Why Horizontal Slices Need Care
A circle and a line can enclose a familiar-looking region, but the correct area integral depends on the region being measured. A horizontal slice runs from a left \(x\)-boundary to a right \(x\)-boundary. Those boundaries can change as the height \(y\) changes, even when the same circle and line enclose the region throughout.
The key steps are to rewrite the circle in terms of \(x\), find the heights where the line meets it, and inspect the region between those heights. In particular, an intersection with the circle’s left branch is not interchangeable with one on its right branch.
Writing a Circle as Two Functions of \(y\)
A circle with center \((h,k)\) and radius \(r>0\) has equation \((x-h)^2+(y-k)^2=r^2\). Solving for \(x\) gives two branches: the left side of the circle and the right side. These formulas are defined only for heights from \(k-r\) to \(k+r\).
The square root is nonnegative, so the minus sign gives the smaller \(x\)-value and the plus sign gives the larger one. At \(y=k-r\) and \(y=k+r\), both branches meet at the bottom and top of the circle. For a line that is not horizontal, it is often useful to write its equation as \(x=\ell(y)\), so it can be compared directly with the circle branches.
Why the Horizontal-Slice Formula Works
Suppose a region extends from \(y=a\) to \(y=b\), and each slice has a left boundary \(L(y)\) and a right boundary \(R(y)\). Divide the vertical interval into short pieces of height \(\Delta y\). Over each piece, the region is approximated by a thin rectangle whose width is approximately \(R(y)-L(y)\) and whose height is \(\Delta y\). Adding the rectangle areas and taking the limit gives the area integral.
This argument applies when the boundaries are continuous over the interval and the chosen slice really does run from the stated left boundary to the stated right boundary. If either boundary changes, split the integral at the change. If the line lies outside the circle’s horizontal span for some heights, that part is not part of the enclosed region.
Worked Example: A Circular Segment Cut by a Vertical Line
Worked Example: The Right-Hand Segment of a Circle
Find the area enclosed by the circle \(x^2+y^2=25\) and the line \(x=3\) on the side of the line inside the circle.
The circle has center \((0,0)\) and radius \(5\). Its branches are \(x=-\sqrt{25-y^2}\) and \(x=\sqrt{25-y^2}\). At the intersection with \(x=3\), substitute into the circle:
At both heights, the line meets the right branch, since \(x=3\) is positive. The desired segment lies to the right of the line and to the left of the circle’s right branch. Thus its width is \(\sqrt{25-y^2}-3\), for \(-4\le y\le4\).
An antiderivative of \(\sqrt{25-y^2}\) is \(\frac{y}{2}\sqrt{25-y^2}+\frac{25}{2}\arcsin(y/5)\). Therefore an antiderivative of the full integrand is
Evaluate at the upper endpoint: \(\sqrt{25-16}=3\), so \(F(4)=6+\frac{25}{2}\arcsin(4/5)-12=\frac{25}{2}\arcsin(4/5)-6\). At the lower endpoint, \(\arcsin(-4/5)=-\arcsin(4/5)\), so \(F(-4)=-6-\frac{25}{2}\arcsin(4/5)+12=6-\frac{25}{2}\arcsin(4/5)\). Subtracting gives
Check the slice widths: at \(y=-4\) and \(y=4\), \(\sqrt{25-y^2}-3=0\), as required at the intersections. At \(y=0\), the width is \(5-3=2>0\), confirming the order of the boundaries. Differentiating \(F\) gives \(\sqrt{25-y^2}-3\), so the antiderivative matches the integrand.
When a Slice Boundary Changes
A line may intersect the circle’s left branch at one height and its right branch at another. This can make a single expression for the slice width incorrect over part of the vertical range. One useful test at each height is to compare the line’s \(x\)-value with both circle branches. If the line is left of the entire circle, a region on the appropriate side may contain the full circle chord at that height. If it lies between the branches, it cuts the chord and the width uses the line as one endpoint.
The next example shows why those cases must be distinguished. The line \(x=y\) passes through the center of the circle \(x^2+y^2=2\). We choose the half-disk on the side \(x\ge y\).
Worked Example: A Line Through the Circle’s Center
Worked Example: The Half-Disk to the Right of \(x=y\)
The circle branches are \(x=-\sqrt{2-y^2}\) and \(x=\sqrt{2-y^2}\), with \(-\sqrt2\le y\le\sqrt2\). Intersections with \(x=y\) satisfy \(y^2+y^2=2\), so \(2y^2=2\) and \(y=-1\) or \(y=1\). These are the heights at which the line meets the circle.
For \(-\sqrt2\le y\le-1\), the line value \(x=y\) is at or to the left of the circle’s left branch. For instance, at \(y=-\sqrt2\), the circle has just one point with \(x=0\), which lies to the right of the line. At these lower heights the chosen region contains the full chord, so its width is \(2\sqrt{2-y^2}\). For \(-1\le y\le1\), the line lies between the circle branches; the part satisfying \(x\ge y\) runs from the line to the right branch. Above \(y=1\), the line is to the right of the circle, so this half-disk has no slice there. The area is therefore
For the first integral, an antiderivative is \(y\sqrt{2-y^2}+2\arcsin(y/\sqrt2)\). At \(y=-1\), its value is \(-1-\frac{\pi}{2}\); at \(y=-\sqrt2\), its value is \(-\pi\). The first integral is therefore \(\frac{\pi}{2}-1\).
