Tutorials › Calculus III › Additivity Over Piecewise Smooth Curves

Home  /  Tutorials  /  Work and Vector Line Integrals

Work and Vector Line Integrals · Tutorial 758 of 999

Additivity Over Piecewise Smooth Curves

See how to compute work along a piecewise smooth path by adding the integrals over its consecutively oriented segments.

Intermediate 10 min read

What You'll Learn

  • Define a piecewise smooth oriented curve and its vector line integral.
  • State the conditions under which additivity applies.
  • Derive the additivity formula by splitting the parameter interval.
  • Compute work along paths with corners and curved segments.
  • Identify why changing the route can change the integral.

Why Additivity Matters

A path may consist of several simpler pieces: a straight segment followed by another segment at a corner, or a curved arc followed by a straight line. The work done by a force along the whole path can be calculated by adding the work along those pieces. This is useful because each piece can be parametrized and integrated separately.

The corner itself causes no difficulty. A curve can fail to have a single tangent direction at a joining point and still have a well-defined vector line integral. What matters is that the path is continuous, that each piece is smooth enough to integrate, and that the force field is defined and continuous along the path.

Key idea: For a path made by traversing oriented segments consecutively, the vector line integral over the entire path equals the sum of the integrals over those segments. The direction and order of the segments are part of the path.

Piecewise Smooth Curves and Their Integrals

Let \(\mathbf{F}:D\to\mathbb{R}^n\) be a continuous vector field, where \(D\) is a region containing the curve. A smooth parametrized curve is a continuously differentiable function \(\mathbf{r}:[a,b]\to D\). Its vector line integral is defined by

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

Here \(\mathbf{r}'(t)\) is the velocity vector, and the dot product measures the component of the field in the direction of motion. In three dimensions, if \(\mathbf{F}=(P,Q,R)\) and \(\mathbf{r}=(x,y,z)\), the same definition can be written as \(\int_C P\,dx+Q\,dy+R\,dz\).

A piecewise smooth curve is a continuous curve for which the parameter interval can be divided into finitely many subintervals, with the curve continuously differentiable on each subinterval. The derivative need not exist at the division points. Those points may be corners, where the direction of travel changes abruptly.

Definition: Suppose \(C_1,\ldots,C_m\) are oriented curves, and the endpoint of each \(C_i\) is the starting point of \(C_{i+1}\). Their concatenation, written \(C=C_1*\cdots*C_m\), is the path that traverses \(C_1\) first, then \(C_2\), and continues in order to \(C_m\).

The endpoint condition is what makes the pieces a continuous path. Each segment's orientation specifies its direction of travel. The concatenation is not simply a collection of geometric pieces: it records the route and the order in which those pieces are traversed.

The Additivity Theorem

Theorem: Let \(\mathbf{F}\) be continuous along a piecewise smooth concatenation \(C=C_1*\cdots*C_m\). Then each segment integral exists and $$ \int_C \mathbf{F}\cdot d\mathbf{r} = \sum_{i=1}^{m}\int_{C_i}\mathbf{F}\cdot d\mathbf{r}. $$ The curve may have corners at the joins. It is enough for the curve to be continuously differentiable on each of its finitely many smooth pieces.

The theorem follows from additivity of ordinary definite integrals. To see this directly, parametrize each segment by \(\mathbf{r}_i:[0,1]\to D\), and define the whole path on \([0,m]\) by using \(\mathbf{r}_i\) on the interval \([i-1,i]\). The join condition ensures that this parametrization is continuous. On the \(i\)th interval, the parametrization is \(\mathbf{r}_i(t-i+1)\).

1
Write the integral on the full parameter interval.
On each smooth piece, the integrand is the field evaluated along the curve, dotted with the curve's derivative.
2
Split at every join.
Additivity of the definite integral splits the integral over \([0,m]\) into integrals over \([i-1,i]\).
3
Use the parameter for each segment.
On \([i-1,i]\), set \(u=t-i+1\). Then \(u\) runs from \(0\) to \(1\), and the \(i\)th integral is exactly the integral along \(C_i\).

