One Path, Several Smooth Pieces
A curve need not have one smooth formula from beginning to end. It may travel along a straight segment, turn at a corner, and then follow another segment. Such a route is still a single continuous path if each piece ends exactly where the next one begins.
When a vector field has a potential, the endpoint theorem makes the line integral along this entire route especially simple: only the path’s overall initial and terminal points determine the result. The join points matter for confirming that the pieces form one path, but they are not additional endpoints in the final formula.
Piecewise Smooth Curves and Joins
A smooth parametrized curve on an interval \([a,b]\) is a differentiable function \(\mathbf{r}:[a,b]\to\mathbb{R}^n\) whose derivative is continuous. A piecewise smooth curve consists of finitely many smooth curve pieces, with the terminal point of each piece equal to the initial point of the next.
A corner is therefore permitted. For example, a path may change direction abruptly at a join, as long as the location at the end of one piece is exactly the location at the start of the next.
For a vector field \(\mathbf{F}\), the line integral along one smooth piece \(\mathbf{r}_i\) is \(\int_{a_i}^{b_i}\mathbf{F}(\mathbf{r}_i(t))\cdot\mathbf{r}_i'(t)\,dt\). The line integral along the piecewise smooth curve is the sum of the integrals along its pieces. This sum rule does not require the pieces to meet; however, the endpoint theorem for one path uses the overall endpoints only when the pieces really do join.
The Endpoint Theorem for Joined Pieces
Suppose \(\mathbf{F}=\nabla\phi\) throughout a domain \(D\), where \(\phi\) is a differentiable scalar potential. If a piecewise smooth curve lies in \(D\) and its pieces join continuously, then the line integral equals the potential at the overall terminal point minus the potential at the overall initial point.
The assumptions matter. The field must have the stated potential on a domain containing the curve, and the curve must be made from finitely many smooth pieces. The joining condition lets us regard the pieces as one continuous path. No matching of derivatives at the joins is required.
Why the Join Values Cancel
On each smooth piece, the chain rule gives \(\frac{d}{dt}\phi(\mathbf{r}_i(t))=\nabla\phi(\mathbf{r}_i(t))\cdot\mathbf{r}_i'(t)\). The fundamental theorem of calculus then turns that piece’s line integral into the change in potential from its start to its finish. Adding the changes makes the potential values at the joins cancel.
At each join, \(\mathbf{r}_i(b_i)=\mathbf{r}_{i+1}(a_{i+1})\), so the end value from one piece is the start value for the next. In the sum, each such value appears once with a plus sign and once with a minus sign. Only the first start value and the last end value remain:
This proves the endpoint formula for a curve with any finite number of smoothly parametrized pieces. The corner at a join causes no difficulty: the calculation applies separately on each smooth interval, and the matching positions make the potential values cancel.
Worked Example: Two Segments in the Plane
Worked Example: Two Segments in the Plane
Let \(\mathbf{F}(x,y)=\langle 2x,2y\rangle\), with potential \(\phi(x,y)=x^2+y^2\). Find the line integral along the two-segment path from \((0,0)\) to \((1,0)\), then from \((1,0)\) to \((1,2)\).
First check the joins: the first segment ends at \((1,0)\), and the second starts at \((1,0)\). Thus the pieces form a continuous path. Its overall initial point is \((0,0)\), and its overall terminal point is \((1,2)\). By the endpoint theorem,
To see the contribution of each piece, the potential at the join is \(\phi(1,0)=1^2+0^2=1\). The first piece contributes \(1-0=1\), and the second contributes \(5-1=4\). Their sum is \(1+4=5\), agreeing with the endpoint calculation.
We can also verify directly. On the first segment, use \(\mathbf{r}_1(t)=(t,0)\), \(0\leq t\leq1\). Then \(\mathbf{r}_1'(t)=(1,0)\), \(\mathbf{F}(\mathbf{r}_1(t))=(2t,0)\), and the dot product is \(2t\). Its integral is \(\int_0^1 2t\,dt=[t^2]_0^1=1\).
On the second segment, use \(\mathbf{r}_2(t)=(1,t)\), \(0\leq t\leq2\). Then \(\mathbf{r}_2'(t)=(0,1)\), \(\mathbf{F}(\mathbf{r}_2(t))=(2,2t)\), and the dot product is \(2t\). Its integral is \(\int_0^2 2t\,dt=[t^2]_0^2=4\). Therefore the total is \(1+4=5\), as required.
Worked Example: A Three-Dimensional Path
Worked Example: A Three-Dimensional Path
Let \(\phi(x,y,z)=x^2+yz\), so \(\mathbf{F}=\nabla\phi=\langle 2x,z,y\rangle\). A path has two pieces: from \((0,1,0)\) to \((1,1,0)\), followed by a piece from \((1,1,0)\) to \((1,2,3)\). Find its line integral.
The first piece ends where the second begins, at \((1,1,0)\). The overall initial point is \((0,1,0)\), and the overall terminal point is \((1,2,3)\). Evaluate the potential at these three points:
The two pieces contribute \(1-0=1\) and \(7-1=6\). Hence the total line integral is \(1+6=7\), which is also the overall endpoint difference \(7-0=7\).
