Work Along Ellipses in Linear Force Fields

Why Ellipses Give a Useful Work Calculation

A closed path can return an object to its starting point and still require nonzero work from a force field. For a linear force field in the plane, work around an ellipse has a particularly simple form: the answer depends on the ellipse’s two semi-axis lengths and on two coefficients of the force field.

The calculation is useful because it separates three features that can otherwise be easy to confuse: the field, the size and shape of the ellipse, and the direction and speed of traversal. In particular, an ellipse need not be a circle, and the point on the ellipse need not move at a uniform rate.

Set-Up and Notation

Write the position as \(\mathbf{r}(t)=(x(t),y(t))\). For a continuously differentiable parametrization on \([u,v]\), the work of a force field \(\mathbf{F}(x,y)=(P(x,y),Q(x,y))\) along the parametrized curve is

W=\int_u^v \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt =\int_u^v \bigl(P(x(t),y(t))x'(t)+Q(x(t),y(t))y'(t)\bigr)\,dt.

Here, \(x'(t)\) and \(y'(t)\) are the components of the velocity. Thus, work accumulates from the component of force in the direction of motion.

Definition: A homogeneous linear planar force field has the form \(\mathbf{F}(x,y)=(\alpha x+\beta y,\gamma x+\delta y)\), where \(\alpha,\beta,\gamma,\delta\) are constants. A translated ellipse with center \((h,k)\) and positive semi-axis lengths \(a,b\) can be parametrized as \(x=h+a\cos\theta,\ y=k+b\sin\theta\). Increasing \(\theta\) traces it counterclockwise; decreasing \(\theta\) traces it clockwise.

For one traversal, the angle changes by \(2\pi\) in magnitude. Its rate of change may vary: \(\theta\) can be any continuously differentiable monotone function that goes from \(0\) to \(2\pi\) for a counterclockwise traversal, or from \(0\) to \(-2\pi\) for a clockwise traversal.

The Work Formula for an Ellipse

Theorem: For the linear field \(\mathbf{F}(x,y)=(\alpha x+\beta y,\gamma x+\delta y)\), work along an ellipse with semi-axis lengths \(a,b\), traversed once, is \(W=\pi ab(\gamma-\beta)\) counterclockwise and \(W=-\pi ab(\gamma-\beta)\) clockwise. The center \((h,k)\) and the speed of traversal do not affect the result.

The coefficients \(\alpha\) and \(\delta\), which multiply \(x\) in the first component and \(y\) in the second, do not contribute to the work around the closed ellipse. The cross-component coefficients \(\beta\) and \(\gamma\) determine the result through their difference. Here is a derivation that also shows where the orientation and axis lengths enter.

Why the Formula Works

First suppose the ellipse is traversed counterclockwise with angle \(\theta\) increasing from \(0\) to \(2\pi\). Along the ellipse, \(dx=-a\sin\theta\,d\theta\) and \(dy=b\cos\theta\,d\theta\). The work integral can be written as

W=\oint \bigl((\alpha x+\beta y)\,dx+(\gamma x+\delta y)\,dy\bigr).

The terms \(\alpha x\,dx\) and \(\delta y\,dy\) integrate to zero on a closed curve, since \(x\,dx=\tfrac12 d(x^2)\) and \(y\,dy=\tfrac12 d(y^2)\). For the remaining terms, substitute the ellipse parametrization:

\begin{aligned} \oint y\,dx &=\int_0^{2\pi}(k+b\sin\theta)(-a\sin\theta)\,d\theta\\ &=-ak\int_0^{2\pi}\sin\theta\,d\theta -ab\int_0^{2\pi}\sin^2\theta\,d\theta\\ &=0-ab\pi=-\pi ab, \end{aligned}

\begin{aligned} \oint x\,dy &=\int_0^{2\pi}(h+a\cos\theta)b\cos\theta\,d\theta\\ &=bh\int_0^{2\pi}\cos\theta\,d\theta +ab\int_0^{2\pi}\cos^2\theta\,d\theta\\ &=0+ab\pi=\pi ab. \end{aligned}

These evaluations use the standard integrals of sine and cosine over a full period: the integrals of \(\sin\theta\) and \(\cos\theta\) are zero, while those of \(\sin^2\theta\) and \(\cos^2\theta\) are \(\pi\). Combining the two cross terms gives

W=\beta\oint y\,dx+\gamma\oint x\,dy =-\beta\pi ab+\gamma\pi ab =\pi ab(\gamma-\beta).

