The Central Idea
Ordinary integration accumulates values along a one-dimensional interval.
A line integral also accumulates along something one-dimensional, but that one-dimensional object may bend through two-dimensional or three-dimensional space.
The curve might be a line segment, circle, helix, or any sufficiently smooth path.
The central strategy is:
\[ \boxed{ \text{curve} \longrightarrow \text{parameter} \longrightarrow \text{ordinary integral} }. \]Curves as Parameterized Paths
A curve \(C\) is commonly represented by a vector-valued function
\[ \mathbf{r}(t) = \langle x(t),y(t),z(t)\rangle, \]where
\[ a\le t\le b. \]As \(t\) moves from \(a\) to \(b\), the point
\[ \mathbf{r}(t) \]traces the curve.
Two-Dimensional Parameterizations
In the plane, we may write
\[ \mathbf{r}(t) = \langle x(t),y(t)\rangle. \]For example, the line segment from \((0,0)\) to \((2,4)\) can be parameterized by
\[ \boxed{ \mathbf{r}(t)=\langle2t,4t\rangle, \qquad 0\le t\le1 }. \]Three-Dimensional Parameterizations
A helix can be parameterized by
\[ \boxed{ \mathbf{r}(t) = \langle\cos t,\sin t,t\rangle }. \]The \(x\)- and \(y\)-coordinates move around a circle while the \(z\)-coordinate increases.
Line integrals work with curves in either two or three dimensions.
The Tangent Vector
Differentiate the parameterization:
\[ \boxed{ \mathbf{r}'(t) = \langle x'(t),y'(t),z'(t)\rangle }. \]This vector is tangent to the path and points in the direction in which the parameterization travels.
Its magnitude is
\[ \|\mathbf{r}'(t)\|. \]The Arc-Length Element
A small change \(dt\) corresponds approximately to a small distance along the curve.
The arc-length differential is
\[ \boxed{ ds = \|\mathbf{r}'(t)\|\,dt }. \]This converts parameter change into actual distance traveled along the curve.
Arc Length as a Line Integral
The length of a curve \(C\) is
\[ \boxed{ L = \int_C ds }. \]Using the parameterization,
\[ \boxed{ L = \int_a^b \|\mathbf{r}'(t)\|\,dt }. \]Thus, ordinary arc length is already a simple line integral.
Example: Length of a Line Segment
Let
\[ \mathbf{r}(t) = \langle2t,4t\rangle, \qquad 0\le t\le1. \]Then
\[ \mathbf{r}'(t) = \langle2,4\rangle, \]so
\[ \|\mathbf{r}'(t)\| = \sqrt{2^2+4^2} = 2\sqrt5. \]Therefore,
\[ L = \int_0^1 2\sqrt5\,dt = \boxed{2\sqrt5}. \]Two Main Kinds of Line Integral
There are two major line-integral forms in this section.
A scalar line integral:
\[ \boxed{ \int_C f\,ds }. \]A vector line integral:
\[ \boxed{ \int_C \mathbf{F}\cdot d\mathbf{r} }. \]These integrate different kinds of fields and have different geometric meanings.
Scalar Line Integrals
Suppose \(f\) is a scalar field and \(C\) is parameterized by
\[ \mathbf{r}(t), \qquad a\le t\le b. \]The scalar line integral is
\[ \boxed{ \int_C f\,ds = \int_a^b f(\mathbf{r}(t)) \|\mathbf{r}'(t)\|\,dt }. \]Why the Speed Factor Appears
The function \(f(\mathbf{r}(t))\) gives the scalar-field value at the current point.
The factor
\[ \|\mathbf{r}'(t)\| \]converts \(dt\) into physical distance \(ds\).
Without this factor, the result could depend incorrectly on how quickly the parameter moves along the same geometric curve.
Example: Scalar Line Integral on a Line Segment
Let
\[ f(x,y)=x+y \]and let \(C\) be the line segment from \((0,0)\) to \((1,2)\).
