Scalar Line Integrals on Circular Arcs

Why Circular Arcs Have a Convenient Parameter

A circular arc is curved, but its points can be described using one changing quantity: the angle. Once the arc is parametrized this way, a scalar line integral becomes an ordinary one-variable integral. The key is to evaluate the scalar field at the points on the circle and use the correct arc-length element.

Angle bounds do not have to be familiar quadrant angles. An interval such as \([7\pi/6,11\pi/6]\) describes a perfectly well-defined arc, even though both angles lie outside the usual first-quadrant range. The same method works as long as the interval describes the intended sweep along the circle.

Definition: If a curve \(C\) is parametrized by \(\mathbf{r}(t)\), where \(a\le t\le b\), and \(f\) is a scalar field defined on the curve, its scalar line integral is \(\int_C f\,ds=\int_a^b f(\mathbf{r}(t))\|\mathbf{r}'(t)\|\,dt\). Here \(ds\) is the element of arc length, so the speed \(\|\mathbf{r}'(t)\|\) is nonnegative.

This definition separates the calculation into two parts. First, substitute the parametrized point into the field to obtain \(f(\mathbf{r}(t))\). Second, multiply by the speed to account for the length of each small piece of the curve.

Parametrizing a Circular Arc

Let a circle have center \((h,k)\) and radius \(R>0\). The standard angle parametrization is

\mathbf{r}(\theta)=(h+R\cos\theta,\ k+R\sin\theta).

At each angle \(\theta\), the point is \(R\) units from \((h,k)\), because

(x-h)^2+(y-k)^2 =R^2\cos^2\theta+R^2\sin^2\theta =R^2.

For angle bounds \(\alpha\le\beta\), this parametrization starts at angle \(\alpha\) and ends at angle \(\beta\). It moves counterclockwise as the angle increases. The endpoints are found by substituting the bounds, not by guessing from the appearance of the arc.

Why the Arc-Length Factor Is the Radius

Differentiate each coordinate with respect to \(\theta\). The derivative vector is tangent to the circle, and its length gives the speed with respect to angle.

Differentiate the parametrization.
\(\mathbf{r}'(\theta)=(-R\sin\theta,\ R\cos\theta)\).

Compute its length.
\(\|\mathbf{r}'(\theta)\|=\sqrt{R^2\sin^2\theta+R^2\cos^2\theta}=R\), since \(R>0\).

Insert the speed in the line-integral definition.
For increasing angle bounds, \(ds=R\,d\theta\), so \(\int_C f\,ds=R\int_\alpha^\beta f(h+R\cos\theta,k+R\sin\theta)\,d\theta\).

This is the central formula for a circular arc. It follows directly from the definition of the scalar line integral and the computed speed; no separate geometric assumption about the field is needed.

Formula: For the counterclockwise circular arc from angle \(\alpha\) to angle \(\beta\), with \(\alpha\le\beta\), \(\displaystyle \int_C f\,ds=R\int_\alpha^\beta f(h+R\cos\theta,k+R\sin\theta)\,d\theta.\) The interval must describe the arc being integrated. In the examples below it describes one sweep, no longer than one full revolution.

If the same arc is described with decreasing angle, the scalar line integral remains nonnegative with respect to length. For example, set \(\theta(u)=\alpha+(\beta-\alpha)u\) for \(0\le u\le1\), allowing \(\beta<\alpha\). The speed with respect to \(u\) is \(R|\beta-\alpha|\). Substitution in the definition, followed by the change of variable from \(u\) to \(\theta\), gives the angle integral with the smaller bound first. This absolute value is why reversing the direction does not make a scalar line integral negative.

Worked Examples

Worked Example: A Coordinate-Dependent Field on a 60-Degree Arc

Evaluate \(\int_C (x+2y)\,ds\) on the circle \(x^2+y^2=4\), from angle \(\pi/6\) to angle \(\pi/2\).

The circle has center \((0,0)\) and radius \(R=2\). Its parametrization is \((x,y)=(2\cos\theta,2\sin\theta)\), so the field along the circle is

x+2y=2\cos\theta+4\sin\theta.

The angle interval has width \(\pi/2-\pi/6=\pi/3\), which is \(60\) degrees. The arc-length element is \(ds=2\,d\theta\), and therefore

\begin{aligned} \int_C(x+2y)\,ds &=\int_{\pi/6}^{\pi/2}(2\cos\theta+4\sin\theta)\,2\,d\theta\\ &=\int_{\pi/6}^{\pi/2}(4\cos\theta+8\sin\theta)\,d\theta\\ &=\left[4\sin\theta-8\cos\theta\right]_{\pi/6}^{\pi/2}\\ &=(4-0)-\left(2-4\sqrt{3}\right)\\ &=2+4\sqrt{3}. \end{aligned}

At the starting angle \(\pi/6\), the point is \((\sqrt{3},1)\); at the ending angle \(\pi/2\), it is \((0,2)\). These points agree with the parametrization and its order. As a check on the integration, differentiating \(4\sin\theta-8\cos\theta\) gives \(4\cos\theta+8\sin\theta\), the integrand after including \(ds\).

Worked Example: A Coordinate-Dependent Field on the Right-Hand Arc

Evaluate \(\int_C x^2\,ds\) on \(x^2+y^2=4\), using angle bounds \(-\pi/3\) to \(\pi/3\).

