Coordinate-Weighted Integrals on Circular Wires

Why Coordinate-Weighted Integrals on Circles Are Useful

A scalar line integral adds up values of a function along a curve, with each small contribution weighted by the length of the curve element. When the function is a coordinate such as $x$ or $y$, the integral records how that coordinate is distributed along the wire. Functions such as $x^2$ and $xy$ provide further information about how points on the wire are arranged relative to the coordinate axes.

For a circular wire, trigonometric parametrization turns these integrals into ordinary one-variable integrals. The formulas also make the role of the circle’s center, radius, and angular interval explicit.

Definition: If a curve $C$ is parametrized by $\mathbf{r}(t)=(x(t),y(t))$, for $t$ in $[\alpha,\beta]$, its scalar line integral of a function $f$ is $$ \int_C f\,ds=\int_\alpha^\beta f(x(t),y(t))\|\mathbf{r}'(t)\|\,dt. $$ Here $ds=\|\mathbf{r}'(t)\|\,dt$ is the arc-length element.

We will consider a circle with center $(a,b)$ and radius $R>0$, parametrized by $\mathbf{r}(t)=(a+R\cos t,\ b+R\sin t)$. On an angular interval $[\alpha,\beta]$ that traces the chosen arc once, $ds=R\,dt$. Thus, for any function $f$, the integral on that arc is $R\int_\alpha^\beta f(a+R\cos t,b+R\sin t)\,dt$.

Formulas for an Arc

Substituting the circle’s coordinates into the four functions gives integrands involving $\cos t$, $\sin t$, and their products. The following formulas result from integrating those expressions. They apply to any specified arc represented by the stated interval.

Arc formulas: For a circle with center $(a,b)$, radius $R$, and angular interval $[\alpha,\beta]$, let $\Delta=\beta-\alpha$. Then $$ \int_C x\,ds =R\left[a\Delta+R(\sin\beta-\sin\alpha)\right], $$ $$ \int_C y\,ds =R\left[b\Delta+R(\cos\alpha-\cos\beta)\right], $$ $$ \int_C x^2\,ds =R\left[a^2\Delta+2aR(\sin\beta-\sin\alpha) +R^2\left(\frac{\Delta}{2}+\frac{\sin 2\beta-\sin 2\alpha}{4}\right)\right], $$ and $$ \int_C xy\,ds =R\left[ab\Delta+aR(\cos\alpha-\cos\beta) +bR(\sin\beta-\sin\alpha) +\frac{R^2}{2}(\sin^2\beta-\sin^2\alpha)\right]. $$

Why the Arc Formulas Work

Here is the derivation, including the trigonometric integrations that produce the squared and mixed-coordinate terms. Since $x=a+R\cos t$ and $ds=R\,dt$, the integral of $x$ is $R\int_\alpha^\beta(a+R\cos t)\,dt$. Using $\int\cos t\,dt=\sin t$ gives the first formula. Similarly, $y=b+R\sin t$, and $\int\sin t\,dt=-\cos t$, which gives the formula for $y$.

For $x^2$, expand the square and use the identity $\cos^2t=(1+\cos 2t)/2$:

\begin{aligned} \int_C x^2\,ds &=R\int_\alpha^\beta(a+R\cos t)^2\,dt\\ &=R\int_\alpha^\beta(a^2+2aR\cos t+R^2\cos^2t)\,dt\\ &=R\left[a^2\Delta+2aR(\sin\beta-\sin\alpha) +R^2\left(\frac{\Delta}{2}+\frac{\sin 2\beta-\sin 2\alpha}{4}\right)\right]. \end{aligned}

For the product $xy$, first expand: $xy=ab+aR\sin t+bR\cos t+R^2\cos t\sin t$. The last term has antiderivative $\sin^2t/2$, because the derivative of $\sin^2t/2$ is $\sin t\cos t$. Integrating each term and multiplying by $R$ gives the stated formula for $xy$. These derivations work because the parametrization has constant speed $R$, so the arc-length factor is constant throughout the interval.

Full-Circle Formulas and Symmetry

For a full circle, take an interval of length $2\pi$, such as $[0,2\pi]$. The sine and cosine endpoint differences in the arc formulas vanish. The resulting expressions are especially simple:

Full-circle formulas: For the full circle with center $(a,b)$ and radius $R$, $$ \int_C x\,ds=2\pi aR,\qquad \int_C y\,ds=2\pi bR, $$ $$ \int_C x^2\,ds=2\pi a^2R+\pi R^3,\qquad \int_C xy\,ds=2\pi abR. $$

For a circle centered at the origin, these reduce to $\int_C x\,ds=0$, $\int_C y\,ds=0$, $\int_C x^2\,ds=\pi R^3$, and $\int_C xy\,ds=0$. There is also a geometric explanation for the zero values: reflecting the circle across the $y$-axis pairs each point with one having the opposite $x$-coordinate, so the contributions to the integral of $x$ cancel. Reflection across the $x$-axis gives the corresponding cancellation for $y$. For $xy$, reflection across either axis changes the sign of the product, so its contributions cancel as well. By contrast, $x^2$ is unchanged by either reflection, so its contributions do not cancel.

Worked Examples

Worked Example: Coordinate Integrals on an Off-Center Full Circle

Find the integrals of $x$, $y$, $x^2$, and $xy$ over the full circle with center $(2,-1)$ and radius $3$.

Here $a=2$, $b=-1$, and $R=3$. Substituting into the full-circle formulas gives $\int_C x\,ds=2\pi(2)(3)=12\pi$ and $\int_C y\,ds=2\pi(-1)(3)=-6\pi$.