Use
\[ \mathbf{r}(t)=\langle t,2t\rangle, \qquad 0\le t\le1. \]Then
\[ f(\mathbf{r}(t)) = t+2t = 3t. \]Also,
\[ \mathbf{r}'(t)=\langle1,2\rangle \]and
\[ \|\mathbf{r}'(t)\|=\sqrt5. \]Therefore,
\[ \int_C f\,ds = \int_0^1 3t\sqrt5\,dt = \boxed{\frac{3\sqrt5}{2}}. \]Mass of a Thin Wire
If a wire lies along a curve \(C\) and has linear density
\[ \rho(x,y,z), \]then its total mass is
\[ \boxed{ m = \int_C \rho\,ds }. \]This is one of the standard physical interpretations of a scalar line integral.
Average Value Along a Curve
If \(C\) has length \(L\), the average value of a scalar field along the curve is
\[ \boxed{ f_{\text{avg}} = \frac{1}{L} \int_C f\,ds }. \]This is analogous to the average value of a one-variable function over an interval.
Vector Line Integrals
Let
\[ \mathbf{F}(x,y,z) \]be a vector field.
The vector line integral along \(C\) is
\[ \boxed{ \int_C \mathbf{F}\cdot d\mathbf{r} }. \]If \(C\) is parameterized by \(\mathbf{r}(t)\), then
\[ \boxed{ \int_C \mathbf{F}\cdot d\mathbf{r} = \int_a^b \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t)\,dt }. \]The Differential Vector \(d\mathbf{r}\)
Along a parameterized curve,
\[ d\mathbf{r} = \mathbf{r}'(t)\,dt. \]In Cartesian coordinates,
\[ d\mathbf{r} = \langle dx,dy,dz\rangle. \]Therefore,
\[ \mathbf{F}\cdot d\mathbf{r} \]measures the component of the field aligned with an infinitesimal displacement.
Component Form
If
\[ \mathbf{F} = \langle P,Q,R\rangle, \]then
\[ \boxed{ \mathbf{F}\cdot d\mathbf{r} = P\,dx+Q\,dy+R\,dz }. \]Hence,
\[ \boxed{ \int_C \mathbf{F}\cdot d\mathbf{r} = \int_C P\,dx+Q\,dy+R\,dz }. \]Work Done by a Force Field
If
\[ \mathbf{F} \]is a force field and a particle moves along \(C\), then the work done by the field is
\[ \boxed{ W = \int_C \mathbf{F}\cdot d\mathbf{r} }. \]Only the component of force in the direction of motion contributes to the instantaneous work.
Why the Dot Product Appears
Since
\[ \mathbf{F}\cdot\mathbf{r}' = \|\mathbf{F}\| \|\mathbf{r}'\| \cos\theta, \]the vector line integral counts the portion of the field acting along the direction of travel.
A field perpendicular to the motion contributes zero instantaneous work.
Positive, Negative, and Zero Work
Locally:
- if the field points generally with the motion, the contribution is positive;
- if the field points generally against the motion, the contribution is negative;
- if the field is perpendicular to the motion, the contribution is zero.
The line integral adds these local contributions along the entire path.
Example: Work Along a Line Segment
Let
\[ \mathbf{F}(x,y) = \langle y,x\rangle \]and let \(C\) run from \((0,0)\) to \((1,2)\) along the line segment
\[ \mathbf{r}(t)=\langle t,2t\rangle, \qquad 0\le t\le1. \]Along the path,
\[ \mathbf{F}(\mathbf{r}(t)) = \langle2t,t\rangle. \]Also,
\[ \mathbf{r}'(t) = \langle1,2\rangle. \]Their dot product is
\[ \langle2t,t\rangle \cdot \langle1,2\rangle = 4t. \]Therefore,
\[ W = \int_0^1 4t\,dt = \boxed{2}. \]Scalar and Vector Line Integrals Use Different Factors
For a scalar line integral:
\[ \int_C f\,ds = \int_a^b f(\mathbf{r}(t)) \boxed{\|\mathbf{r}'(t)\|} \,dt. \]For a vector line integral:
\[ \int_C \mathbf{F}\cdot d\mathbf{r} = \int_a^b \mathbf{F}(\mathbf{r}(t)) \boxed{\cdot\mathbf{r}'(t)} \,dt. \]Confusing these two structures is a common source of errors.
Orientation Matters for Vector Line Integrals
A parameterization gives a curve an orientation.
Reversing the direction of travel changes
\[ d\mathbf{r} \]to its negative.
