A Work and Circulation Problem-Solving Workshop

A Reliable Plan for Work and Circulation

A vector line-integral problem is not only an integration exercise. A correct solution must represent the path in the requested direction, evaluate the field along that path, and pair the field with the path’s direction of travel. If the path has corners or crosses a boundary where the field changes formula, the integral must be split into the relevant pieces.

Work and circulation use the same vector line integral. Work describes the accumulated effect of a force as an object moves along a path. Circulation is the corresponding integral around a closed, oriented curve. The context determines which name is appropriate; the calculation depends on the same dot product.

Definition: Let a path be parametrized by a piecewise-smooth function $\mathbf{r}(t)$, $a\le t\le b$, and let $\mathbf{F}$ be a vector field defined along the path. The work integral along the oriented path is $$ \int_C \mathbf{F}\cdot d\mathbf{r} = \int_a^b \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt. $$ If $C$ is a closed, oriented curve, this integral is called the circulation of $\mathbf{F}$ around $C$.

Here, $\mathbf{r}'(t)$ gives the direction and rate of motion along the parametrized curve. The field must be evaluated at the current point $\mathbf{r}(t)$. The dot product selects the component of the field parallel to the direction of travel; a component perpendicular to the path contributes no work.

Why the Parametrized Formula Works

Over a short parameter interval of length $\Delta t$, the displacement along the path is approximately $\mathbf{r}'(t)\Delta t$. The work over that short movement is therefore approximately the force at the point, dotted with the displacement:

\mathbf{F}(\mathbf{r}(t))\cdot\bigl(\mathbf{r}'(t)\Delta t\bigr) = \bigl(\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\bigr)\Delta t.

Adding these contributions along the path and taking the limit gives the integral in the definition. This also explains why replacing the dot product by the magnitude of the force, or by an ordinary scalar integral, generally gives the wrong quantity.

The orientation is part of the integral. To see how it affects the answer, traverse the same path backwards by setting $\mathbf{q}(s)=\mathbf{r}(a+b-s)$, where $a\le s\le b$. Then $\mathbf{q}'(s)=-\mathbf{r}'(a+b-s)$. Substitution into the integral gives:

\int_a^b \mathbf{F}(\mathbf{q}(s))\cdot\mathbf{q}'(s)\,ds = -\int_a^b \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt.

Thus, reversing the direction changes the sign, not the magnitude, of the line integral. This fact is useful both for interpreting answers and for checking a calculation.

A Practical Setup Checklist

Before integrating, identify the path’s endpoints and direction, then choose a parametrization that covers the path once. Substitute the parametrization into the field and compute the derivative of the path. For a piecewise-smooth path, or a path crossing between field formulas, split the integral and use the correct formula on each piece. Finally, check that the sign and size of the answer make sense for the motion or orientation.

Set the orientation.
Record the starting and ending points, or the direction of travel around a closed curve.

Parametrize each smooth piece.
Choose simple parameter limits that cover the piece exactly once in the required direction.

Form the integrand.
Calculate $\mathbf{F}(\mathbf{r}(t))$, $\mathbf{r}'(t)$, and their dot product.

Integrate and check.
Evaluate each piece, add the results, and examine whether the sign agrees with the orientation and physical interpretation.

Worked Examples

Worked Example: Work Along a Parabolic Path

Find the work done by $\mathbf{F}(x,y)=\langle y,x\rangle$ as a particle moves from $(0,0)$ to $(2,4)$ along the parabola $y=x^2$.

The path is given as a graph, and $x$ increases from $0$ to $2$. Set $x=t$, so $y=t^2$, with $0\le t\le2$. This parametrization begins at $(0,0)$ and ends at $(2,4)$:

\mathbf{r}(t)=\langle t,t^2\rangle, \qquad \mathbf{r}'(t)=\langle 1,2t\rangle.

Evaluate the field along the path, then take its dot product with the path velocity:

\mathbf{F}(\mathbf{r}(t)) = \langle t^2,t\rangle, \qquad \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t) = t^2(1)+t(2t) = 3t^2.

Therefore, the work is:

W=\int_0^2 3t^2\,dt = \left[t^3\right]_0^2 = 8.

As a check, using $x$ directly gives $d\mathbf{r}=\langle 1,2x\rangle\,dx$ along $y=x^2$. The field is $\langle x^2,x\rangle$, so its dot product with $d\mathbf{r}/dx$ is $x^2+2x^2=3x^2$. Integrating from $x=0$ to $x=2$ again gives $[x^3]_0^2=8$. The parametrization and limits therefore match the stated direction and produce consistent results.

Worked Example: Work on a Route Crossing a Field Boundary

A particle travels from $(0,0)$ to $(2,0)$, then from $(2,0)$ to $(2,2)$. The force field is $\mathbf{F}(x,y)=\langle x,0\rangle$ for $y<1$, and $\mathbf{F}(x,y)=\langle 0,x\rangle$ for $y\ge1$. Find the total work.

