Worked Integrals over a Disk Clipped by a Line

Introduction

A line can divide a disk into two regions with very different-looking boundaries. In Cartesian coordinates, one side may be simple to describe, but a polar integral requires more than substituting $x=r\cos\theta$ and $y=r\sin\theta$. The radial limits and the angles over which they apply must match the region.

The key step is to find where the line meets the circle. Those intersection points determine the transition angles. On the side of the line farther from the disk’s center, a ray begins at the line and ends at the circle. On the other side, it is often simpler to integrate over the whole disk and subtract that cap.

Central idea: For a circular disk centered at the origin, describe the line by its distance from the origin and the direction perpendicular to it. The line-circle intersections give the angular limits; the line itself gives a radial limit.

Polar Bounds for a Circular Cap

Let the disk have radius $R>0$, and let a line lie at perpendicular distance $d$ from the origin, where $0<d<R$. Let $\alpha$ be the angle of a unit normal pointing from the origin toward the line. The line and the half-plane on its farther side can be written as

x\cos\alpha+y\sin\alpha=d, \qquad x\cos\alpha+y\sin\alpha\ge d.

The part of the disk in that half-plane is a circular cap. In polar coordinates, $x=r\cos\theta$ and $y=r\sin\theta$, so the line inequality becomes

r\cos(\theta-\alpha)\ge d.

For a fixed angle, a ray can reach this cap only if $\cos(\theta-\alpha)>0$. Along such a ray, the line is at radius $r=d/\cos(\theta-\alpha)$, and the circle is at radius $r=R$. The ray crosses the cap only when the line is reached no farther out than the circle:

\frac{d}{\cos(\theta-\alpha)}\le R \quad\Longleftrightarrow\quad \cos(\theta-\alpha)\ge\frac dR.

Define $\beta=\arccos(d/R)$, so $0<\beta<\pi/2$. The angles satisfying this condition are, for every integer $k$,

\alpha-\beta+2k\pi \le \theta \le \alpha+\beta+2k\pi.

Because polar angles repeat every $2\pi$, these intervals describe the same angular sector modulo a full turn. For a single integral, we can choose the representative interval around $\alpha$, namely $[\alpha-\beta,\alpha+\beta]$. The radial coordinate runs from the line to the circle.

Polar bounds for the cap: If $0<d<R$, then in the representative angular interval $\alpha-\beta\le\theta\le\alpha+\beta$, where $\beta=\arccos(d/R)$, the cap has bounds $$ \frac{d}{\cos(\theta-\alpha)}\le r\le R. $$ The denominator is positive throughout this interval.

For an integrand $f(x,y)$, the polar double-integral formula includes the Jacobian factor $r$: $dA=r\,dr\,d\theta$. Thus the cap integral is set up as

\int_{\alpha-\beta}^{\alpha+\beta} \int_{d/\cos(\theta-\alpha)}^{R} f(r\cos\theta,r\sin\theta)\,r\,dr\,d\theta.

The formula assumes the disk is centered at the origin and that the cap is the side of the line away from the origin. If the desired region is the other side of the line, it is the disk with this cap removed. An integral over that region can therefore be found by subtracting the cap integral from the integral over the full disk, whenever the integrals exist.

Why the Transition Angles Work

The transition angles are the directions of the two rays from the origin through the line-circle intersections. At each intersection, the line’s radial distance equals the circle’s radius. Substituting $r=R$ into the line equation gives the transition condition directly.

Write the line along a ray.
For angle $\theta$, its radial equation is $r\cos(\theta-\alpha)=d$.

Require the ray to meet the disk beyond the line.
The intersection radius $d/\cos(\theta-\alpha)$ must be at most $R$.

Solve for the angular interval.
The condition becomes $\cos(\theta-\alpha)\ge d/R$, giving the angles $\alpha-\beta+2k\pi\le\theta\le\alpha+\beta+2k\pi$, where $k\in\mathbb Z$.