For the square-coordinate integral, $\int_C x^2\,ds=2\pi(2^2)(3)+\pi(3^3)=24\pi+27\pi=51\pi$. For the product, $\int_C xy\,ds=2\pi(2)(-1)(3)=-12\pi$.

As a check, the circle’s circumference is $2\pi R=6\pi$, so its average $x$-coordinate is $12\pi/(6\pi)=2$, its center’s $x$-coordinate. Its average $y$-coordinate is $-6\pi/(6\pi)=-1$, also the center’s coordinate. The values are consistent with symmetry about the circle’s center.

Worked Example: Four Integrals on a Quarter Circle

Evaluate the integrals of $x$, $y$, $x^2$, and $xy$ on the first-quadrant quarter of the circle $x^2+y^2=4$.

Use $a=b=0$, $R=2$, $\alpha=0$, and $\beta=\pi/2$. The parametrization is $x=2\cos t$, $y=2\sin t$, with $ds=2\,dt$. Directly, $\int_C x\,ds=2\int_0^{\pi/2}2\cos t\,dt=4[\sin t]_0^{\pi/2}=4$. Likewise, $\int_C y\,ds=2\int_0^{\pi/2}2\sin t\,dt=4[-\cos t]_0^{\pi/2}=4$.

For $x^2$, $\int_C x^2\,ds=2\int_0^{\pi/2}4\cos^2t\,dt =8\int_0^{\pi/2}(1+\cos 2t)/2\,dt =4[\ t+\sin(2t)/2\ ]_0^{\pi/2}=2\pi$. For $xy$, $\int_C xy\,ds=2\int_0^{\pi/2}4\cos t\sin t\,dt =8[\sin^2t/2]_0^{\pi/2}=4$.

Checking with the arc formulas gives $2(2)=4$ for each of the first two integrals. For $x^2$, the formula gives $2\cdot4(\pi/4)=2\pi$; for $xy$, it gives $2\cdot(4/2)(1-0)=4$. Each result agrees with direct integration.

Worked Example: A Right Semicircle with a Shifted Center

Find all four integrals on the right semicircle of the circle with center $(1,2)$ and radius $2$. Use $-\pi/2\le t\le\pi/2$.

Here $\Delta=\pi$, $\sin\beta-\sin\alpha=1-(-1)=2$, and $\cos\alpha-\cos\beta=0-0=0$. Thus $\int_C x\,ds=2[1\cdot\pi+2\cdot2]=2\pi+8$, and $\int_C y\,ds=2[2\pi+2\cdot0]=4\pi$.

For $x^2$, $\sin(2\beta)-\sin(2\alpha)=\sin\pi-\sin(-\pi)=0$, so $\int_C x^2\,ds =2[1^2\pi+2(1)(2)(2)+2^2(\pi/2)] =2[\pi+8+2\pi]=6\pi+16$.

For $xy$, the squared-sine endpoint difference is $1-1=0$. Therefore $\int_C xy\,ds =2[(1)(2)\pi+(1)(2)(0)+(2)(2)(2)+\frac{2^2}{2}(0)] =4\pi+16$.

As a check, the arc has length $R\Delta=2\pi$. Dividing the first two integrals by this length gives average coordinates $1+4/\pi$ and $2$, respectively. The semicircle is symmetric about the horizontal line $y=2$, so its average $y$-coordinate is indeed $2$. The nonzero shift in the average $x$-coordinate reflects that this is the right half of the circle.

A Common Pitfall: Forgetting the Arc-Length Factor

The parametrization supplies both the coordinate functions and the speed. The speed must be included: on a circle of radius $R$, $\|\mathbf{r}'(t)\|=R$, so $ds=R\,dt$. For example, on the first-quadrant arc of radius $2$, a mistaken calculation for $\int_C x\,ds$ might omit the speed and write $\int_0^{\pi/2}2\cos t\,dt=2$. That is the integral with respect to $dt$, not arc length.

The correction is to include $ds=2\,dt$: $\int_C x\,ds=\int_0^{\pi/2}(2\cos t)(2)\,dt=4$. Substituting the result into the definition confirms the correction: the coordinate value is $2\cos t$, and every parameter increment contributes a length factor $2\,dt$. A useful check is that omitting this factor gives the wrong units as well as the wrong value: an integral of a coordinate with respect to arc length has units of length squared.

Using Symmetry Before Calculating

On a full circle centered at the origin, symmetry can identify the zero integrals immediately. It can also simplify work on arcs: for example, the right semicircle in the previous example is symmetric about $y=2$, which explains its average $y$-coordinate. But symmetry must match both the curve and the integrand. On an off-center circle, $x$ and $y$ do not generally cancel over the full circle; their integrals depend on the center coordinates. The full-circle formulas keep this distinction explicit.

For a specified arc, the endpoints matter. A cancellation that holds over a full circle need not hold on a quarter circle or semicircle. When symmetry does not settle the integral, use the arc formula with the actual angular endpoints and the circle’s center and radius.

Check Your Understanding

Use the parametrization and formulas in this tutorial to answer the following questions.

What is $ds$ for a circle of radius $5$ parametrized by $x=a+5\cos t$, $y=b+5\sin t$?
Find $\int_C x\,ds$ over the full circle centered at $(-2,3)$ with radius $1$.
Why does $\int_C xy\,ds=0$ for a full circle centered at the origin?
Evaluate $\int_C y\,ds$ on the upper semicircle of $x^2+y^2=9$, using $0\le t\le\pi$.
On a radius-$2$ circle, what factor must multiply $dt$ to obtain $ds$, and what error results if it is omitted?