There are two path segments, and the vertical segment crosses the field boundary $y=1$. Evaluate each segment separately. On the horizontal segment, use $\mathbf{r}_1(t)=\langle t,0\rangle$, $0\le t\le2$. The field there is $\langle t,0\rangle$, and $\mathbf{r}_1'(t)=\langle1,0\rangle$. Its work is:

W_1=\int_0^2 \langle t,0\rangle\cdot\langle1,0\rangle\,dt =\int_0^2 t\,dt =\left[\frac{t^2}{2}\right]_0^2 =2.

For the vertical segment, use $\mathbf{r}_2(y)=\langle2,y\rangle$, $0\le y\le2$, so $\mathbf{r}_2'(y)=\langle0,1\rangle$. Split at $y=1$. Below the boundary, the field is $\langle2,0\rangle$, whose dot product with $\langle0,1\rangle$ is zero. At and above the boundary, it is $\langle0,2\rangle$, whose dot product is $2$. Thus:

W_2 = \int_0^1 \langle2,0\rangle\cdot\langle0,1\rangle\,dy + \int_1^2 \langle0,2\rangle\cdot\langle0,1\rangle\,dy = 0+(2)(2-1) = 2.

The total work is $W_1+W_2=2+2=4$. As a check, the horizontal motion contributes positive work because the force points in the direction of travel. On the first part of the vertical motion, the force is perpendicular to the displacement and contributes zero; on the final part, both point upward, giving positive work. The value at the single boundary point does not change these integrals.

Worked Example: Circulation Around a Triangle

Compute the counterclockwise circulation of $\mathbf{F}(x,y)=\langle-y,x\rangle$ around the triangle with vertices $(0,0)$, $(2,0)$, and $(0,1)$, in that order.

Break the closed curve into its three directed sides. For the first side, from $(0,0)$ to $(2,0)$, set $\mathbf{r}_1(t)=\langle2t,0\rangle$, $0\le t\le1$. Then $\mathbf{r}_1'(t)=\langle2,0\rangle$ and $\mathbf{F}(\mathbf{r}_1(t))=\langle0,2t\rangle$, so the dot product is zero and the first integral is $0$.

For the second side, from $(2,0)$ to $(0,1)$, use $\mathbf{r}_2(t)=\langle2-2t,t\rangle$, $0\le t\le1$. The derivative is $\langle-2,1\rangle$, and the field along the side is $\langle-t,2-2t\rangle$. Therefore:

\mathbf{F}(\mathbf{r}_2(t))\cdot\mathbf{r}_2'(t) = (-t)(-2)+(2-2t)(1) = 2.

The integral on this side is $\int_0^1 2\,dt=2$. For the last side, from $(0,1)$ back to $(0,0)$, set $\mathbf{r}_3(t)=\langle0,1-t\rangle$, $0\le t\le1$. Here $\mathbf{r}_3'(t)=\langle0,-1\rangle$ and $\mathbf{F}(\mathbf{r}_3(t))=\langle-(1-t),0\rangle$, giving dot product zero. Adding the three contributions, the counterclockwise circulation is $0+2+0=2$.

As an orientation check, traversing the same triangle clockwise reverses each directed side and changes the total to $-2$. The positive answer for the stated direction is consistent with the calculation: the field contributes positively on the diagonal side and contributes nothing on the other two sides.

Orientation and Physical Interpretation

A signed answer is meaningful, not an error to remove by taking an absolute value. Positive work means the field’s tangential component contributes in the direction of travel overall; negative work means it contributes against that direction overall. A zero result can arise from perpendicular force along the path, or from positive and negative contributions that cancel.

Circulation is likewise signed according to the chosen direction around the closed curve. If the orientation is reversed, the integral changes sign. For a piecewise path, the total is the sum of the signed contributions along its pieces. A negative contribution on one segment must not be discarded simply because another segment contributes positively.

Common Pitfall: Ignoring the Path Direction

Suppose a path goes from $(2,0)$ to $(0,0)$ along the $x$-axis in the field $\mathbf{F}(x,y)=\langle x,0\rangle$. A common mistake is to use the limits $0$ to $2$, calculate $\int_0^2 x\,dx=2$, and report positive work. Those limits describe travel from $(0,0)$ to $(2,0)$, not the requested direction.

For the actual motion, set $\mathbf{r}(t)=\langle2-t,0\rangle$, $0\le t\le2$. Then $\mathbf{r}'(t)=\langle-1,0\rangle$, while $\mathbf{F}(\mathbf{r}(t))=\langle2-t,0\rangle$. The correct result is:

W=\int_0^2 -(2-t)\,dt = -\left[2t-\frac{t^2}{2}\right]_0^2 = -2.

The negative sign is expected because the force points toward increasing $x$, while the particle moves toward decreasing $x$. Reversing the limits of the original integral gives the same check: $\int_2^0 x\,dx=-2$.

Check Your Understanding

For each question, identify the path direction and the field contribution before carrying out the calculation.

What information does $\mathbf{r}'(t)$ provide in the work-integral formula?
A path crosses a boundary where the field changes formula. What should you do before integrating?
Why does reversing the direction of a path change the sign of its work integral?
If the force is perpendicular to the direction of motion at every point, what is the work?
How does the circulation around a closed curve change if its orientation is reversed?