For a clockwise traversal, the angle runs through the same values in the opposite direction. Reversing the direction of a work integral changes its sign, so the answer is \(-\pi ab(\gamma-\beta)\). This also explains why a nonuniform parametrization does not alter the result: the integral depends on the path and its orientation, not on how quickly the path is traversed. In the substitution above, the factor \(\theta'(t)\) in the velocity is accounted for by integrating with respect to \(\theta\).

Worked Examples

Worked Example: An Unequal, Translated Ellipse

Find the work of \(\mathbf{F}(x,y)=(2x+3y,-x+4y)\) along the counterclockwise ellipse \(x=1+2\cos t,\ y=-1+\sin t\), for \(0\leq t\leq 2\pi\).

The semi-axis lengths are \(a=2\) and \(b=1\). The cross-component coefficients are \(\beta=3\) and \(\gamma=-1\). The formula therefore predicts

W=\pi(2)(1)(-1-3)=-8\pi.

To check by direct integration, put \(c=\cos t\) and \(s=\sin t\). The position, velocity, and force along the curve are

\begin{aligned} x&=1+2c, & y&=-1+s, & (x',y')&=(-2s,c),\\ P&=2(1+2c)+3(-1+s)=-1+4c+3s,\\ Q&=-(1+2c)+4(-1+s)=-5-2c+4s. \end{aligned}

The integrand is

\begin{aligned} Px'+Qy' &=(-1+4c+3s)(-2s)+(-5-2c+4s)c\\ &=2s-8cs-6s^2-5c-2c^2+4sc\\ &=2s-5c-4cs-6s^2-2c^2. \end{aligned}

Over \(0\leq t\leq 2\pi\), the integrals of \(s\), \(c\), and \(cs\) are zero, and the integrals of \(s^2\) and \(c^2\) are each \(\pi\). Hence

W=0-0-0-6\pi-2\pi=-8\pi.

The direct calculation agrees with the formula, including the negative sign.

Worked Example: Nonuniform Counterclockwise Traversal

Let \(\mathbf{F}(x,y)=(-2y,5x)\) and let the ellipse be \(x=3\cos\theta,\ y=2\sin\theta\), where \(\theta(t)=t+\tfrac12\sin t\) for \(0\leq t\leq 2\pi\). Find the work.

First check that this parametrization makes one counterclockwise traversal. Its angle starts at \(0\), ends at \(2\pi\), and satisfies \(\theta'(t)=1+\tfrac12\cos t\geq\tfrac12>0\). Thus, the angle increases throughout the traversal, though not at a constant rate. The field has \(\beta=-2\) and \(\gamma=5\), so the formula gives

W=\pi(3)(2)\bigl(5-(-2)\bigr)=42\pi.

For a direct check, differentiate the position, keeping the factor \(\theta'(t)\):

\begin{aligned} x'&=-3\sin\theta\,\theta'(t),& y'&=2\cos\theta\,\theta'(t),\\ P&=-2y=-4\sin\theta,& Q&=5x=15\cos\theta. \end{aligned}

Therefore,

\begin{aligned} Px'+Qy' &=(-4\sin\theta)(-3\sin\theta\,\theta') +(15\cos\theta)(2\cos\theta\,\theta')\\ &=\bigl(12\sin^2\theta+30\cos^2\theta\bigr)\theta'. \end{aligned}

Since \(d\theta=\theta'(t)\,dt\), change variables to the angle. The limits are \(0\) and \(2\pi\), so

\begin{aligned} W &=\int_0^{2\pi}\bigl(12\sin^2\theta+30\cos^2\theta\bigr)\,d\theta\\ &=12\pi+30\pi=42\pi. \end{aligned}

The direct computation confirms that the nonuniform speed changes the integrand but not the total work.