This derivation also explains why using the entire range $0\le\theta<2\pi$ with the cap’s radial bounds would be invalid. Outside the transition interval, a ray does not reach the cap at all: its intersection with the line is outside the disk, or it points away from the cap.

Worked Integrals

Worked Example: Area of a Cap Cut from a Disk

Find the area of the part of the disk $x^2+y^2\le25$ where $x\ge3$. Here $R=5$, $d=3$, and $\alpha=0$. The transition angle is $\beta=\arccos(3/5)$. Since $\cos\beta=3/5$ and $\sin\beta=4/5$, $\tan\beta=4/3$. The polar bounds are $3\sec\theta\le r\le5$ and $-\beta\le\theta\le\beta$. For area, the integrand is $1$, and the polar area factor is $r$:

\begin{aligned} A &=\int_{-\beta}^{\beta}\int_{3\sec\theta}^{5}r\,dr\,d\theta\\ &=\frac12\int_{-\beta}^{\beta} \left(25-9\sec^2\theta\right)\,d\theta\\ &=\frac12\left[25\theta-9\tan\theta\right]_{-\beta}^{\beta}\\ &=25\beta-9\tan\beta\\ &=25\arccos(3/5)-12. \end{aligned}

As a check, the result is positive: $\arccos(3/5)$ is approximately $0.9273$, so the area is approximately $25(0.9273)-12=11.18$. The cap is a proper portion of a disk of area $25\pi$, as expected.

Worked Example: Integrating the Horizontal Coordinate over a Cap

For the same disk and cap, evaluate $\iint_{\text{cap}}x\,dA$. In polar coordinates, $x=r\cos\theta$, so including the area factor gives the integrand $r^2\cos\theta$. Use the bounds found above:

\begin{aligned} \iint_{\text{cap}}x\,dA &=\int_{-\beta}^{\beta} \int_{3\sec\theta}^{5}r^2\cos\theta\,dr\,d\theta\\ &=\frac13\int_{-\beta}^{\beta} \left(125\cos\theta-27\sec^2\theta\right)\,d\theta\\ &=\frac13\left[125\sin\theta-27\tan\theta\right]_{-\beta}^{\beta}\\ &=\frac13\left(250\sin\beta-54\tan\beta\right)\\ &=\frac13\left(250\cdot\frac45-54\cdot\frac43\right)\\ &=\frac13(200-72)\\ &=\frac{128}{3}. \end{aligned}

We can check this value by integrating vertically in Cartesian coordinates. At a fixed $x$ between $3$ and $5$, the cap extends from $y=-\sqrt{25-x^2}$ to $y=\sqrt{25-x^2}$. Therefore

\begin{aligned} \iint_{\text{cap}}x\,dA &=2\int_3^5x\sqrt{25-x^2}\,dx\\ &=-\frac23\left[(25-x^2)^{3/2}\right]_3^5\\ &=\frac23(16)^{3/2}\\ &=\frac23\cdot64\\ &=\frac{128}{3}. \end{aligned}

The independent calculation agrees with the polar result.

Worked Example: A Radial Integrand over the Complement of a Cap

Evaluate $\iint_E(x^2+y^2)\,dA$, where $E$ is the part of $x^2+y^2\le25$ satisfying $x\le3$. This is the full disk with the cap $x\ge3$ removed. First evaluate the integral over the full disk. Since $x^2+y^2=r^2$, the polar integrand, including the area factor, is $r^3$:

\begin{aligned} \iint_{\text{disk}}(x^2+y^2)\,dA &=\int_0^{2\pi}\int_0^5r^3\,dr\,d\theta\\ &=2\pi\left[\frac{r^4}{4}\right]_0^5\\ &=\frac{625\pi}{2}. \end{aligned}

Now subtract the cap integral. Its bounds are $3\sec\theta\le r\le5$ and $-\beta\le\theta\le\beta$, with $\beta=\arccos(3/5)$:

\begin{aligned} \iint_{\text{cap}}(x^2+y^2)\,dA &=\int_{-\beta}^{\beta}\int_{3\sec\theta}^{5}r^3\,dr\,d\theta\\ &=\frac14\int_{-\beta}^{\beta} \left(625-81\sec^4\theta\right)\,d\theta. \end{aligned}

