Why Setup Errors Matter
A volume integral can be evaluated correctly and still give the wrong answer if the integral does not represent the solid. The most common mistakes happen before integration: a length is identified incorrectly, boundaries are put in the wrong order, part of the region is omitted, or the integration variable does not match the direction of the slices.
A reliable setup begins with the geometry. For each slice, identify its position, its cross-sectional area, and the interval over which the slices fill the solid. Only then write the integral. This tutorial focuses on diagnosing errors in those choices; it does not replace the earlier material on disks, washers, and cross-sectional areas.
A Setup Check Before Integrating
For slices perpendicular to the \(x\)-axis, the thickness is \(dx\), and the volume is written as \(V=\int_a^b A(x)\,dx\). For slices perpendicular to the \(y\)-axis, the thickness is \(dy\), and the volume is \(V=\int_c^d A(y)\,dy\). Here \(A\) is the cross-sectional area, and the bounds describe the full range of slice positions.
For rotation about an axis, a disk has area \(\pi r^2\), where \(r\) is the radius. A washer with outer radius \(R\) and inner radius \(r\) has area \(\pi(R^2-r^2)\), with \(R\ge r\ge0\). The radii are distances from the axis, not simply the names or values of the boundary functions.
Decide whether the slices are perpendicular to the \(x\)-axis or to the \(y\)-axis. This determines whether the thickness is \(dx\) or \(dy\).
Write its radius, its inner and outer radii, or its side lengths in terms of the slice variable.
Use the extent of the region in the direction of slicing. Check endpoints and any intersections where the boundary or radius ordering may change.
Check that areas are nonnegative, the bounds include the whole solid, and the integrand has units of area so that multiplying by a length gives volume.
Why Cross-Section Integrals Represent Volume
The cross-section method is justified by approximating the solid with thin slabs. Suppose slices perpendicular to the \(x\)-axis have area \(A(x)\) over \(a\le x\le b\). Divide this interval into subintervals of widths \(\Delta x_i\). A thin slab near a sample point \(x_i^*\) has approximate volume \(A(x_i^*)\Delta x_i\). Adding the slab volumes gives an approximation to the solid’s volume.
As the subdivisions become finer, the slab approximation approaches the volume. When \(A\) is integrable, the limit of these sums is the definite integral. The same reasoning applies to slices perpendicular to the \(y\)-axis, using widths \(\Delta y_i\).
This explains why a mismatched variable is not a minor notation issue. The differential records the thickness of the slabs, so it must measure distance in the direction in which the slices are stacked.
Mistake 1: Using a Diameter as a Radius
A disk’s area uses the square of its radius, not the square of its diameter. If a boundary curve gives the distance from the axis to the edge of a disk, that distance is the radius. If the curve gives the full width across the disk, it is a diameter and must be halved first.
Worked Example: A Disk Radius Mistaken for Half Its Value
Take the region \(0\le y\le 2x\), \(0\le x\le1\), and rotate it about the \(x\)-axis. At a position \(x\), the segment from the axis \(y=0\) to \(y=2x\) sweeps out a disk. Its radius is \(2x\), so the correct setup is:
Integrate and evaluate at the bounds:
A common wrong setup treats \(2x\) as a diameter, uses radius \(x\), and obtains \(\pi\int_0^1x^2\,dx=\pi/3\). But the axis-to-curve distance is already the radius; it is not the full width of the disk. As a check, at \(x=1\) the radius is \(2\), and the cross-sectional area is \(4\pi\), consistent with the correct integrand \(\pi(2x)^2\). Also, differentiating the antiderivative \(4\pi x^3/3\) gives \(4\pi x^2\), the integrand used.
Mistake 2: Reversing the Washer Radii
For a washer, the outer radius is the greater distance from the axis, and the inner radius is the lesser distance. Reversing them makes the cross-sectional area negative. That sign is a warning that the setup does not describe a physical area.
Worked Example: Putting the Inner Radius First
Consider the region \(x\le y\le2\), \(0\le x\le2\), rotated about the \(x\)-axis. At each \(x\), the outer radius is \(R=2\) and the inner radius is \(r=x\). Therefore:
Evaluating step by step gives:
If the radii are reversed, the proposed area is \(\pi(x^2-4)\). At \(x=0\), that expression is \(-4\pi\), even though a cross-sectional area cannot be negative. The correction is to compare distances from the axis: \(2\ge x\) throughout \(0\le x\le2\), so subtract the inner disk’s area from the outer disk’s area. At the endpoint \(x=2\), the radii agree and the washer area is zero, as the corrected formula gives. The antiderivative \(4x-x^3/3\) differentiates to \(4-x^2\), confirming the integration.
