Start with the Solid’s Description
A volume problem may provide a planar region and ask you to rotate it, or it may describe the shapes of slices through a solid directly. These are different geometric instructions, and they point to different cross-sectional area formulas.
If the problem says that a region is rotated around a line, the slices perpendicular to that line form circles or rings. A circle produces a disk; a ring with a hole produces a washer. If the problem instead says that slices are squares, semicircles, equilateral triangles, or another shape, use that stated shape to find each slice’s area. Do not assume a rotation when none is described.
The first decision is therefore not an integration technique. It is to ask what the solid’s slices actually are.
A Decision Guide
| What the problem describes | Area of a typical slice | Volume setup |
|---|---|---|
| Squares, triangles, semicircles, or other specified cross sections | The area formula for the specified shape, using the slice’s base length | \(\int A(x)\,dx\) or \(\int A(y)\,dy\), according to the slice orientation |
| A plane region rotated about an axis, with slices perpendicular to the axis meeting the axis | Disk area, \(\pi R^2\) | Integrate the disk area along the axis |
| A plane region rotated about an axis, with slices perpendicular to the axis leaving a hole | Washer area, \(\pi(R^2-r^2)\) | Integrate the washer area along the axis |
Here \(A\) denotes the area of a slice, \(R\) is the outer radius of a washer, and \(r\) is its inner radius. A disk is the special case of a washer with \(r=0\). For vertical slices, the slice position is usually described by \(x\), so the volume is integrated with respect to \(x\). For horizontal slices, the position is described by \(y\), so the volume is integrated with respect to \(y\).
Why Adding Slice Areas Gives Volume
The basic principle is the same for each method. Divide the solid into many thin slices. If a slice at position \(x\) has cross-sectional area \(A(x)\), then a thin slice of thickness \(\Delta x\) has approximate volume \(A(x)\Delta x\). Adding these approximate volumes gives a sum. As the slices become thinner and more numerous, the sum approaches the exact volume.
This expression requires that the cross-sectional area be defined across the interval and that the resulting integral exists. With horizontal slices, the corresponding expression is \(V=\int_c^d A(y)\,dy\). The variable records where the slice is located; it is not itself the slice’s area.
For a rotation, geometry supplies the area formula. A slice perpendicular to the rotation axis sweeps out a circle if it reaches the axis, so its area is \(\pi R^2\). If the slice lies between an outer and an inner radius, the cross section is a large circle with a smaller circle removed. Its area is \(\pi R^2-\pi r^2=\pi(R^2-r^2)\). Integrating these areas adds the volumes of the thin slices.
The radii are distances from the axis of rotation, not merely the function values by definition. Their expressions depend on the region and the axis. The bounds must cover the whole solid in the direction of the slices.
Worked Example: Prescribed Square Cross Sections
Worked Example: Prescribed Square Cross Sections
A solid has base region bounded by \(y=0\), \(y=2-x\), and \(0\le x\le2\). Slices perpendicular to the \(x\)-axis are squares. Find the volume.
Step 1: Read the slice instruction. The problem specifies squares, so this is a cross-section problem, not a rotation problem. At a fixed \(x\), the base segment runs vertically from \(y=0\) to \(y=2-x\). Its length, which is the side of the square, is
Step 2: Find the area of one slice. A square with side length \(s\) has area \(s^2\). Therefore
Step 3: Integrate across the base. The slices run from \(x=0\) to \(x=2\), so
An antiderivative is \(4x-2x^2+\frac{x^3}{3}\). Substituting the bounds gives
Check: At \(x=0\), the square side is \(2\), and its area is \(4\). At \(x=2\), the side and area are \(0\). The area decreases across the interval, consistent with the base narrowing to a point. The volume is positive, as required.
Worked Example: Rotation Producing Washers
Worked Example: Rotation Producing Washers
The region between \(y=\sqrt{x}\) and \(y=\frac{x}{4}\), for \(0\le x\le16\), is rotated about the \(x\)-axis. Find the volume.