Again \(R=2\), and \(x=2\cos\theta\). Thus \(x^2=4\cos^2\theta\), while \(ds=2\,d\theta\). The bounds span \(2\pi/3\), and describe the right-hand arc. The integral is

\begin{aligned} \int_C x^2\,ds &=\int_{-\pi/3}^{\pi/3}4\cos^2\theta\cdot2\,d\theta\\ &=8\int_{-\pi/3}^{\pi/3}\cos^2\theta\,d\theta\\ &=\left[4\theta+2\sin(2\theta)\right]_{-\pi/3}^{\pi/3}\\ &=\left(\frac{4\pi}{3}+\sqrt{3}\right) -\left(-\frac{4\pi}{3}-\sqrt{3}\right)\\ &=\frac{8\pi}{3}+2\sqrt{3}. \end{aligned}

At the two endpoints, the points are \((2,-\sqrt{3})\) and \((2,\sqrt{3})\), respectively. They are both on the right side of the circle, as required by these bounds. To check the antiderivative, differentiate \(4\theta+2\sin(2\theta)\): the result is \(4+4\cos(2\theta)=8\cos^2\theta\), matching the integrand.

Worked Example: A Radial Integrand on an Arc With Nonstandard Bounds

Evaluate \(\int_C \sqrt{x^2+y^2}\,ds\) on the circle \(x^2+y^2=9\), from angle \(7\pi/6\) to angle \(11\pi/6\).

The radius is \(3\), and every point on this circle satisfies \(\sqrt{x^2+y^2}=3\). The angle difference is \(11\pi/6-7\pi/6=2\pi/3\); these bounds describe a \(120\)-degree arc through the bottom of the circle. Since \(ds=3\,d\theta\),

\begin{aligned} \int_C\sqrt{x^2+y^2}\,ds &=\int_{7\pi/6}^{11\pi/6}3\cdot3\,d\theta\\ &=9\left(\frac{11\pi}{6}-\frac{7\pi}{6}\right)\\ &=9\cdot\frac{2\pi}{3}\\ &=6\pi. \end{aligned}

The endpoints are \((-3\sqrt{3}/2,-3/2)\) and \((3\sqrt{3}/2,-3/2)\), respectively, confirming that the arc runs from the lower-left point to the lower-right point. The integrand is \(3\) everywhere on the circle, and the arc length is \(3(2\pi/3)=2\pi\). Their product is \(6\pi\), which independently checks the result.

Radial Fields and the Arc Angle

The last example illustrates a useful simplification. When the circle is centered at the origin and the field depends only on the distance from the origin, its value is constant along the circle. More precisely, if \(f(x,y)=g(\sqrt{x^2+y^2})\), then on the circle of radius \(R\), \(f=g(R)\). For increasing bounds spanning angle \(\beta-\alpha\), the integral becomes

\int_C f\,ds =R\int_\alpha^\beta g(R)\,d\theta =g(R)R(\beta-\alpha).

This reduction is appropriate only when the field really depends on the distance from the origin and the circle is centered there. On a circle centered elsewhere, the distance \(\sqrt{x^2+y^2}\) generally changes along the arc. Likewise, a field such as \(x+y\) is not radial: its value can vary as the angle changes, so it must be substituted into the angle integral.

A Common Mistake: Omitting the Radius

A frequent error is to replace \(ds\) by \(d\theta\). Angle change is not arc length: on a circle of radius \(R\), the arc-length element is \(R\,d\theta\). For example, in the right-hand arc calculation, omitting the factor \(2\) would give the incorrect result

\int_{-\pi/3}^{\pi/3}4\cos^2\theta\,d\theta =\frac{4\pi}{3}+\sqrt{3},

which is half the correct value \(\frac{8\pi}{3}+2\sqrt{3}\). The correction is to multiply by the radius \(2\) before integrating. A separate warning applies when the angle decreases: using signed bounds without accounting for the speed can produce a negative answer. A scalar line integral uses the nonnegative speed, so reversal of direction does not change its value.

Why This Method Is Useful

Circular arcs appear whenever a curve follows part of a circle, including paths around a fixed center and boundaries made of curved pieces. The angle parameter gives a consistent way to express the coordinates, while the radius converts angle change into actual distance along the curve. Together, these facts make it possible to account for both the field’s variation and the length of the path.

For any calculation, check the setup before evaluating the ordinary integral: confirm the center and radius, substitute the angle bounds to identify the endpoints, evaluate the field along the parametrization, and include \(R\) in \(ds\). These checks catch errors that may otherwise leave an integral looking plausible while representing the wrong arc or the wrong length.

Check Your Understanding

Use the parametrization and arc-length factor to answer these questions.

What parametrization describes the circle centered at \((2,-1)\) with radius \(4\)?
What is \(ds\) for an angle parameter on a circle of radius \(5\)?
For the origin-centered circle of radius \(3\), what angle interval describes a \(60\)-degree arc starting at angle \(-\pi/4\) and moving counterclockwise?
Why is \(\sqrt{x^2+y^2}\) constant on an origin-centered circle, and what is its value on a circle of radius \(R\)?
What error occurs if an angle integral for a circle of radius \(R\) uses \(d\theta\) in place of \(R\,d\theta\)?