Therefore,
\[ \boxed{ \int_{-C} \mathbf{F}\cdot d\mathbf{r} = - \int_C \mathbf{F}\cdot d\mathbf{r} }. \]Orientation Does Not Change a Scalar \(ds\) Integral
Because
\[ ds \]measures positive distance,
\[ \boxed{ \int_{-C} f\,ds = \int_C f\,ds }. \]Scalar line integrals with respect to arc length do not change sign when orientation is reversed.
Line Integrals Around Closed Curves
If a path starts and ends at the same point, it is a closed curve.
A line integral around a closed curve is often written
\[ \boxed{ \oint_C \mathbf{F}\cdot d\mathbf{r} }. \]This notation emphasizes that the path forms a loop.
Circulation
For a vector field, the closed line integral
\[ \boxed{ \oint_C \mathbf{F}\cdot d\mathbf{r} } \]measures circulation around the oriented curve.
Positive circulation means the field tends to align with the chosen orientation around the loop.
Example: Circulation Around a Circle
Consider
\[ \mathbf{F}(x,y) = \langle-y,x\rangle \]around the unit circle oriented counterclockwise.
Parameterize:
\[ \mathbf{r}(t) = \langle\cos t,\sin t\rangle, \qquad 0\le t\le2\pi. \]Then
\[ \mathbf{F}(\mathbf{r}(t)) = \langle-\sin t,\cos t\rangle, \]while
\[ \mathbf{r}'(t) = \langle-\sin t,\cos t\rangle. \]Therefore,
\[ \mathbf{F}\cdot\mathbf{r}' = \sin^2t+\cos^2t = 1. \]Hence,
\[ \oint_C \mathbf{F}\cdot d\mathbf{r} = \int_0^{2\pi}1\,dt = \boxed{2\pi}. \]Clockwise Orientation
If the same circle is traversed clockwise, the orientation reverses.
Therefore,
\[ \boxed{ \oint_{-C} \mathbf{F}\cdot d\mathbf{r} = -2\pi }. \]Orientation is part of the data of a vector line integral.
Piecewise Smooth Curves
A path may consist of several smooth pieces.
If
\[ C=C_1\cup C_2\cup\cdots\cup C_n, \]then
\[ \boxed{ \int_C \mathbf{F}\cdot d\mathbf{r} = \sum_{k=1}^n \int_{C_k} \mathbf{F}\cdot d\mathbf{r} }. \]The same additive principle applies to scalar line integrals.
Example: A Two-Segment Path
Suppose a path travels from \((0,0)\) to \((1,0)\), then from \((1,0)\) to \((1,2)\).
We may represent it as two pieces:
\[ C_1: \mathbf{r}_1(t)=\langle t,0\rangle, \qquad 0\le t\le1, \]and
\[ C_2: \mathbf{r}_2(t)=\langle1,2t\rangle, \qquad 0\le t\le1. \]Compute the line integral on each piece and add the results.
Parameterization Independence
A properly defined line integral depends on the oriented geometric curve, not on the particular regular parameterization used to describe it.
For scalar \(ds\) integrals, suitable reparameterization preserves the value.
For vector line integrals, an orientation-preserving reparameterization preserves the value, while reversing orientation changes its sign.
Choosing a Good Parameterization
A useful parameterization should make three things manageable:
- the point \(\mathbf{r}(t)\);
- the derivative \(\mathbf{r}'(t)\);
- the parameter interval.
Standard geometric shapes usually have standard parameterizations worth recognizing immediately.
Standard Line-Segment Parameterization
The line segment from point \(\mathbf{A}\) to point \(\mathbf{B}\) can be parameterized by
\[ \boxed{ \mathbf{r}(t) = (1-t)\mathbf{A}+t\mathbf{B}, \qquad 0\le t\le1 }. \]Equivalently,
\[ \mathbf{r}(t) = \mathbf{A} + t(\mathbf{B}-\mathbf{A}). \]Standard Circle Parameterization
A circle of radius \(R\) centered at \((h,k)\), oriented counterclockwise, can be parameterized by
\[ \boxed{ \mathbf{r}(t) = \langle h+R\cos t,\; k+R\sin t \rangle, \qquad 0\le t\le2\pi }. \]Standard Helix Parameterization
A helix of radius \(R\) with vertical rate \(c\) can be parameterized by
\[ \boxed{ \mathbf{r}(t) = \langle R\cos t,\; R\sin t,\; ct \rangle }. \]This is a common test case for three-dimensional line integrals.
