Scalar and Vector Line Integrals

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Two different questions

A wire with density $\delta$ has mass $\int_C\delta\,ds$ . A force field $\mathbf F$ does work $\int_C\mathbf F\cdot d\mathbf r$ . Both use a curve, but reversing its direction changes work and leaves mass unchanged. Keep the parameter interval with the parametrization.

Visual guide

VISUAL GUIDEWork follows the tangent component

Unit circle

The field F = (−y, x) points counterclockwise around the unit circle. The tangent direction matches it everywhere, producing positive circulation 2π. Reversing the traversal reverses work but not the circle’s length.

Set up the integral

For a piecewise smooth curve $\mathbf r(t)$ , $a\le t\le b$ ,

\int_C f\,ds=\int_a^b f(\mathbf r(t))\|\mathbf r'(t)\|\,dt.

\int_C\mathbf F\cdot d\mathbf r=\int_a^b\mathbf F(\mathbf r(t))\cdot\mathbf r'(t)\,dt.

The first uses speed, a nonnegative scalar. The second retains the direction of the tangent. Split a path with corners into smooth pieces and add their contributions.

Worked example: wire mass

On $\mathbf r(t)=(t,t)$ , $0\le t\le1$ , let $\delta=x+y$ . Then $ds=\sqrt2\,dt$ and $\delta=2t$ . The mass is $\int_0^1 2t\sqrt2\,dt=\sqrt2$ . Substituting the density without the speed factor would give the wrong mass.

PAUSE & THINKA quick check, not a grade

Try it yourself.

Choose an answer and explain your reasoning to yourself. Use a hint if you get stuck.

Reversing a path’s direction changes which integral’s sign?

The scalar length integral ∫ f ds

Both always

The work integral ∫ F·dr

Hint 1 · Find a starting point

dr is oriented; ds is a nonnegative length element.

Hint 2 · Take the next step

Reverse the path’s tangent vector.

Show the reasoning

Answer: The work integral ∫ F·dr

Work changes sign under reversal, while a scalar integral with ds is unchanged.

Before moving on: what would make one of the other answers wrong? Saying why is part of understanding.

Worked example: circulation

Take $\mathbf F=(-y,x)$ on the unit circle $\mathbf r(t)=(\cos t,\sin t)$ , $0\le t\le2\pi$ . The dot product with $\mathbf r'=(-\sin t,\cos t)$ is $1$ . Counterclockwise work is $2\pi$ ; clockwise work is $-2\pi$ . The field always points along the counterclockwise tangent.

Choosing a path

Write down start, end and orientation before computing. A different parametrization of the same oriented curve gives the same integral. A different geometric path can give different work. Path independence requires additional properties of the field, studied next.

Practice

Find the length of the line segment from $(0,0)$ to $(3,4)$ using an integral.
Find the work of $(2,0)$ along that segment.
Reverse the segment. Which answer changes?

Show worked solutions

Use $\mathbf r(t)=(3t,4t)$ on $[0,1]$ . Its speed is $5$ , so the length is $5$ .
The dot product is $6$ , so the work is $6$ .
The length remains $5$ ; the work becomes $-6$ . Scalar arc length has no preferred direction.

Further study

MIT OpenCourseWare: Multivariable Calculus provides a full university course with additional lectures and exercises.

MAKE IT YOURS

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