In equations, the split and change of variable give

$$ \begin{aligned} \int_C\mathbf{F}\cdot d\mathbf{r} &=\sum_{i=1}^{m}\int_{i-1}^{i} \mathbf{F}\bigl(\mathbf{r}_i(t-i+1)\bigr)\cdot \mathbf{r}_i'(t-i+1)\,dt\\ &=\sum_{i=1}^{m}\int_0^1 \mathbf{F}\bigl(\mathbf{r}_i(u)\bigr)\cdot\mathbf{r}_i'(u)\,du\\ &=\sum_{i=1}^{m}\int_{C_i}\mathbf{F}\cdot d\mathbf{r}. \end{aligned} $$

This proves the formula. At a corner, the derivative of the full parametrization might not exist. That does not invalidate the argument: the integral is evaluated piece by piece, and the finitely many joining points do not affect the definite integrals over those pieces.

Worked Example: A Two-Segment Path with a Corner

Worked Example: A Two-Segment Path with a Corner

Let \(\mathbf{F}(x,y)=(0,x)\). The path \(C\) goes from \((0,0)\) to \((2,0)\), then from \((2,0)\) to \((2,1)\). Compute the integral along \(C\).

For the first segment, use \(\mathbf{r}_1(t)=(t,0)\), \(0\le t\le2\). Then \(\mathbf{r}_1'(t)=(1,0)\), and \(\mathbf{F}(\mathbf{r}_1(t))=(0,t)\). Thus

$$ \int_{C_1}\mathbf{F}\cdot d\mathbf{r} =\int_0^2(0,t)\cdot(1,0)\,dt =\int_0^2 0\,dt =0. $$

For the second segment, use \(\mathbf{r}_2(t)=(2,t)\), \(0\le t\le1\). Then \(\mathbf{r}_2'(t)=(0,1)\), and \(\mathbf{F}(\mathbf{r}_2(t))=(0,2)\). Therefore

$$ \int_{C_2}\mathbf{F}\cdot d\mathbf{r} =\int_0^1(0,2)\cdot(0,1)\,dt =\int_0^1 2\,dt =2. $$

The segments meet at \((2,0)\), so their concatenation is continuous. By additivity,

$$ \int_C\mathbf{F}\cdot d\mathbf{r} = \int_{C_1}\mathbf{F}\cdot d\mathbf{r} + \int_{C_2}\mathbf{F}\cdot d\mathbf{r} =0+2=2. $$

Check: the first parametrization runs from \((0,0)\) to \((2,0)\), and the second from \((2,0)\) to \((2,1)\), as required. The integrand on the vertical piece is the constant \(2\), so its integral over an interval of length \(1\) is \(2\). The corner does not require a tangent vector shared by both pieces.

Worked Example: A Curved Segment Followed by a Line

Worked Example: A Curved Segment Followed by a Line

Let \(\mathbf{F}(x,y)=(x,2y)\). The path first follows the quarter-circle from \((1,0)\) to \((0,1)\), then follows the vertical segment from \((0,1)\) to \((0,0)\). Compute the integral.

Parametrize the quarter-circle by \(\mathbf{r}_1(\theta)=(\cos\theta,\sin\theta)\), \(0\le\theta\le\frac{\pi}{2}\). Its derivative is \(\mathbf{r}_1'(\theta)=(-\sin\theta,\cos\theta)\), and the field along it is \((\cos\theta,2\sin\theta)\). The dot product is

$$ (\cos\theta,2\sin\theta)\cdot(-\sin\theta,\cos\theta) =-\cos\theta\sin\theta+2\sin\theta\cos\theta =\sin\theta\cos\theta. $$

Hence the arc contributes

$$ \int_{C_1}\mathbf{F}\cdot d\mathbf{r} =\int_0^{\pi/2}\sin\theta\cos\theta\,d\theta =\left[\frac12\sin^2\theta\right]_0^{\pi/2} =\frac12. $$

For the line segment, set \(\mathbf{r}_2(t)=(0,1-t)\), \(0\le t\le1\). Then \(\mathbf{r}_2'(t)=(0,-1)\) and \(\mathbf{F}(\mathbf{r}_2(t))=(0,2(1-t))\). Thus

$$ \int_{C_2}\mathbf{F}\cdot d\mathbf{r} =\int_0^1(0,2(1-t))\cdot(0,-1)\,dt =\int_0^1-2(1-t)\,dt =\left[-2t+t^2\right]_0^1 =-1. $$

The arc ends where the line begins, at \((0,1)\). Additivity gives

$$ \int_C\mathbf{F}\cdot d\mathbf{r} =\frac12+(-1)=-\frac12. $$

Check: the arc's derivative and the line's derivative are each used only on their own parameter interval. Their directions at the joining point differ, which makes a corner, but the two segment contributions still add to \(-\frac12\).