Line Integrals Along Graphs
If a planar curve is given by
\[ y=g(x), \qquad a\le x\le b, \]we may use \(x\) itself as the parameter:
\[ \mathbf{r}(x) = \langle x,g(x)\rangle. \]Then
\[ ds = \sqrt{1+[g'(x)]^2}\,dx. \]Scalar Line Integral Along a Graph
For \(y=g(x)\),
\[ \boxed{ \int_C f\,ds = \int_a^b f(x,g(x)) \sqrt{1+[g'(x)]^2}\,dx }. \]This is the familiar arc-length structure with an additional scalar-field weight.
Vector Line Integral Along a Graph
If
\[ \mathbf{F} = \langle P,Q\rangle \]and \(y=g(x)\), then
\[ dy=g'(x)\,dx. \]Therefore,
\[ \boxed{ \int_C P\,dx+Q\,dy = \int_a^b \left[ P(x,g(x)) + Q(x,g(x))g'(x) \right]dx }. \]Path Dependence
For a general vector field,
\[ \int_C \mathbf{F}\cdot d\mathbf{r} \]may depend on the actual route taken between two endpoints.
Two paths with the same starting and ending points can produce different values.
Such an integral is path dependent.
Path Independence
Some vector fields have the special property that the line integral depends only on the endpoints.
Then
\[ \int_C \mathbf{F}\cdot d\mathbf{r} \]is path independent.
Gradient fields provide the central example.
The Fundamental Theorem for Line Integrals
If
\[ \mathbf{F}=\nabla f \]and \(C\) runs from point \(A\) to point \(B\), then
\[ \boxed{ \int_C \mathbf{F}\cdot d\mathbf{r} = f(B)-f(A) }. \]The entire path disappears from the final calculation.
Why the Fundamental Theorem Works
If
\[ \mathbf{F}=\nabla f \]and the path is \(\mathbf{r}(t)\), then
\[ \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) = \nabla f(\mathbf{r}(t)) \cdot \mathbf{r}'(t). \]By the multivariable chain rule,
\[ \nabla f(\mathbf{r}(t)) \cdot \mathbf{r}'(t) = \frac{d}{dt} f(\mathbf{r}(t)). \]Therefore,
\[ \int_a^b \frac{d}{dt} f(\mathbf{r}(t))\,dt = f(\mathbf{r}(b)) - f(\mathbf{r}(a)). \]Example: Using a Potential Instead of Parameterizing
Let
\[ \mathbf{F}(x,y) = \langle2x,2y\rangle. \]A potential is
\[ f(x,y)=x^2+y^2. \]Suppose \(C\) runs from
\[ A=(1,1) \]to
\[ B=(3,2). \]Then
\[ \int_C \mathbf{F}\cdot d\mathbf{r} = f(3,2)-f(1,1). \]Thus,
\[ \boxed{ 13-2=11 }. \]Closed Curves in Conservative Fields
If a conservative field has potential \(f\), then for a closed curve, the starting point and ending point are the same.
Therefore,
\[ \boxed{ \oint_C \mathbf{F}\cdot d\mathbf{r} = 0 }. \]This is an important signature of path independence.
Circulation and Curl
The previous lesson introduced curl as local circulation density.
The line integral
\[ \oint_C \mathbf{F}\cdot d\mathbf{r} \]measures total circulation around a finite loop.
Green's Theorem and Stokes' Theorem will later connect this boundary circulation to curl accumulated across the enclosed region or surface.
Work and Potential Energy
In many conservative force models,
\[ \mathbf{F} = -\nabla U, \]where \(U\) is potential energy.
Then
\[ W = \int_C \mathbf{F}\cdot d\mathbf{r} = -[U(B)-U(A)]. \]Therefore,
\[ \boxed{ W = U(A)-U(B) }. \]Units of Scalar Line Integrals
In
\[ \int_C f\,ds, \]the units are
\[ \boxed{ [f]\times[\text{length}] }. \]If \(f\) is linear density in kilograms per meter, then multiplying by \(ds\) produces kilograms.