Worked Example: Clockwise Traversal of a Translated Ellipse

Find the work of \(\mathbf{F}(x,y)=(x-2y,3x+y)\) along the clockwise ellipse \(x=1+\cos t,\ y=-2-3\sin t\), for \(0\leq t\leq2\pi\).

The semi-axis lengths are \(a=1\), \(b=3\). The angle is \(-t\), so the traversal is clockwise. Here \(\beta=-2\) and \(\gamma=3\). Therefore,

W=-\pi(1)(3)\bigl(3-(-2)\bigr)=-15\pi.

To check, let \(c=\cos t\), \(s=\sin t\). Differentiating the given parametrization gives \(x'=-s\) and \(y'=-3c\). The force components on the curve are

\begin{aligned} P&=(1+c)-2(-2-3s)=5+c+6s,\\ Q&=3(1+c)+(-2-3s)=1+3c-3s. \end{aligned}

Thus,

\begin{aligned} Px'+Qy' &=(5+c+6s)(-s)+(1+3c-3s)(-3c)\\ &=-5s-cs-6s^2-3c-9c^2+9sc\\ &=-5s-3c+8sc-6s^2-9c^2. \end{aligned}

The sine, cosine, and sine-cosine terms integrate to zero over a full period. The remaining terms give

W=-6\pi-9\pi=-15\pi.

This verifies the formula and shows explicitly how the clockwise orientation produces a negative result in this example.

A Common Error: Using the Wrong Area or Sign

For an ellipse with semi-axis lengths \(a\) and \(b\), the factor in the work formula is \(\pi ab\), not \(\pi a^2\). The latter is the area factor for a circle of radius \(a\). For example, the counterclockwise ellipse in the nonuniform-traversal example has \(a=3\), \(b=2\), and \(\gamma-\beta=7\). Using \(\pi a^2\) would incorrectly give \(7\pi(3^2)=63\pi\). The correct calculation is

W=7\pi ab=7\pi(3)(2)=42\pi.

A second sign error is to use the counterclockwise formula for a clockwise parametrization. Always check whether the angle increases or decreases over the traversal. The formula’s sign records orientation, not speed.

When the Result Applies

The stated formula applies to a homogeneous linear force field and a single traversal of an ellipse parametrized smoothly with the specified orientation. The ellipse can be translated and its axis lengths can differ. A constant force added to the field would not change the closed-curve work: for a constant vector \(\mathbf{c}\), its contribution is \(\mathbf{c}\cdot\int_u^v\mathbf{r}'(t)\,dt =\mathbf{c}\cdot(\mathbf{r}(v)-\mathbf{r}(u))=0\) when the path is closed. Thus, the formula also holds for an affine field formed by adding a constant vector to the stated linear field.

The result matters because it lets you compute work without tracking every detail of the motion. The field’s two cross-component coefficients determine the signed contribution, the product \(ab\) accounts for both dimensions of the ellipse, and the direction of traversal determines the sign. The integral itself remains the check: substitute the force and velocity, integrate over the parameter interval, and compare with the formula.

Check Your Understanding

Use the ellipse-work formula and the work-integral definition to test your understanding.

For a counterclockwise ellipse with semi-axis lengths \(4\) and \(1\), and a field \(\mathbf{F}(x,y)=(2x-y,3x+5y)\), what is the work?
What changes in the work formula when the same ellipse is traversed clockwise?
Why do the coefficients multiplying \(x\) in the first force component and \(y\) in the second not affect the closed-ellipse work?
For \(\theta(t)=t+\tfrac14\sin t\) on \(0\leq t\leq2\pi\), verify that the ellipse is traversed counterclockwise and nonuniformly.
For an ellipse with semi-axis lengths \(a=5\), \(b=2\) and \(\gamma-\beta=-3\), compute the counterclockwise work. What is the clockwise work?