From a Disk to a Washer
A disk is a solid circular cross section. A washer is an annular cross section: it has a circular outer edge and a circular hole in the middle. When a solid has a hole running along its length, a slice perpendicular to that length can look like a washer rather than a filled disk.
The washer method is based on two ideas. First, find the area of one annular slice by subtracting the hole’s disk area from the outer disk area. Then add the volumes of thin slices along the solid. The resulting definite integral gives the volume of the hollow solid.
Deriving the Washer Area
The area of a disk with radius \(s\) is \(\pi s^2\). The outer boundary of a washer encloses a disk of radius \(R\), while the hole removes a disk of radius \(r\). Subtracting the removed area gives the area of the material that remains:
This is the washer area formula. The radii must be measured from the same center, and \(R\) and \(r\) are distances, so they are nonnegative. The condition \(r\leq R\) ensures that the hole fits inside the outer disk.
Notice that the formula subtracts the squares of the radii. It does not square the difference between the radii. The washer consists of the part of the outer disk left after the entire inner disk is removed.
Worked Example: Find the Area of a Washer
Worked Example: Find the Area of a Washer
A washer has outer radius \(5\text{ cm}\) and inner radius \(3\text{ cm}\). Find its area.
Use \(R=5\text{ cm}\) and \(r=3\text{ cm}\). Subtract the inner disk area from the outer disk area:
Check: the outer disk has area \(25\pi\text{ cm}^2\), and the hole has area \(9\pi\text{ cm}^2\). Their difference is \(16\pi\text{ cm}^2\), as required. The units are square centimeters because area is measured in square units.
Why Washer Areas Produce Volume
Imagine a solid extending along a coordinate \(t\), from \(t=a\) to \(t=b\). At each value of \(t\), suppose a slice perpendicular to that coordinate is a washer. Let its outer radius be \(R(t)\) and its inner radius be \(r(t)\), where \(0\leq r(t)\leq R(t)\). The cross-sectional area at \(t\) is therefore
A very thin slice of thickness \(\Delta t\) has volume approximately equal to its cross-sectional area times its thickness. If the area changes slightly across the slice, choosing a sample point \(t_i^*\) in that small interval gives the approximation \(A(t_i^*)\Delta t_i\). Adding these approximations across the solid gives a sum of thin-slice volumes.
As the slices become thinner, the approximation approaches the exact volume. This is the same cross-sectional-area principle used for other solids: when the area function is continuous, or is piecewise continuous with finitely many pieces, the limit of these sums is its definite integral.
At position \(t\), subtract the inner disk from the outer disk to obtain \(A(t)=\pi(R(t)^2-r(t)^2)\).
For a small thickness \(\Delta t_i\), the slice volume is approximately \(A(t_i^*)\Delta t_i\).
Sum the approximations over a partition of the interval from \(a\) to \(b\).
As the largest slice thickness approaches zero, the sums approach the definite integral of the cross-sectional area.
Derivation of the Washer Volume Integral
To make the limiting argument precise, partition \([a,b]\) into subintervals with widths \(\Delta t_i\). Choose a sample point \(t_i^*\) in each subinterval. The volume of the \(i\)th thin slice is approximated by \(A(t_i^*)\Delta t_i\), so the entire solid is approximated by the sum of these slice volumes.
When \(A(t)\) is continuous, these sums converge to the definite integral of \(A\) as the partition is refined. Substituting the washer area formula into that integral gives
This derivation explains why the method works: each thin slice contributes its washer area multiplied by its thickness, and the definite integral adds those contributions across the solid. The radius functions must describe the actual outer and inner boundaries of every slice on the interval being integrated. If they or the cross-sectional geometry change in a way that prevents one formula from applying throughout, the interval must be handled in pieces.
Worked Example: Calculate Washer Areas Along a Solid
Worked Example: Calculate Washer Areas Along a Solid
Suppose the outer radius is constantly \(3\text{ cm}\) along a solid, while the inner radius is \(1+\frac{t}{2}\text{ cm}\), for \(0\leq t\leq2\text{ cm}\). Find the cross-sectional areas at the two endpoints.
