From Two Disks to One Washer
A cross-section of a solid of revolution can be a disk, or it can have a hole in its middle. The latter shape is a washer: imagine a circular disk with a smaller circular disk removed from its centre. Its area is not a new geometric quantity that must be memorized independently. It follows directly by subtracting the area of the hole from the area of the full disk.
That subtraction also explains how to find volume. If the washer-shaped cross-sections vary along the axis of rotation, add their areas across the interval. The resulting integral is the washer formula. The essential details are to identify the outer and inner radii, keep them as distances from the axis, and use the area of the annulus at each position.
Definition and Area Formula
Consider a cross-section perpendicular to the axis of rotation. Suppose its outer circular boundary has radius \(R\), and its inner circular boundary has radius \(r\), where \(R\ge r\ge 0\). The region between the two circles is a washer. The radii are measured from the axis of rotation; \(R\) is the distance to the farther boundary and \(r\) the distance to the nearer boundary.
The subtraction applies to areas, not directly to radii. A washer with \(r=0\) has no hole, so the formula reduces to the disk area \(A=\pi R^2\). If \(R=r\), its area is zero because there is no region between the two circles.
Worked Example: Find a Washer’s Area by Subtraction
A washer has outer radius \(R=3\) units and inner radius \(r=1\) unit. Find its area.
First find the area of the outer disk:
Then find the area of the inner disk that is removed:
Subtract the inner area from the outer area:
As a check, the area is positive because the outer disk is larger than the inner disk. The result is also less than the \(9\pi\ \text{units}^2\) area of the full outer disk, as it must be when a hole has been removed.
Why Subtraction Gives the Area
The inner disk lies entirely inside the outer disk. The outer disk consists of two non-overlapping parts: the washer and the inner disk. Areas add when regions are joined without overlap, so the outer disk area equals the washer area plus the inner disk area. Rearranging that equality gives the washer area as the difference.
Using the disk area formula \(A=\pi(\text{radius})^2\), the two disk areas are \(\pi R^2\) and \(\pi r^2\). Therefore \(A=\pi R^2-\pi r^2\). This argument also shows why the order matters: subtracting the smaller, inner disk from the larger, outer disk leaves exactly the washer.
From Washer Area to Volume
Now suppose a region rotates around an axis and produces washers perpendicular to that axis. Use \(s\) to denote position along the axis, over an interval \([a,b]\). At each position \(s\), let \(R(s)\) be the outer radius and \(r(s)\) the inner radius. These are distances from the axis, and they must satisfy \(R(s)\ge r(s)\ge0\).
Each cross-section has area \(\pi(R(s)^2-r(s)^2)\). To obtain volume, accumulate these cross-sectional areas along the interval. For continuous radius functions, the definite integral of the cross-sectional area gives the volume.
The condition \(R(s)\ge r(s)\) ensures that each slice has a nonnegative area. Continuity ensures that the area function can be integrated over the interval. If the radii are measured in units of length, the integrand has units of area, and integrating with respect to a length gives units of volume.
Why the Integral Gives the Volume
Divide the interval from \(a\) to \(b\) into short subintervals. In a subinterval of width \(\Delta s\), the solid is approximately a short prism whose base is the washer at a representative position \(s_i\). Its approximate volume is the cross-sectional area times the width:
Adding these approximations gives an approximate total volume. As the widths become smaller and the partition becomes finer, the sum approaches the definite integral of the cross-sectional area. Thus the volume is the integral in the theorem. The crucial point is that the quantity being accumulated is the washer area, not a radius or a circumference.
At position \(s\), identify its outer radius \(R(s)\) and inner radius \(r(s)\), both measured from the rotation axis.
Write the cross-sectional area as \(\pi R(s)^2-\pi r(s)^2\).
Integrate the cross-sectional area from the starting position \(a\) to the ending position \(b\).
Applying the Formula
In the next examples, the radius functions and interval are given. This lets us focus on how subtraction produces the integrand and how integrating that area produces the volume. In a geometric problem, the radii must first be interpreted as distances from the axis; they are not automatically the boundary coordinates themselves.
Worked Example: Integrate a Washer Area with Constant Outer Radius
Suppose a solid has outer radius \(R(s)=3\) units and inner radius \(r(s)=1+s\) units for \(0\le s\le1\). Find its volume.
First check which boundary is outer throughout the interval. Since \(1\le1+s\le2\) and \(R(s)=3\), we have \(R(s)>r(s)\ge0\) for every \(s\in[0,1]\).
Substitute the radii into the washer formula:
Expand the square and simplify the integrand:
Therefore,
Evaluate at both endpoints:
Check the antiderivative by differentiating: \(\frac{d}{ds}(8s-s^2-s^3/3)=8-2s-s^2\), which is the simplified integrand. Also, the cross-sectional area decreases from \(8\pi\) at \(s=0\) to \(5\pi\) at \(s=1\); a volume of \(20\pi/3\) over an interval of length \(1\) is consistent with areas between those values.
Worked Example: Subtract Two Varying Radius Functions
A solid has outer radius \(R(s)=4-s\) and inner radius \(r(s)=1+s\) for \(0\le s\le1\). Find its volume.
Check the radii on this interval. The outer radius decreases from \(4\) to \(3\), while the inner radius increases from \(1\) to \(2\). Thus \(R(s)\ge3>2\ge r(s)\ge1\), so the stated outer and inner radii remain in the correct order.
Set up the volume integral:
Expand each square separately:
Subtract the inner square from the outer square:
Now integrate and evaluate:
Differentiate the antiderivative \(15s-5s^2\) to check: its derivative is \(15-10s\), as required. A second check uses the endpoint areas: they are \(15\pi\) and \(5\pi\). Since the area function \(15\pi-10\pi s\) is linear, its average on \([0,1]\) is \(10\pi\), giving volume \(10\pi\) over an interval of length \(1\), in agreement with the integral.
A Common Subtraction Error
A frequent mistake is to subtract the radii first and then square the result. For \(R=3\) and \(r=1\), that incorrect calculation gives
But \(R-r\) is the washer’s radial thickness, not the radius of a disk whose area equals the washer. The washer consists of the outer disk with the inner disk removed, so the correct calculation is
The two expressions differ because \(R^2-r^2\) is not generally equal to \((R-r)^2\). In fact, \(R^2-r^2=(R-r)(R+r)\). The subtraction method keeps the two disk areas separate and avoids confusing the washer’s thickness with its area.
What to Check Before Integrating
Before using the washer formula, check that the slices are perpendicular to the axis of rotation and that each slice really is an outer disk with an inner disk removed. Then verify that both radii are distances from the axis and that their order is correct across the interval. The formula requires \(R(s)\ge r(s)\ge0\); if that description fails, the chosen radii or interval may not describe the intended washers.
Finally, keep the square on each radius inside the subtraction, include the factor \(\pi\), and integrate with respect to position along the axis. These checks connect the geometry to the calculation: the integrand must be the area of one washer, and the integral must accumulate those areas across the solid.
Check Your Understanding
Use subtraction of disk areas and the washer integral to answer the following.
- A washer has outer radius \(5\) and inner radius \(3\). Write its area as a difference of disk areas and simplify.
- For radii \(R(s)=3\) and \(r(s)=s\) on \(0\le s\le2\), write the volume integral. Do not evaluate it.
- Why does \(\pi(R-r)^2\) not give the area of a washer in general?
- For \(R(s)=5-s\) and \(r(s)=2\) on \(0\le s\le1\), verify that the radii are correctly ordered throughout the interval and write the volume integral.
- What units result when the radii and the integration variable are both measured in units of length?