Since an antiderivative of $\sec^4\theta$ is $\tan\theta+\tan^3\theta/3$, and $\tan\beta=4/3$,

\begin{aligned} \int_{-\beta}^{\beta}\sec^4\theta\,d\theta &=2\left(\tan\beta+\frac{\tan^3\beta}{3}\right)\\ &=2\left(\frac43+\frac{64}{81}\right) =\frac{344}{81}. \end{aligned}

It follows that the cap integral is

\begin{aligned} \iint_{\text{cap}}(x^2+y^2)\,dA &=\frac14\left(1250\beta-81\cdot\frac{344}{81}\right)\\ &=\frac{625\beta}{2}-86. \end{aligned}

Subtracting this from the full-disk value gives

\begin{aligned} \iint_E(x^2+y^2)\,dA &=\frac{625\pi}{2}-\left(\frac{625\beta}{2}-86\right)\\ &=\frac{625}{2}\left(\pi-\arccos(3/5)\right)+86. \end{aligned}

As a check, $E$ contains the entire left half of the disk, so its integral must be positive and less than the full-disk value $625\pi/2$. The displayed result has these properties because $0<\beta<\pi/2$, and it was obtained by subtracting the positive cap integral from the full-disk integral.

Common Errors and Their Corrections

A frequent mistake is to reverse the radial limits. In the cap $x\ge3$, a ray with angle $\theta=0$ enters the region at the line $r=3$ and leaves at the circle $r=5$. The bounds must therefore be $3\le r\le5$ on this ray. The incorrect bounds $0\le r\le3$ describe points between the origin and the line, which do not satisfy $x\ge3$. For example, the origin has $x=0$, so it is not in the cap.

Another error is to state the angle interval as $\alpha-\beta\le\theta\le\alpha+\beta$ for all real $\theta$ without addressing periodicity. The complete solution is $\alpha-\beta+2k\pi\le\theta\le\alpha+\beta+2k\pi$, for $k\in\mathbb Z$. For a single integral, choose one representative interval, such as $[\alpha-\beta,\alpha+\beta]$, and use it consistently. In particular, $\theta=\alpha+2\pi$ represents the same direction as $\theta=\alpha$ and must not be excluded by a nonperiodic statement of the solution.

Finally, the line’s radial expression is not itself a valid radial bound at every angle. The expression $d/\cos(\theta-\alpha)$ is relevant to the cap only where the cosine is positive and where the line lies inside the disk along that ray. The transition-angle condition ensures both facts are handled.

Why the Setup Matters

The line-circle intersections determine where the radial description changes. Once those angles are established, the direction of the inequality identifies whether the line is an inner or outer boundary. For a cap away from the origin, it is an inner boundary: the radius starts at the line and ends at the circle. For the complementary region, using the full disk minus the cap avoids introducing separate radial cases.

The method applies directly when $0<d<R$. If $d=R$, the line is tangent to the circle and the cap has zero area. If $d>R$, the line does not meet the disk on that side, so there is no cap of positive area. If the line passes through the center, the cut is a diameter and the cap-side region is a half-disk. Checking these geometric conditions before integrating prevents bounds from being used in a situation where their transition angles do not exist.

Check Your Understanding

Use the disk-and-line method to answer these questions. Keep track of which side of the line is included and which radial boundary is reached first.

For a disk of radius $6$ clipped by the line $x=2$ on the side $x\ge2$, what is the transition angle $\beta$, and what are the polar bounds?
For the line $x\cos\alpha+y\sin\alpha=d$, why must the transition angles be stated modulo $2\pi$?
On a ray with $\theta=\alpha$, which comes first for the cap: the line or the circle? State the radial interval.
How could you set up an integral over the side of the disk satisfying $x\le3$ without finding separate radial bounds for every angle?
What happens to the cap when $d=R$, and why does the transition-angle formula no longer describe an interval of positive width?