For the second integral, an antiderivative is \(\frac{y}{2}\sqrt{2-y^2}+\arcsin(y/\sqrt2)-\frac{y^2}{2}\). At \(y=1\), this is \(\frac{\pi}{4}\). At \(y=-1\), it is \(-1-\frac{\pi}{4}\). The second integral is \(1+\frac{\pi}{2}\). Adding the two parts gives
At \(y=-1\), the lower-interval full-chord width is \(2\sqrt{2-1}=2\), while the upper-interval width is \(\sqrt{2-1}-(-1)=2\); the formulas agree where the slice description changes. At \(y=1\), the second width is \(1-1=0\), as it must be at the intersection. The result also agrees with symmetry: a line through the center divides a circle of area \(2\pi\) into two equal halves. Differentiating each stated antiderivative recovers its corresponding integrand.
Worked Example: A Shifted Circle and a Vertical Line
Worked Example: The Segment to the Right of \(x=3\)
Consider the circle \((x-1)^2+(y+1)^2=9\) and the line \(x=3\). Find the area of the part of the circle to the right of the line. The circle branches are \(x=1\pm\sqrt{9-(y+1)^2}\). Substitute \(x=3\) into the circle equation to find the intersection heights:
At these intersections, \(x=3\) is on the right branch, because the horizontal displacement from the center is \(3-1=2>0\). Between the two intersection heights, the right branch lies to the right of the line. The slice width is \(1+\sqrt{9-(y+1)^2}-3=\sqrt{9-(y+1)^2}-2\). Thus
Set \(u=y+1\); then \(du=dy\), and the limits become \(-\sqrt5\) and \(\sqrt5\). An antiderivative is
At \(u=\sqrt5\), \(\sqrt{9-u^2}=2\), so \(G(\sqrt5)=\sqrt5+\frac92\arcsin(\sqrt5/3)-2\sqrt5=\frac92\arcsin(\sqrt5/3)-\sqrt5\). At \(u=-\sqrt5\), the square-root term changes sign and the inverse-sine term is odd, giving \(G(-\sqrt5)=-\frac92\arcsin(\sqrt5/3)+\sqrt5\). Therefore
Check the integrand at the midpoint \(u=0\): the width is \(3-2=1>0\). At each endpoint, \(\sqrt{9-5}-2=0\), as required. Differentiating \(G\) gives \(\sqrt{9-u^2}-2\), verifying the integration. In particular, the line subtracts \(2\) from the circle’s right-branch coordinate \(1+\sqrt{9-u^2}\); subtracting \(1\) would use the wrong boundary.
Common Pitfall: Using One Width Across Every Height
For the circle \(x^2+y^2=2\) and line \(x=y\), it is tempting to write the width as \(\sqrt{2-y^2}-y\) throughout the full circle height range. That formula is only appropriate when the line lies between the circle branches. At \(y=-\sqrt2\), it gives \(\sqrt0-(-\sqrt2)=\sqrt2\), even though the circle has a single point at that height and the actual horizontal width is zero. The line is outside the chord there, so the formula has included points that do not form a slice of the chosen half-disk.
The correction is to identify where the line meets each branch and compare its position with the branches. For the region \(x\ge y\), the full chord is used from \(y=-\sqrt2\) to \(y=-1\), and the line-to-right-branch width is used from \(y=-1\) to \(y=1\). The resulting two integrals meet at a height where their widths agree. This boundary check is often the fastest way to catch a missing or extra part of a region.
A Reliable Setup Checklist
Express the circle’s left and right boundaries as functions of \(y\), and record its vertical range.
Substitute the line into the circle equation and solve for the intersection heights.
State which enclosed piece is intended, then compare the line with both circle branches at representative heights.
Use right boundary minus left boundary, split where the boundary description changes, and check that widths are nonnegative and vanish at the relevant intersections.
Horizontal slices are especially useful when the boundaries are naturally expressed as \(x\)-values depending on \(y\). For circle-line enclosures, the main challenge is usually not integration but accurately describing which part of each horizontal chord belongs to the selected region. Correct bounds and boundary order make the integral meaningful before any antiderivative is evaluated.
Check Your Understanding
For each question, identify the region and justify the slice width before evaluating any integral.
- For the circle \((x-2)^2+y^2=16\), write its left and right branches and give their common vertical domain.
- Find the intersection heights of \(x=2\) and \((x-2)^2+y^2=16\). Which circle branch contains the intersections?
- For \(x^2+y^2=9\) and \(x=0\), set up the area integral for the half-disk to the right of the line using horizontal slices.
- For \(x^2+y^2=2\) and \(x=y\), why does the width \(\sqrt{2-y^2}-y\) fail to describe the selected half-disk at every height from \(-\sqrt2\) to \(\sqrt2\)?
- What two checks can help detect a reversed boundary order or an incorrect vertical bound in a circle-line area setup?