Units of Work Integrals
In
\[ \int_C \mathbf{F}\cdot d\mathbf{r}, \]if \(\mathbf{F}\) is measured in newtons and displacement in meters, then the result has units
\[ \text{newton}\cdot\text{meter} = \text{joule}. \]Unit analysis helps verify the interpretation of a line integral.
Common Error: Forgetting to Parameterize the Field
In a line integral, the field must be evaluated on the curve.
If
\[ \mathbf{r}(t) = \langle x(t),y(t),z(t)\rangle, \]then replace
\[ x,y,z \]by
\[ x(t),y(t),z(t). \]The final integrand should normally depend only on the parameter.
Common Error: Forgetting \(\|\mathbf{r}'(t)\|\) in a Scalar Line Integral
The correct scalar formula is
\[ \boxed{ \int_C f\,ds = \int_a^b f(\mathbf{r}(t)) \|\mathbf{r}'(t)\|\,dt }. \]Omitting the speed factor means you are no longer integrating with respect to arc length.
Common Error: Using Magnitude Instead of a Dot Product
For work or circulation,
\[ \boxed{ \int_C \mathbf{F}\cdot d\mathbf{r} = \int_a^b \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t)\,dt }. \]Do not replace the dot product with
\[ \|\mathbf{F}\| \|\mathbf{r}'\| \]unless the angular factor has been accounted for.
Common Error: Ignoring Orientation
Vector line integrals are oriented.
Reversing the path changes the sign:
\[ \boxed{ \int_{-C} \mathbf{F}\cdot d\mathbf{r} = - \int_C \mathbf{F}\cdot d\mathbf{r} }. \]Common Error: Using the Wrong Parameter Bounds
The parameter interval must trace exactly the requested portion of the curve in the requested direction.
A correct formula with incorrect bounds can traverse too much of the curve, too little of it, or the right curve in the wrong orientation.
Common Error: Mixing \(ds\) and \(d\mathbf{r}\)
Remember:
\[ ds = \|\mathbf{r}'(t)\|dt \]is scalar, while
\[ d\mathbf{r} = \mathbf{r}'(t)dt \]is vector-valued.
They belong to different line-integral structures.
A Reliable Procedure for Scalar Line Integrals
- Parameterize the curve as \(\mathbf{r}(t)\).
- Determine the parameter interval.
- Compute \(\mathbf{r}'(t)\).
- Compute \(\|\mathbf{r}'(t)\|\).
- Substitute the parameterization into \(f\).
- Form \(f(\mathbf{r}(t))\|\mathbf{r}'(t)\|\).
- Integrate over the parameter interval.
- Interpret the result with units when appropriate.
A Reliable Procedure for Vector Line Integrals
- Parameterize the oriented curve.
- Determine the parameter interval.
- Compute \(\mathbf{r}'(t)\).
- Substitute the curve into the vector field.
- Compute \(\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\).
- Simplify to a one-variable integrand.
- Integrate over the parameter interval.
- Check whether a potential function could have provided a shortcut.
A Reliable Procedure for Conservative-Field Problems
- Determine whether the field is known or shown to be conservative.
- Find a potential function \(f\) if one is not already given.
- Identify the initial point \(A\).
- Identify the terminal point \(B\).
- Compute \(f(B)-f(A)\).
- Do not parameterize the path unless the problem specifically requires it.
Example: Full Scalar Line-Integral Workflow
Evaluate
\[ \int_C (x+2y)\,ds \]where \(C\) is the line segment from \((0,0)\) to \((2,1)\).
Parameterize:
\[ \mathbf{r}(t) = \langle2t,t\rangle, \qquad 0\le t\le1. \]Then
\[ \mathbf{r}'(t) = \langle2,1\rangle, \]so
\[ \|\mathbf{r}'(t)\| = \sqrt5. \]Along the curve,
\[ x+2y = 2t+2t = 4t. \]Therefore,
\[ \int_C (x+2y)\,ds = \int_0^1 4t\sqrt5\,dt = \boxed{2\sqrt5}. \]Example: Full Vector Line-Integral Workflow
Evaluate
\[ \int_C \mathbf{F}\cdot d\mathbf{r} \]for
\[ \mathbf{F}(x,y) = \langle x+y,\;x-y\rangle \]along the line segment from \((0,0)\) to \((2,1)\).