Mistake 3: Choosing Bounds That Miss Part of the Region
Bounds must cover the entire region in the direction of slicing, not just a convenient subinterval. When curves bound the region, their intersections often determine the endpoints. Check those intersections and test the boundary ordering across the interval.
Worked Example: Stopping at the Wrong Intersection
Take the region between \(y=2x\) and \(y=x^2\) in the first quadrant, and rotate it about the \(x\)-axis. The curves intersect where \(2x=x^2\), so \(x(x-2)=0\), giving \(x=0\) and \(x=2\). On this interval, \(2x\ge x^2\), so the outer radius is \(2x\) and the inner radius is \(x^2\).
Evaluate both terms at both endpoints:
A mistaken upper bound of \(1\) would give \(\pi(4/3-1/5)=17\pi/15\). That integral only includes the part from \(x=0\) to \(x=1\), despite the second intersection being at \(x=2\). The correct interval includes all the way to that second intersection. As an ordering check, at \(x=1\), \(2x=2\) exceeds \(x^2=1\); at \(x=2\), both radii are \(4\), so the washer closes at the region’s endpoint. Differentiating \(4x^3/3-x^5/5\) gives \(4x^2-x^4\), as required.
Mistake 4: Using a Slice Direction That Does Not Match the Geometry
The direction of the slices determines both the variable and the cross-sectional shape described by the problem. For example, if square cross sections are specified perpendicular to the \(y\)-axis, each square’s side must be measured horizontally across the base region, and the integral must use \(dy\). Merely changing \(dy\) to \(dx\) does not turn those slices into a valid description of the same solid.
Worked Example: Square Cross Sections Perpendicular to the \(y\)-Axis
Let the base be \(0\le y\le2\), \(0\le x\le y^2\). The solid has square cross sections perpendicular to the \(y\)-axis. At a fixed \(y\), the horizontal width of the base is \(y^2-0=y^2\), so the square’s area is \(A(y)=(y^2)^2=y^4\). The correct variable and bounds are therefore:
A setup such as \(\int_0^2 y^4\,dx\) is inconsistent: \(y\) is the slice-position variable in this description, while \(dx\) represents thickness in the \(x\)-direction. It also fails to specify a valid \(x\)-interval for each slice. The correction is not to change the differential casually, but to use the prescribed horizontal slices and their full range \(0\le y\le2\). The antiderivative \(y^5/5\) differentiates to \(y^4\), and evaluating it at \(2\) and \(0\) gives \(32/5-0\).
How to Diagnose a Suspicious Setup
Before evaluating a volume integral, inspect the setup independently of the antiderivative. The following checks catch many errors early:
- Radius check: Is the stated length a radius measured from the axis, or a diameter that still needs to be halved?
- Washer check: Is the outer radius at least as large as the inner radius throughout the interval?
- Bounds check: Do the bounds reach the full extent of the region in the chosen slice direction? Have all relevant intersections been included?
- Variable check: Does the differential match the direction in which the slices are stacked?
- Area and unit check: Is the cross-sectional area nonnegative and measured in square units? After multiplying by a length differential, does the integral have cubic units?
These checks are more useful than relying on the final numerical value alone. A positive answer, for instance, does not prove that washer radii were ordered correctly at every point or that the bounds captured the whole solid. A setup should be justified by its geometry, then checked algebraically.
Check Your Understanding
For each question, focus on whether the proposed setup represents the geometry before attempting to integrate.
- A disk’s diameter at position \(x\) is \(6x\). What radius should be squared in its area formula?
- A washer has distances \(5\) and \(x+1\) from its axis, where \(0\le x\le2\). Which distance is the outer radius?
- Why should intersections of boundary curves be checked when setting bounds?
- A problem specifies triangular cross sections perpendicular to the \(x\)-axis. Which differential records the slice thickness?
- If a washer integrand is negative at some point, what geometric feature of the setup should be checked first?