Worked Example: Three Oriented Segments

Worked Example: Three Oriented Segments

Let \(\mathbf{F}(x,y)=(x+y,x-y)\), and let \(C\) travel from \((0,0)\) to \((1,0)\), then to \((1,1)\), and finally to \((0,1)\). Compute the integral, keeping the orientation of each segment.

On the first segment, let \(\mathbf{r}_1(t)=(t,0)\), \(0\le t\le1\). Then \(\mathbf{r}_1'(t)=(1,0)\), \(\mathbf{F}(\mathbf{r}_1(t))=(t,t)\), and their dot product is \(t\). Therefore

$$ \int_{C_1}\mathbf{F}\cdot d\mathbf{r} =\int_0^1t\,dt =\left[\frac{t^2}{2}\right]_0^1 =\frac12. $$

On the second segment, let \(\mathbf{r}_2(t)=(1,t)\), \(0\le t\le1\). Then \(\mathbf{r}_2'(t)=(0,1)\), \(\mathbf{F}(\mathbf{r}_2(t))=(1+t,1-t)\), and their dot product is \(1-t\). Thus

$$ \int_{C_2}\mathbf{F}\cdot d\mathbf{r} =\int_0^1(1-t)\,dt =\left[t-\frac{t^2}{2}\right]_0^1 =\frac12. $$

The third segment runs left, from \((1,1)\) to \((0,1)\). Use \(\mathbf{r}_3(t)=(1-t,1)\), \(0\le t\le1\). Then \(\mathbf{r}_3'(t)=(-1,0)\), \(\mathbf{F}(\mathbf{r}_3(t))=(2-t,-t)\), and the dot product is \(-(2-t)\). Hence

$$ \int_{C_3}\mathbf{F}\cdot d\mathbf{r} =\int_0^1(-2+t)\,dt =\left[-2t+\frac{t^2}{2}\right]_0^1 =-\frac32. $$

The endpoints match at both joins. The theorem gives

$$ \int_C\mathbf{F}\cdot d\mathbf{r} =\frac12+\frac12-\frac32 =-\frac12. $$

Check: adding the three evaluated integrals gives \(\frac12+\frac12-\frac32=-\frac12\). In particular, the third segment's negative contribution is consistent with its leftward orientation; using a rightward derivative there would describe a different traversal.

A Common Pitfall: Replacing the Route

Additivity allows a path to be split into its actual consecutive pieces. It does not allow those pieces to be replaced by a different route with the same endpoints. For a variable field, the integral can depend on the route. The two-segment path in the first example and the straight line from \((0,0)\) to \((2,1)\) have the same starting and ending points, but they are not the same curve.

Common mistake: Treating the L-shaped route in the first example as though it were the straight line joining its endpoints gives the integral for a different path.

To see the error numerically, parametrize that straight line by \(\mathbf{s}(t)=(2t,t)\), \(0\le t\le1\). For \(\mathbf{F}(x,y)=(0,x)\), we have \(\mathbf{F}(\mathbf{s}(t))=(0,2t)\) and \(\mathbf{s}'(t)=(2,1)\). Its integral is

$$ \int_0^1(0,2t)\cdot(2,1)\,dt =\int_0^1 2t\,dt =1. $$

The actual L-shaped route has integral \(2\), as computed above. The value \(1\) is not a correction to that calculation; it is the value for the straight-line route. Additivity says to sum the integrals of the segments that the path actually traverses.

When to Use Additivity

Before splitting an integral into segment contributions, check that the pieces form the intended continuous, oriented path and that the field is defined and continuous along them. Parametrize each piece in its direction of travel, calculate its integral using \(\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\), and then add the results. No matching tangent directions are needed at the joins.

This approach turns a complicated route into manageable calculations while preserving its geometry and orientation. It applies to any finite number of smooth pieces, including paths with corners. If either the path or the field has additional complications, their conditions must be checked separately rather than assumed from additivity alone.

Check Your Understanding

Use the definition and the additivity theorem to answer these questions.

  1. What conditions on the segments ensure that their concatenation is a continuous piecewise smooth curve?
  2. Why does a corner at a joining point not prevent the vector line integral from existing?
  3. If a path consists of four consecutively oriented pieces, how is its integral expressed in terms of the four segment integrals?
  4. For \(\mathbf{F}(x,y)=(0,x)\), why does replacing the L-shaped route from \((0,0)\) to \((2,1)\) with the straight line change the integral?
  5. When parametrizing a segment that travels left, what must the direction of its derivative reflect?

Previous

Next

Back to Tutorials