Introduction
A circle is not usually the graph of one function \(y=f(x)\): a vertical line through most of the circle meets it twice. But its upper and lower halves are graphs, each described by a square-root expression. The same issue arises when a region is bounded by other radical curves: the expression under the square root restricts the domain, and the sign of the branch determines which boundary is represented.
To find area reliably, first identify the relevant branches and their domains. Then choose vertical or horizontal slices so that each slice runs from one well-defined boundary to another. The area-between-curves formulas from earlier material apply once those boundaries and their order are clear.
Circle Boundaries as Radical Branches
A circle of radius \(r>0\) and center \((h,k)\) has equation \((x-h)^2+(y-k)^2=r^2\). Solving for \(y\) gives two branches. The plus sign gives the upper half, and the minus sign gives the lower half. Solving instead for \(x\) gives the right and left halves.
These formulas are defined only when the expression under the square root is nonnegative. For the \(y\)-branches, this requires \(|x-h|\le r\). For the \(x\)-branches, it requires \(|y-k|\le r\). The square-root symbol means the nonnegative square root; the two signs written outside it distinguish the branches.
Choosing the Slice Direction
With vertical slices, the area is the integral of the top boundary minus the bottom boundary over the relevant \(x\)-interval. With horizontal slices, it is the integral of the right boundary minus the left boundary over the relevant \(y\)-interval. A circle's upper and lower branches are naturally used with vertical slices; its left and right branches are naturally used with horizontal slices.
The best orientation is the one that describes the region with the fewest boundary changes and the simplest valid functions. A region may be bounded by only one half of a circle and a line, or it may include both halves. Do not infer the needed branches from the circle equation alone: identify which parts of the circle actually form the boundary of the specified region.
Why the Circle Formulas Give the Correct Branches
Start with the equation of a circle and isolate one coordinate. The algebra gives a squared quantity, and taking square roots produces two possibilities. For example, isolate \(y\) as follows:
The radicand must be nonnegative, which gives \(h-r\le x\le h+r\). For a fixed \(x\) strictly between these endpoints, the square root is positive. The plus choice then gives \(y>k\), and the minus choice gives \(y<k\). At the endpoints the square root is zero, so the branches meet at the circle's leftmost and rightmost points. This establishes both the branch formulas and the interval on which they describe real points of the circle.
For a horizontal slice, isolating \(x\) instead gives \((x-h)^2=r^2-(y-k)^2\). The same reasoning yields the right branch with a plus sign, the left branch with a minus sign, and the domain \(k-r\le y\le k+r\). Choosing the sign is therefore not a matter of algebraic preference: it determines which side of the circle is being described.
Worked Example: Area Under an Upper Semicircle
Worked Example: Area Under an Upper Semicircle
Find the area bounded by the upper half of the circle \(x^2+y^2=9\) and the \(x\)-axis.
Solving for \(y\) gives \(y=\pm\sqrt{9-x^2}\). The upper semicircle is the plus branch, \(y=\sqrt{9-x^2}\). The radicand is nonnegative for \(-3\le x\le3\), and the \(x\)-axis is \(y=0\). Thus the top-minus-bottom area integral is
An antiderivative on the interior of this interval is
To evaluate, use \(\arcsin(1)=\pi/2\) and \(\arcsin(-1)=-\pi/2\):
Check the antiderivative by differentiating. For \(-3<x<3\),
This matches the integrand. The resulting area is also half the area \(\pi(3)^2=9\pi\) of the full circle, as expected.
Other Radical Boundaries
Not every square-root graph comes from a circle. A graph such as \(y=\sqrt{x+1}\) is still restricted to values for which its radicand is nonnegative, here \(x\ge-1\). Its nonnegative branch is the function specified by the square-root symbol. When using it as an area boundary, verify where it meets the other boundaries and which curve lies above the other on the interval.
Worked Example: A Radical Curve and a Horizontal Line
Find the area bounded by \(x=-1\), \(y=3\), and \(y=\sqrt{x+1}\).
