Choosing the Right Slice Model
A volume problem begins with a geometric question: what does a thin slice through the solid look like? In some problems, the cross-sectional shape is specified directly, such as a square or semicircle. In others, a plane region is revolved around an axis, so each slice perpendicular to that axis is a disk or washer.
Both methods use the same basic principle: the volume is found by adding the volumes of thin slices. The important difference is how the area of each slice is determined. For prescribed cross-sections, use the stated shape and its dimensions. For a solid of revolution, use the distance from the axis to the boundary to determine the radii.
Area Functions and Volume Integrals
Suppose a solid extends along the \(x\)-axis from \(x=a\) to \(x=b\), and a cross-section perpendicular to that axis has area \(A(x)\). If the areas vary continuously, the volume is the definite integral of the cross-sectional area:
For slices perpendicular to the \(y\)-axis, the corresponding expression is \(V=\int_c^d A(y)\,dy\). In either case, the area must describe the actual slice of the solid at that position.
For a solid formed by revolving a region around an axis, a perpendicular slice is a disk if there is no hole, and a washer if there is a hole. If the outer radius is \(R\) and the inner radius is \(r\), the area is the area of the outer circle minus the area of the inner circle:
A disk is the special case \(r=0\), with area \(\pi R^2\). Radii are distances from the axis of rotation, not necessarily the values of the boundary functions themselves. Their values depend on the axis and the region’s boundaries.
Why Adding Slice Areas Gives Volume
The integral formula expresses a geometric approximation. Divide the interval \([a,b]\) into thin subintervals. Over a short subinterval of width \(\Delta x\), the solid is approximately a thin slab whose volume is its cross-sectional area times its thickness. If \(x_i^*\) is a point in that subinterval, the slab’s approximate volume is \(A(x_i^*)\Delta x\). Adding these approximations gives a Riemann sum.
As the subinterval widths approach zero, the approximation becomes exact when the cross-sectional area is integrable. By the definition of the definite integral, the limiting sum is \(\int_a^b A(x)\,dx\). The same reasoning applies to slices measured in \(y\). For rotation, the area inserted into that sum is the disk or washer area.
A Practical Decision Process
If the problem specifies shapes for slices, use those shapes. If a plane region is revolved around an axis, use disks or washers perpendicular to that axis.
Use slices perpendicular to the direction along which the solid is being measured. For rotation, they must also be perpendicular to the axis of rotation.
For prescribed shapes, find the required side length, diameter, or other dimension. For washers, identify the outer and inner distances from the axis.
Integrate the cross-sectional area over the full interval. Check the bounds, the slice geometry, the evaluation, and the volume units.
Worked Example: Square Cross-Sections
Worked Example: Square Cross-Sections
The base of a solid is the region under \(y=\sqrt{x}\), above the \(x\)-axis, and between \(x=0\) and \(x=4\). Cross-sections perpendicular to the \(x\)-axis are squares. Find the volume.
At a fixed \(x\), the vertical segment in the base runs from \(y=0\) to \(y=\sqrt{x}\). This segment is the side of the square, so its length is \(s(x)=\sqrt{x}\). The area of a square of side \(s\) is \(s^2\), hence
Integrate from the left endpoint \(0\) to the right endpoint \(4\):
To check the integration, differentiate the antiderivative: \(\frac{d}{dx}(x^2/2)=x\), which is the integrand. The area is nonnegative throughout the interval, and it grows from \(0\) to \(4\), as expected from the increasing side length. If lengths are measured in units, the volume is \(8\) cubic units.
Worked Example: Semicircular Cross-Sections
Worked Example: Semicircular Cross-Sections
The base is the region between \(y=x\) and \(y=x^2\) for \(0\leq x\leq1\). Cross-sections perpendicular to the \(y\)-axis are semicircles whose diameters lie in the base. Find the volume.
Because the slices are horizontal, express the left and right boundaries as \(x\)-values in terms of \(y\). From \(y=x\), we obtain \(x=y\). From \(y=x^2\), we obtain \(x=\sqrt{y}\) on this region. For \(0<y<1\), \(y<\sqrt{y}\), so the diameter is
A semicircle with diameter \(d\) has radius \(d/2\), so its area is \(\frac12\pi(d/2)^2=\frac{\pi}{8}d^2\). Therefore,
The \(y\)-values in the base run from \(0\) to \(1\). Integrating gives
For a check, the derivative of the bracketed antiderivative is \(y-2y^{3/2}+y^2\), as required. The diameter is zero at both endpoints and positive between them, so the slice areas have the same behavior. The resulting volume, \(\pi/240\) cubic units, is positive.
Worked Example: A Washer Solid
Worked Example: A Washer Solid
The region bounded by \(y=2\), \(y=x\), and \(x=0\) is revolved around the \(x\)-axis. Find the volume.
The region runs from \(x=0\) to the intersection \(x=2\), where \(x=2\) meets \(y=2\). A vertical slice perpendicular to the axis of rotation becomes a washer. The outer radius is the distance from the \(x\)-axis to \(y=2\), so \(R=2\). The inner radius is the distance from the axis to \(y=x\), so \(r=x\). Thus
Integrate over \(0\leq x\leq2\):
Differentiate \(4x-x^3/3\) to check: its derivative is \(4-x^2\), matching the integrand after the factor \(\pi\) is taken outside. At \(x=0\), the inner radius is zero, so the washer begins as a disk; at \(x=2\), the radii agree and the slice area is zero. These endpoint checks fit the region’s geometry. The volume is \(16\pi/3\) cubic units.
A Common Mistake: Confusing Diameter and Radius
In the semicircle example, the distance between the boundary curves is the diameter, not the radius. A tempting but incorrect setup is to use the full-circle area \(\pi d^2\). That would give
This is wrong for two reasons: the slices are semicircles rather than full circles, and \(d\) is a diameter rather than a radius. A semicircle’s area in terms of its diameter is \(\frac{\pi}{8}d^2\), so the correct result is \(\frac{\pi}{8}\cdot\frac{1}{30}=\frac{\pi}{240}\). The correction is not an arbitrary factor; it follows from using radius \(d/2\) and taking half the area of a full circle.
Final Checks Before Accepting a Volume
Before finalizing a setup, verify that the integral describes the solid rather than merely using familiar formulas. Check that the slice direction matches the construction, that the interval spans the entire solid, and that the area is nonnegative. In a washer setup, confirm which boundary is farther from the axis; in a cross-section problem, confirm which segment supplies the specified dimension.
Also check the calculation independently. The antiderivative should differentiate to the integrand, and the evaluated volume should be positive for a solid with positive thickness somewhere. If the dimensions are lengths, area has squared-length units and integration with respect to length produces cubic-length units.
Check Your Understanding
For each question, identify the slice geometry before calculating.
- A solid has square cross-sections perpendicular to the \(x\)-axis, with side length \(3-x\) for \(0\leq x\leq2\). What area function and integral give its volume?
- A semicircle has diameter \(d\). Derive its area in terms of \(d\), and explain why \(\pi d^2\) is not the correct area.
- When a region is revolved around the \(x\)-axis, what do the outer and inner radii of a washer measure?
- For the region between \(y=3\) and \(y=x\), \(0\leq x\leq3\), revolved around the \(x\)-axis, identify the washer radii and write the volume integral.
- Why does integrating cross-sectional area with respect to \(y\) require slices perpendicular to the \(y\)-axis?