Step 1: Identify the method. The instruction says to rotate a region about the \(x\)-axis. Vertical slices are perpendicular to that axis. For \(0<x<16\), the slice reaches from the axis up to \(y=\sqrt{x}\) and leaves a hole up to \(y=\frac{x}{4}\), so the rotated slice is a washer.
The curve order can be checked by solving \(\sqrt{x}=\frac{x}{4}\). Squaring gives \(x=\frac{x^2}{16}\), so \(16x=x^2\), or \(x(x-16)=0\). The curves meet at \(x=0\) and \(x=16\). For example, at \(x=4\), \(\sqrt{4}=2\) and \(\frac{4}{4}=1\), so the outer radius is \(\sqrt{x}\) and the inner radius is \(\frac{x}{4}\) throughout the interior of the interval.
Step 2: Write the washer area.
Step 3: Integrate and evaluate.
Check: At both endpoints the two radii are equal, so the washer area is zero. Between the endpoints, \(R(x)>r(x)\), so the washer area is positive. The computed volume is positive and uses the full interval between the intersections.
Worked Example: Rotation Producing Disks
Worked Example: Rotation Producing Disks
The rectangle \(0\le x\le3\), \(0\le y\le2\) is rotated about the \(x\)-axis. Find the volume.
Step 1: Identify the slices. The region is rotated, and vertical slices are perpendicular to the \(x\)-axis. Each slice runs from the axis \(y=0\) to \(y=2\). Its rotation fills a circle, so the cross section is a disk, not a washer with a hole.
Step 2: Determine the radius and area. The distance from the axis \(y=0\) to the top of the rectangle is \(R=2\). Thus each disk has area
Step 3: Integrate over the rectangle’s length.
Check: The rotated solid is a cylinder with radius \(2\) and length \(3\). Its volume is base area times length, \(\pi(2)^2(3)=12\pi\), which agrees with the integral.
A Common Washer Mistake
A washer is the area inside the outer circle but outside the inner circle. That means subtract the two circle areas. It does not mean square the difference between the radii.
Worked Example: Correcting a Washer-Area Error
Suppose a slice rotates around an axis to make a washer with outer radius \(3\) and inner radius \(1\). A common incorrect calculation is
This treats the cross section as a disk whose radius is the radial gap. But the washer’s area is the large disk’s area minus the small disk’s area:
Check: The correct result is greater than \(4\pi\), because the washer includes the area across radii from \(1\) to \(3\), not just a disk of radius \(2\). In general, \(\pi(R^2-r^2)\) is the washer area; \(\pi(R-r)^2\) is not.
A Reliable Selection Process
Before calculating, identify what determines the cross-sectional area. The following sequence prevents a common mix-up between a prescribed shape and a shape created by rotation.
If the problem names the shapes of slices, use those shapes. If it says to rotate a region, use the geometry of the rotation.
For slices perpendicular to the \(x\)-axis, describe area as a function of \(x\) and integrate with \(dx\). For slices perpendicular to the \(y\)-axis, use \(A(y)\) and \(dy\).
For a prescribed shape, obtain its dimensions from the base segment. For rotation, decide whether the slice makes a disk or a washer and express its radii as distances from the axis.
Bounds should span the solid in the slice direction. Areas should be nonnegative, and the radii of a washer should satisfy \(R\ge r\).
Writing the area function first makes the final integral a direct statement of how the solid is built. It also makes errors easier to find: an incorrect shape or dimension will usually appear before any integration begins.
Check Your Understanding
For each situation, decide which area formula is appropriate and identify the variable of integration.
- A region is rotated about the \(x\)-axis, and vertical slices extend from the axis to one curve. Are the cross sections disks or washers?
- A solid has horizontal slices that are equilateral triangles. What geometric information is needed to express the area of a slice?
- A washer has outer radius \(R\) and inner radius \(r\). Write its area and explain why the radii must be measured from the rotation axis.
- A region is rotated about the \(y\)-axis, and you use slices perpendicular to that axis. Which variable describes slice position?
- A student writes \(\pi(R-r)^2\) for the area of a washer. State the correction.
