Component Form of Work Integrals

From a Vector Field to Coordinate Integrals

A force field in three-dimensional space can have a different force at every point. To calculate the work done along a specified path, we combine the field with the small displacements made along that path. The result is a vector line integral, and its component form shows exactly how each coordinate direction contributes.

Suppose a path is parametrized by $\mathbf{r}(t)=(x(t),y(t),z(t))$, for $a\leq t\leq b$, and a vector field is $\mathbf{F}(x,y,z)=(P(x,y,z),Q(x,y,z),R(x,y,z))$. The functions $P,Q,R$ are the field's $x$-, $y$-, and $z$-components. The work along the parametrized path is the integral of the field, evaluated on the path, dotted with the path's velocity.

Definition: If $P,Q,R$ are continuous along a continuously differentiable curve $\mathbf{r}(t)=(x(t),y(t),z(t))$, its work integral is $$ \int_C \mathbf{F}\cdot d\mathbf{r} = \int_a^b \mathbf{F}(\mathbf{r}(t))\cdot \mathbf{r}'(t)\,dt. $$ Here $C$ is traversed as $t$ increases from $a$ to $b$.

Since $\mathbf{r}'(t)=(x'(t),y'(t),z'(t))$, the dot product gives a sum of three products. This is the component form that we will use throughout the tutorial.

\int_C \mathbf{F}\cdot d\mathbf{r} = \int_a^b \left[ P(x(t),y(t),z(t))x'(t) + Q(x(t),y(t),z(t))y'(t) + R(x(t),y(t),z(t))z'(t) \right]dt.

Equivalently, the coordinate differentials are $dx=x'(t)\,dt$, $dy=y'(t)\,dt$, and $dz=z'(t)\,dt$. Thus the same formula is often written as

\int_C P\,dx+Q\,dy+R\,dz = \int_C P\,dx+\int_C Q\,dy+\int_C R\,dz.

In each integral on the left, the field component is evaluated at the point $(x,y,z)$ on the curve. In an actual calculation, one substitutes $x(t),y(t),z(t)$ into that component and replaces the corresponding differential by its parameter expression. A derivative such as $x'(t)$ can be positive, negative, or zero, and its sign is part of the contribution.

Why the Component Formula Works

The formula follows directly from the dot product and the chain rule. A small change in the curve has coordinate changes $dx,dy,dz$. The work contribution from the field $(P,Q,R)$ is the dot product with that displacement, namely $P\,dx+Q\,dy+R\,dz$. Along a parametrized curve, each coordinate change per unit parameter is its derivative.

Derivation: The chain rule gives $dx=x'(t)\,dt$, $dy=y'(t)\,dt$, and $dz=z'(t)\,dt$. Therefore $$ P\,dx+Q\,dy+R\,dz = \bigl(Px'(t)+Qy'(t)+Rz'(t)\bigr)\,dt, $$ where each field component is evaluated at $\mathbf{r}(t)$. Integrating this equality from $t=a$ to $t=b$ gives the component formula. In particular, the three coordinate integrals add because the dot product itself is the sum of its three component products.

This also explains why the parameter derivatives cannot be omitted: they convert a change in the parameter into the actual displacement in each coordinate. The field values alone do not describe how the path moves.

Worked Examples

Worked Example: A Polynomial Curve

Let $\mathbf{F}(x,y,z)=(x^2,xy,z)$ and $\mathbf{r}(t)=(t,t^2,t^3)$, for $0\leq t\leq1$. Compute the work integral.

Here $P=x^2$, $Q=xy$, and $R=z$. Along the curve, $x=t$, $y=t^2$, and $z=t^3$, so $dx=dt$, $dy=2t\,dt$, and $dz=3t^2\,dt$. Substituting into each component contribution gives

\begin{aligned} \int_C P\,dx &=\int_0^1 (t^2)(1)\,dt =\left[\frac{t^3}{3}\right]_0^1 =\frac13,\\ \int_C Q\,dy &=\int_0^1 (t\cdot t^2)(2t)\,dt =\int_0^1 2t^4\,dt =\left[\frac{2t^5}{5}\right]_0^1 =\frac25,\\ \int_C R\,dz &=\int_0^1 (t^3)(3t^2)\,dt =\int_0^1 3t^5\,dt =\left[\frac{t^6}{2}\right]_0^1 =\frac12. \end{aligned}

Adding the three contributions, the work is $\frac13+\frac25+\frac12=\frac{10+12+15}{30}=\frac{37}{30}$. To check the calculation, the combined integrand is $t^2+2t^4+3t^5$, whose antiderivative is $\frac{t^3}{3}+\frac{2t^5}{5}+\frac{t^6}{2}$. Its value from $0$ to $1$ is again $\frac{37}{30}$.

Worked Example: A Curve with a Trigonometric Coordinate

Let $\mathbf{F}(x,y,z)=(x+y,yz,xz)$ and $\mathbf{r}(t)=(\cos t,\sin t,t)$, for $0\leq t\leq\frac{\pi}{2}$. Find the work integral.