For a direct check, parametrize the first piece by \(\mathbf{r}_1(t)=(t,1,0)\), \(0\leq t\leq1\). Then \(\mathbf{r}_1'(t)=(1,0,0)\), and \(\mathbf{F}(\mathbf{r}_1(t))=(2t,0,1)\). The dot product is \(2t\), so this piece contributes \(\int_0^1 2t\,dt=1\).
Parametrize the second piece by \(\mathbf{r}_2(t)=(1,1+t,3t)\), \(0\leq t\leq1\). Then \(\mathbf{r}_2'(t)=(0,1,3)\) and \(\mathbf{F}(\mathbf{r}_2(t))=(2,3t,1+t)\). The dot product is \(0+3t+3(1+t)=3+6t\). Thus the contribution is \(\int_0^1(3+6t)\,dt=[3t+3t^2]_0^1=6\). Adding the direct integrals gives \(1+6=7\), confirming the result.
Worked Example: Three Pieces and a Corner
Worked Example: Three Pieces and a Corner
Let \(\mathbf{F}(x,y)=\langle y,x\rangle\), with potential \(\phi(x,y)=xy\). Consider the path with successive vertices \((0,0)\), \((2,0)\), \((2,1)\), and \((1,2)\). Find its line integral.
Each piece ends at the next piece’s starting vertex, so the route is continuous, including at its corners. The overall endpoints are \((0,0)\) and \((1,2)\). The endpoint theorem gives \(\phi(1,2)-\phi(0,0)=(1)(2)-(0)(0)=2\).
Check this by calculating each segment. On the first, let \(\mathbf{r}_1(t)=(2t,0)\), \(0\leq t\leq1\). The derivative is \((2,0)\), the field is \((0,2t)\), and the dot product is \(0\). The integral is \(0\).
On the second, let \(\mathbf{r}_2(t)=(2,t)\), \(0\leq t\leq1\). The derivative is \((0,1)\), the field is \((t,2)\), and the dot product is \(2\). The integral is \(\int_0^1 2\,dt=2\).
On the third, let \(\mathbf{r}_3(t)=(2-t,1+t)\), \(0\leq t\leq1\). This starts at \((2,1)\) and ends at \((1,2)\). Its derivative is \((-1,1)\), and the field along it is \((1+t,2-t)\). The dot product is \(-(1+t)+(2-t)=1-2t\). The integral is \(\int_0^1(1-2t)\,dt=[t-t^2]_0^1=0\). The three contributions sum to \(0+2+0=2\), matching the endpoint result.
A Common Mistake: Assuming the Pieces Join
The endpoint formula for one piecewise path cannot be used just because several curve segments have been listed. Their endpoints must match in order. Otherwise there is no single continuous path with the claimed overall starting and ending points.
Worked Example: Correcting a False Join
Take \(\mathbf{F}(x,y)=\langle 2x,2y\rangle=\nabla(x^2+y^2)\). Consider a segment from \((0,0)\) to \((1,0)\), followed in a list by a separate segment from \((2,0)\) to \((2,1)\). A tempting but wrong calculation is to treat these as one path and write \(\phi(2,1)-\phi(0,0)=5-0=5\).
This is wrong because the first segment ends at \((1,0)\), whereas the second begins at \((2,0)\). There is a gap, so the join condition fails. The two segment integrals can still be added as integrals over separate curves: the first equals \(\phi(1,0)-\phi(0,0)=1-0=1\), and the second equals \(\phi(2,1)-\phi(2,0)=5-4=1\). Their sum is \(1+1=2\), not \(5\).
The correction is to keep the two integrals separate, or to specify an additional connecting curve and include its integral. Only after all pieces have matching joins can the endpoint theorem be applied using the start of the first piece and the finish of the last.
What to Check Before Using the Formula
A reliable calculation separates two questions: do the pieces form the stated path, and does the field have a potential on a domain containing that path? Once both conditions hold, find the overall endpoints—not merely a convenient pair of vertices—and evaluate the potential there.
Confirm that \(\mathbf{F}=\nabla\phi\) on a domain containing every piece of the curve.
For every adjacent pair, verify that the earlier piece’s endpoint equals the next piece’s starting point.
Use the first point of the first piece and the last point of the final piece.
Evaluate terminal potential minus initial potential. If desired, check by summing the changes on individual pieces.
The endpoint method matters because it replaces a collection of potentially different parametrized integrals with one subtraction. It also makes the role of each point precise: join points confirm continuity, while only the overall initial and terminal points remain in the final value.
Check Your Understanding
Use the endpoint theorem where its conditions hold, and distinguish the join points from the overall endpoints.
- What equality must hold between two consecutive smooth pieces for them to form a continuous piecewise smooth curve?
- A potential is \(\phi(x,y)=x^2+y^2\). A continuous piecewise smooth path starts at \((1,0)\) and ends at \((0,2)\). What is its line integral for \(\mathbf{F}=\nabla\phi\)?
- Why do the potential values at the join points cancel when the contributions from all pieces are added?
- Can two consecutive pieces have different tangent vectors at their join and still form a piecewise smooth curve? Explain.
- For pieces listed from \((0,0)\) to \((1,0)\) and then from \((2,0)\) to \((2,1)\), why is the overall endpoint formula for a single joined path not applicable?