrize the curve \( C \)

\( \hookrightarrow \) write an equation for it (line here)

\( \vec{r}(t) = \vec{r}_0 + t\vec{v} \)

\( \vec{v} = \langle 4, -1 \rangle \quad \vec{r}_0 = \langle 0, 0 \rangle \)

\( \vec{r}(t) = \langle 0, 0 \rangle + t \langle 4, -1 \rangle \)

\( \vec{r}(t) = \langle 4t, -t \rangle \quad 0 \le t \le 1 \)

\( \vec{r}' = \langle 4, -1 \rangle \)

\( ds = |\vec{r}'(t)| \, dt = |\langle 4, -1 \rangle| \, dt = \sqrt{4^2 + (-1)^2} \, dt = \sqrt{17} \, dt \)

\[ \int_{C} xe^y \, ds = \int_{0}^{1} (4t)e^{-t} \sqrt{17} \, dt = 4\sqrt{17} \int_{0}^{1} te^{-t} \, dt = \dots = \boxed{2\sqrt{17}(e-1)} \]

Note: The final result is boxed in the original notes.

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note \( \vec{r}(t) = \langle 4t, -t \rangle \quad 0 \le t \le 1 \) is not the only way to parametrize \( C \)

also correct:

  • \( \vec{r}(t) = \langle 8t, -2t \rangle \quad 0 \le t \le \frac{1}{2} \)
  • or \( \vec{r}(t) = \langle 2t, -\frac{1}{2}t \rangle \quad 0 \le t \le 2 \)

does \( \int_C f(x,y) ds \) depend on how we parametrize?

let's check, using \( \vec{r}(t) = \langle \underbrace{8t}_{x}, \underbrace{-2t}_{y} \rangle \quad 0 \le t \le \frac{1}{2} \)

\( \vec{r}'(t) = \langle 8, -2 \rangle \quad ds = |\vec{r}'(t)| dt = \sqrt{68} dt \)

\[ \int_C x e^y ds = \int_0^{\frac{1}{2}} 8t e^{-2t} \sqrt{68} dt \]\[ = 8\sqrt{68} \int_0^{\frac{1}{2}} t e^{-2t} dt = \dots = \text{same} \]

we don't expect it to change: if \( f(x,y) = 1 \), then \( \int_C ds = \text{length} \) and should not be affected by choice of parametrization

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example

\( \int_C (x + \sqrt{y}) ds \)

\( C \): from \( (0,0) \) to \( (1,1) \) along \( y = x^2 \)
then from \( (1,1) \) to \( (0,0) \) along \( y = x \)

Coordinate graph showing path C1 along y=x^2 from (0,0) to (1,1) and path C2 along y=x from (1,1) back to (0,0).

we need to parametrize the two portions separately

  • \( C_1: (0,0) \to (1,1) \) along \( y = x^2 \)
  • \( C_2: (1,1) \to (0,0) \) along \( y = x \)

\( C_1 \): \( y = x^2 \)

so if we let \( x = t, \quad 0 \le t \le 1 \)

then \( \vec{r}(t) = \langle t, t^2 \rangle \) describes \( C_1 \)

  • or \( \vec{r}(t) = \langle \sqrt{t}, t \rangle \quad 0 \le t \le 1 \)
  • or \( \vec{r}(t) = \langle 2t, 4t^2 \rangle \quad 0 \le t \le \frac{1}{2} \)

for simplicity, let's choose \( \vec{r}(t) = \langle t, t^2 \rangle \quad 0 \le t \le 1 \) for \( C_1 \)

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Line Integral Parameterization and Calculation

Parameterizing Curve \( C_2 \)

\( C_2 \): from \( (1, 1) \) to \( (0, 0) \) along \( y = x \)