At \(t=0\), the inner radius is \(1+\frac{0}{2}=1\text{ cm}\). Thus
At \(t=2\text{ cm}\), the inner radius is \(1+\frac{2}{2}=2\text{ cm}\). Therefore
Check: the inner radius increases from \(1\text{ cm}\) to \(2\text{ cm}\), while the outer radius stays at \(3\text{ cm}\). The hole therefore occupies more of each outer disk at the second endpoint, so the area decreases from \(8\pi\text{ cm}^2\) to \(5\pi\text{ cm}^2\). Both computed areas are positive and consistent with the radii.
Worked Example: Integrate the Areas to Find a Volume
Worked Example: Integrate the Areas to Find a Volume
For the solid in the previous example, use the radius functions on \(0\leq t\leq2\text{ cm}\) to find the volume.
The outer radius is \(3\text{ cm}\), and the inner radius is \(1+\frac{t}{2}\text{ cm}\). Using the washer volume integral,
Check the antiderivative: \(\frac{d}{dt}(8t-\frac{t^2}{2}-\frac{t^3}{12})=8-t-\frac{t^2}{4}\), the integrand. The endpoint evaluation gives \(16-2-\frac{2}{3}=\frac{40}{3}\), since \(16-2-\frac{2}{3}=14-\frac{2}{3}=\frac{40}{3}\). The volume is positive, and the units are cubic centimeters because area was multiplied by a length.
A Second Volume Check
Consider a solid extending through \(0\leq t\leq3\text{ m}\) with outer radius \(4\text{ m}\) and inner radius \(1+t\text{ m}\). The inner radius reaches the outer radius at the endpoint, so the final cross section has zero area. This does not affect the integral: a single endpoint contributes no positive volume by itself.
As a check, the cross-sectional area is \(15\pi\text{ m}^2\) at \(t=0\), and it is \(0\text{ m}^2\) at \(t=3\). The area function \(15-2t-t^2\) is nonnegative on the interval and decreases to zero. Differentiating the antiderivative gives \(15-2t-t^2\), confirming the integration.
Common Mistake: Subtracting Radii Before Squaring
A tempting but incorrect formula is \(\pi(R-r)^2\). It treats the washer as though its area were the area of a disk whose radius is the radial thickness. But a washer is not such a disk: its area is the outer disk area minus the inner disk area.
Worked Example: Correct a Washer-Area Error
For \(R=5\text{ cm}\) and \(r=3\text{ cm}\), the incorrect calculation would be
This is wrong because it squares the radial difference instead of subtracting the two disk areas. The correct calculation is
Check by the geometric definition: the outer disk has area \(25\pi\text{ cm}^2\); removing its \(9\pi\text{ cm}^2\) inner disk leaves \(16\pi\text{ cm}^2\). Thus \(4\pi\text{ cm}^2\) cannot be the washer area.
Why the Method Matters
The washer method turns a three-dimensional volume problem into a one-dimensional integral of cross-sectional areas. Its structure is simple, but keeping the geometry visible matters: identify the two disk areas in each slice, subtract the hole, then integrate that remaining area over the length of the solid.
Before evaluating a washer integral, check that the radii are distances, the inner radius does not exceed the outer radius on the interval, and the resulting area is nonnegative. Also check dimensions: radii have units of length, the washer area has square units, and integrating with respect to a length coordinate produces cubic units.
Check Your Understanding
Use the washer-area and volume ideas to answer these questions.
- A washer has outer radius \(7\) units and inner radius \(4\) units. What is its area?
- Explain why the washer area is \(\pi(R^2-r^2)\), not \(\pi(R-r)^2\).
- If each perpendicular slice has area \(A(t)\), what definite integral gives the volume from \(t=a\) to \(t=b\)?
- For outer radius \(5\text{ cm}\) and inner radius \(3\text{ cm}\), what are the cross-sectional area’s units?
- Why do sums of thin-slice volumes approach an integral as the slice thicknesses approach zero?