Use
\[ \mathbf{r}(t) = \langle2t,t\rangle, \qquad 0\le t\le1. \]Then
\[ \mathbf{F}(\mathbf{r}(t)) = \langle3t,t\rangle. \]Also,
\[ \mathbf{r}'(t) = \langle2,1\rangle. \]Therefore,
\[ \mathbf{F}\cdot\mathbf{r}' = 6t+t = 7t. \]Hence,
\[ \int_C \mathbf{F}\cdot d\mathbf{r} = \int_0^1 7t\,dt = \boxed{\frac72}. \]Example: Three-Dimensional Line Integral
Let
\[ \mathbf{F}(x,y,z) = \langle y,z,x\rangle \]and
\[ \mathbf{r}(t) = \langle t,t^2,t^3\rangle, \qquad 0\le t\le1. \]Along the path,
\[ \mathbf{F}(\mathbf{r}(t)) = \langle t^2,t^3,t\rangle. \]Also,
\[ \mathbf{r}'(t) = \langle1,2t,3t^2\rangle. \]The dot product is
\[ t^2+2t^4+3t^3. \]Therefore,
\[ \int_C \mathbf{F}\cdot d\mathbf{r} = \int_0^1 \left( t^2+3t^3+2t^4 \right)dt. \]Thus,
\[ \boxed{ \frac13+\frac34+\frac25 = \frac{89}{60} }. \]Practice 1: Parameterize a Segment
Parameterize the line segment from
\[ (1,2) \]to
\[ (5,4). \]Practice 2: Arc-Length Element
For
\[ \mathbf{r}(t) = \langle3t,4t\rangle, \]find \(ds\).
Practice 3: Scalar Line Integral
Evaluate
\[ \int_C x\,ds \]where
\[ \mathbf{r}(t) = \langle t,2t\rangle, \qquad 0\le t\le1. \]Practice 4: Work Integral
Let
\[ \mathbf{F}(x,y) = \langle2x,y\rangle \]and
\[ \mathbf{r}(t) = \langle t,t^2\rangle, \qquad 0\le t\le1. \]Find
\[ \int_C \mathbf{F}\cdot d\mathbf{r}. \]Practice 5: Reverse Orientation
Suppose
\[ \int_C \mathbf{F}\cdot d\mathbf{r} = 7. \]Find the value along the same curve with the opposite orientation.
Practice 6: Circulation Around a Circle
Let
\[ \mathbf{F}(x,y) = \langle-y,x\rangle. \]Find the circulation around the circle
\[ x^2+y^2=4 \]oriented counterclockwise.
Practice 7: Conservative Shortcut
Let
\[ \mathbf{F}(x,y) = \langle2x,4y\rangle. \]Find the line integral from \((0,0)\) to \((3,2)\) along any path.
Practice 8: Closed Conservative Curve
Suppose
\[ \mathbf{F}=\nabla f \]on a region containing a closed curve \(C\).