\( \vec{r}(t) = \langle -t, -t \rangle \quad t = 1 \text{ to } t = 0 \)

doesn't work, because we don't start at \( (1, 1) \)

but \( \vec{r}(t) = \langle t, t \rangle \quad t = 1 \text{ to } t = 0 \) does

how about \( \vec{r}(t) = \langle 1-t, 1-t \rangle \quad 0 \le t \le 1 \)

As long as \( y = x \) (curve), and start/end are right, the parametrization should work.

let's go with \( \vec{r}(t) = \langle 1-t, 1-t \rangle \quad 0 \le t \le 1 \) for \( C_2 \)

Setting up the Line Integral

\[ \int_C (x + \sqrt{y}) \, ds = \int_{C_1} (x + \sqrt{y}) \, ds + \int_{C_2} (x + \sqrt{y}) \, ds \]

For \( C_1 \):

\( \vec{r}(t) = \langle t, t^2 \rangle \quad 0 \le t \le 1 \)

\( \vec{r}' = \langle 1, 2t \rangle \)

\( ds = |\vec{r}'| \, dt = \sqrt{1 + 4t^2} \, dt \)

For \( C_2 \):

\( \vec{r}(t) = \langle 1-t, 1-t \rangle \quad 0 \le t \le 1 \)

\( \vec{r}' = \langle -1, -1 \rangle \)

\( ds = \sqrt{2} \, dt \)

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Integral Evaluation

\[ \underbrace{\int_0^1 (t + \sqrt{t^2}) \sqrt{1 + 4t^2} \, dt}_{C_1} + \underbrace{\int_0^1 (1-t + \sqrt{1-t}) \sqrt{2} \, dt}_{C_2} \]

Note for \( C_1 \): \( x = t \), \( y = t^2 \), and \( ds = \sqrt{1 + 4t^2} \, dt \).

\[ = \dots = \frac{1}{6}(5\sqrt{5} - 1) + \frac{7}{3\sqrt{2}} \]
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Line Integral Example

\[ \int_C (x + y + z) \, ds \]

Where \( C \) is the intersection of \( z = x^2 + y^2 \) (paraboloid) and \( x^2 + y^2 = 1 \) (cylinder) going counterclockwise when viewed from above.

The curve \( C \) is a circle of radius 1 (part of the cylinder \( x^2 + y^2 = 1 \)).

At \( z = 1 \) from \( z = x^2 + y^2 \), the circle is radius 1, so \( x^2 + y^2 = 1 \) and \( z = 1 \).

3D plot showing a paraboloid and cylinder intersecting at a circle at height z=1.
2D projection of the unit circle in the xy-plane with counterclockwise orientation arrows.

Now we parametrize this:

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Parametrization Options

\[ \vec{r}(t) = \langle \cos t, \sin t, t \rangle, \quad 0 \le t \le 1 \]

Note: can't have changing z-level

\[ \vec{r}(t) = \langle \cos t, \sin t, 1 \rangle, \quad 0 \le t \le 2\pi \]
\[ \vec{r}(t) = \langle \cos(2\pi t), \sin(2\pi t), 1 \rangle, \quad 0 \le t \le 1 \]
\[ \vec{r}(t) = \langle \cos(\pi t), \sin(\pi t), 1 \rangle, \quad 0 \le t \le 2 \]

Calculation

Using the chosen parametrization:

  • \( x = \cos(2\pi t), \quad y = \sin(2\pi t), \quad z = 1 \)
  • \( \vec{r}' = \langle -2\pi \sin(2\pi t), 2\pi \cos(2\pi t), 0 \rangle \)
  • \( ds = |\vec{r}'| \, dt = \sqrt{4\pi^2 \sin^2(2\pi t) + 4\pi^2 \cos^2(2\pi t)} \, dt = \sqrt{4\pi^2} = 2\pi \, dt \)
\[ \int_C (x + y + z) \, ds = \int_0^1 (\cos(2\pi t) + \sin(2\pi t) + 1)(2\pi) \, dt = \dots = \boxed{2\pi} \]