The field components are $P=x+y$, $Q=yz$, and $R=xz$. On the curve, $x=\cos t$, $y=\sin t$, and $z=t$. The coordinate derivatives are $x'=-\sin t$, $y'=\cos t$, and $z'=1$. Therefore the three contributions are

\begin{aligned} \int_C P\,dx &=\int_0^{\pi/2}(\cos t+\sin t)(-\sin t)\,dt\\ &=-\int_0^{\pi/2}\cos t\sin t\,dt -\int_0^{\pi/2}\sin^2t\,dt =-\frac12-\frac{\pi}{4},\\ \int_C Q\,dy &=\int_0^{\pi/2}(t\sin t)\cos t\,dt =\frac{\pi}{8},\\ \int_C R\,dz &=\int_0^{\pi/2}(t\cos t)(1)\,dt =\left[t\sin t+\cos t\right]_0^{\pi/2} =\frac{\pi}{2}-1. \end{aligned}

For the middle integral, $\sin t\cos t=\frac12\sin(2t)$, and integration by parts gives $\int_0^{\pi/2}t\sin t\cos t\,dt=\frac{\pi}{8}$. For the first integral, $\int_0^{\pi/2}\cos t\sin t\,dt=\frac12$, while $\int_0^{\pi/2}\sin^2t\,dt=\frac{\pi}{4}$. Adding all three results yields $-\frac12-\frac{\pi}{4}+\frac{\pi}{8}+\frac{\pi}{2}-1=\frac{\pi}{8}-1$. The combined integral is therefore $\int_0^{\pi/2}[-\cos t\sin t-\sin^2t+t\sin t\cos t+t\cos t]\,dt$, and the evaluated contributions above verify its value $\frac{\pi}{8}-1$.

Worked Example: A Coordinate That Does Not Change

Let $\mathbf{F}(x,y,z)=(y,z,x)$ and $\mathbf{r}(t)=(t,1,t^2)$, for $0\leq t\leq1$. Compute the work integral and identify the contribution from $dy$.

Along the curve, $x=t$, $y=1$, and $z=t^2$. Thus $dx=dt$, $dy=0\,dt$, and $dz=2t\,dt$. Since $P=y$, $Q=z$, and $R=x$, substitution gives

\begin{aligned} \int_C P\,dx &=\int_0^1 1\cdot 1\,dt=1,\\ \int_C Q\,dy &=\int_0^1 t^2\cdot 0\,dt=0,\\ \int_C R\,dz &=\int_0^1 t\cdot 2t\,dt =\left[\frac{2t^3}{3}\right]_0^1 =\frac23. \end{aligned}

The total is $1+0+\frac23=\frac53$. As a check, the full integrand from the dot product is $1+2t^2$, whose integral from $0$ to $1$ is $1+\frac23=\frac53$. The $Q\,dy$ term vanishes because $y$ is constant along this path, not because the field component $Q$ is zero.

A Reliable Calculation Procedure

For a parametrized curve and a field given in components, organizing the substitutions prevents most errors. Keep the field components, coordinate derivatives, and resulting integrals distinct until each contribution is clear.

Record the parameter interval.
Write the curve as $\mathbf{r}(t)=(x(t),y(t),z(t))$ for $a\leq t\leq b$.

Differentiate the coordinates.
Find $x'(t)$, $y'(t)$, and $z'(t)$, so that the coordinate differentials are $x'(t)\,dt$, $y'(t)\,dt$, and $z'(t)\,dt$.

Evaluate the field on the curve.
Substitute $x(t),y(t),z(t)$ into $P,Q,R$, respectively.

Form and integrate each term.
Multiply $P(\mathbf{r}(t))$ by $x'(t)$, $Q(\mathbf{r}(t))$ by $y'(t)$, and $R(\mathbf{r}(t))$ by $z'(t)$, then integrate each over $[a,b]$.

Add and check.
Add the three values, and compare with the integral of their combined integrand.

A Common Mistake: Dropping the Coordinate Derivatives

A tempting but incorrect calculation is to substitute the field along the curve and then integrate its components directly with respect to $t$, without multiplying by the coordinate derivatives. For instance, in the polynomial example above, this mistake would give $\int_0^1(t^2+t^3+t^3)\,dt$: it uses the field values $P=t^2$, $Q=t^3$, and $R=t^3$, but treats each coordinate differential as though it were $dt$. The resulting value is $\frac13+\frac14+\frac14=\frac56$, which is not the work integral.

The correction is to use the differential associated with each component. In that example, $dx=dt$, but $dy=2t\,dt$ and $dz=3t^2\,dt$. Thus the correct integrand is $t^2+2t^4+3t^5$, and the work is $\frac{37}{30}$. A quick diagnostic is to check whether every term has the appropriate coordinate derivative: $P$ pairs with $x'$, $Q$ with $y'$, and $R$ with $z'$.

Interpreting the Sum

The component form separates a single work integral into coordinate contributions, but those contributions are not separate paths or separate problems. They are the three terms in the same dot product, evaluated along the same parametrization and over the same parameter interval. A component can contribute positively, negatively, or zero, depending on the field value and the corresponding coordinate derivative.

This form is useful when the field and curve are given by explicit coordinate formulas: it turns the geometric integral into ordinary one-variable integrals. It also makes the calculation auditable. If a result seems wrong, one can inspect whether the field was evaluated on the curve, whether each component was paired with the correct coordinate derivative, and whether the three evaluated contributions were added correctly.

Check Your Understanding

Use the component formula to answer the following questions.

For $\mathbf{F}=(P,Q,R)$ and $\mathbf{r}(t)=(x(t),y(t),z(t))$, which coordinate derivative multiplies $Q$ in the parameter integral?
For $\mathbf{r}(t)=(t,1,t^2)$, which coordinate differential is zero, and why?
If $P(\mathbf{r}(t))=t$ and $x'(t)=-2$, what is the $P\,dx$ contribution to the parameter integrand?
For $\mathbf{F}=(x,y,z)$ along $\mathbf{r}(t)=(t,t,t)$, $0\leq t\leq1$, write the three component integrals and compute their sum.
Why is it incorrect in general to integrate $P(\mathbf{r}(t))+Q(\mathbf{r}(t))+R(\mathbf{r}(t))$ with respect to $t$ without using the coordinate derivatives?