Find
\[ \oint_C \mathbf{F}\cdot d\mathbf{r}. \]Practice 9: Scalar versus Vector Structure
State whether each expression requires \(\|\mathbf{r}'(t)\|\) or a dot product with \(\mathbf{r}'(t)\):
- \(\int_C \rho\,ds\)
- \(\int_C \mathbf{F}\cdot d\mathbf{r}\)
- \(\int_C 1\,ds\)
- \(\oint_C \mathbf{F}\cdot d\mathbf{r}\)
Practice 10: Three-Dimensional Path
Let
\[ \mathbf{F}(x,y,z) = \langle x,y,z\rangle \]and
\[ \mathbf{r}(t) = \langle t,2t,3t\rangle, \qquad 0\le t\le1. \]Evaluate
\[ \int_C \mathbf{F}\cdot d\mathbf{r}. \]Practice Answers
Practice 1
Use
\[ \mathbf{r}(t) = (1-t)\langle1,2\rangle + t\langle5,4\rangle. \]Thus,
\[ \boxed{ \mathbf{r}(t) = \langle1+4t,\;2+2t\rangle, \qquad 0\le t\le1 }. \]Practice 2
\[ \mathbf{r}'(t) = \langle3,4\rangle. \]Hence,
\[ \|\mathbf{r}'(t)\| = 5, \]so
\[ \boxed{ds=5\,dt}. \]Practice 3
Here
\[ x=t \]and
\[ \|\mathbf{r}'(t)\| = \sqrt5. \]Therefore,
\[ \int_C x\,ds = \int_0^1 t\sqrt5\,dt = \boxed{\frac{\sqrt5}{2}}. \]Practice 4
Along the path,
\[ \mathbf{F}(\mathbf{r}(t)) = \langle2t,t^2\rangle. \]Also,
\[ \mathbf{r}'(t) = \langle1,2t\rangle. \]Therefore,
\[ \mathbf{F}\cdot\mathbf{r}' = 2t+2t^3. \]Hence,
\[ \int_0^1 (2t+2t^3)\,dt = 1+\frac12 = \boxed{\frac32}. \]Practice 5
\[ \boxed{-7}. \]Practice 6
Parameterize the radius-\(2\) circle:
\[ \mathbf{r}(t) = \langle2\cos t,2\sin t\rangle. \]Then
\[ \mathbf{F}(\mathbf{r}(t)) = \langle-2\sin t,2\cos t\rangle \]and
\[ \mathbf{r}'(t) = \langle-2\sin t,2\cos t\rangle. \]Their dot product is
\[ 4. \]Therefore,
\[ \oint_C \mathbf{F}\cdot d\mathbf{r} = \int_0^{2\pi}4\,dt = \boxed{8\pi}. \]Practice 7
A potential is
\[ f(x,y)=x^2+2y^2. \]Therefore,
\[ \int_C \mathbf{F}\cdot d\mathbf{r} = f(3,2)-f(0,0). \]Thus,
\[ \boxed{ 9+8=17 }. \]Practice 8
A closed curve has the same initial and terminal point, so
\[ \boxed{ \oint_C \mathbf{F}\cdot d\mathbf{r} = 0 }. \]Practice 9
- \(\int_C \rho\,ds\): use \(\|\mathbf{r}'(t)\|\).
- \(\int_C \mathbf{F}\cdot d\mathbf{r}\): use a dot product with \(\mathbf{r}'(t)\).
- \(\int_C 1\,ds\): use \(\|\mathbf{r}'(t)\|\).
- \(\oint_C \mathbf{F}\cdot d\mathbf{r}\): use a dot product with \(\mathbf{r}'(t)\).
Practice 10
Along the path,
\[ \mathbf{F}(\mathbf{r}(t)) = \langle t,2t,3t\rangle. \]Also,
\[ \mathbf{r}'(t) = \langle1,2,3\rangle. \]Therefore,
\[ \mathbf{F}\cdot\mathbf{r}' = t+4t+9t = 14t. \]Hence,
\[ \int_0^1 14t\,dt = \boxed{7}. \]What to Remember
A scalar line integral is
\[ \boxed{ \int_C f\,ds = \int_a^b f(\mathbf{r}(t)) \|\mathbf{r}'(t)\|\,dt }. \]A vector line integral is
\[ \boxed{ \int_C \mathbf{F}\cdot d\mathbf{r} = \int_a^b \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t)\,dt }. \]- Parameterization converts a curve problem into a one-variable integral.
- \(ds=\|\mathbf{r}'(t)\|dt\) measures distance along the curve.
- Scalar line integrals can accumulate density, mass, or other scalar quantities.
- Vector line integrals measure work or circulation.
- Orientation matters for vector line integrals.
- Piecewise curves are handled by adding the integrals over each piece.
- Conservative fields produce path-independent line integrals.
- If \(\mathbf{F}=\nabla f\), then \(\int_C\mathbf{F}\cdot d\mathbf{r}=f(B)-f(A)\).
Where We Go Next
Line integrals accumulate scalar or vector-field effects along curves. Vector calculus also needs an analogous way to accumulate quantities across two-dimensional surfaces embedded in three-dimensional space.
The next lesson, Surface Integrals, develops surface parameterization, surface-area elements, scalar surface integrals, oriented surfaces, and the geometric machinery needed for flux.