The radical curve is real for \(x\ge-1\). It meets the line \(y=3\) where
On \([-1,8]\), the radical branch increases from \(\sqrt{0}=0\) to \(\sqrt{9}=3\), so the line is above the radical curve. Therefore
An antiderivative is \(3x-\frac{2}{3}(x+1)^{3/2}\). Evaluating both endpoints gives
Check by differentiating the antiderivative: \(3-\frac{2}{3}\cdot\frac{3}{2}(x+1)^{1/2}=3-\sqrt{x+1}\), which is the integrand. At the left endpoint the curve is below the line, and at the right endpoint the two meet, consistent with the chosen top-minus-bottom order.
When Horizontal Slices Are More Natural
If the region lies to one side of a circle, solving for \(x\) may give simpler boundaries than solving for \(y\). The following example uses the right semicircle and the vertical line through its center. The horizontal slices run from that line directly to the curved boundary.
Worked Example: The Right Half of a Shifted Circle
Find the area between the circle \((x-1)^2+(y+2)^2=16\) and the line \(x=1\), taking the region to the right of the line and inside the circle.
Solving for \(x\) gives \(x=1\pm\sqrt{16-(y+2)^2}\). The required boundary is the right branch, \(x=1+\sqrt{16-(y+2)^2}\); the line \(x=1\) is the left boundary of this half-disk. The radicand is nonnegative for \(-6\le y\le2\), so
Set \(u=y+2\), so \(du=dy\) and the limits become \(-4\) and \(4\). An antiderivative of \(\sqrt{16-u^2}\) is \(\frac{u}{2}\sqrt{16-u^2}+8\arcsin(u/4)\). At the endpoints its values are
To check, differentiating the antiderivative gives \(\sqrt{16-u^2}\) for \(-4<u<4\), the transformed integrand. Geometrically, this is half a disk of radius \(4\), so its area is \(\frac12\pi(4)^2=8\pi\), agreeing with the integral.
A Common Branch Error
A frequent error is to keep only the plus branch when a region includes both the upper and lower halves of a circle. For \(x^2+y^2=25\), the upper boundary is \(y=\sqrt{25-x^2}\), but the lower boundary is \(y=-\sqrt{25-x^2}\). If the goal is the full disk, integrating just the upper branch above the \(x\)-axis gives only the upper half:
not the full disk's area. For the full disk, the height of a vertical slice is upper minus lower, so the correct setup is
The correction is to identify both boundaries of each slice before integrating. A plus sign does not mean “the whole circle”; it describes only the upper branch when solving for \(y\).
A Reliable Setup
Before evaluating an area integral involving radical boundaries, check the geometry and algebra in this order:
Use vertical slices for top and bottom boundaries, or horizontal slices for right and left boundaries. Prefer the direction that gives valid, simple boundary functions.
For a circle, retain the sign corresponding to the actual boundary: upper or lower for \(y\), right or left for \(x\).
Require every radicand to be nonnegative, and find where the region's boundaries meet or where a stated endpoint limits the region.
Check that top is at least bottom, or right is at least left, throughout the interval. Then integrate their difference; split the interval if the order or formulas change.
These checks matter because a correct-looking integral can still describe the wrong half of a circle, extend beyond a radical's domain, or subtract the boundaries in the wrong order. Once the branches and bounds are correct, the area calculation follows the usual definite-integral method.
Check Your Understanding
Use branch choice, domain, and slice orientation to justify each setup before evaluating it.
- For the circle \((x-2)^2+(y+1)^2=16\), write the upper and lower branches as functions of \(x\), and state their common \(x\)-domain.
- For \(x^2+(y-3)^2=9\), write the right and left branches as functions of \(y\), and state the \(y\)-domain.
- What is the area under the upper semicircle \(y=\sqrt{16-x^2}\) and above the \(x\)-axis?
- For \(y=\sqrt{x+4}\), find the \(x\)-coordinate where the curve meets \(y=2\), and state the curve's domain.
- Why does integrating only \(\sqrt{25-x^2}\) from \(-5\) to \(5\) give half the area of the circle \(x^2+y^2=25\)?
