Why Volume Setups Go Wrong
A volume integral can be evaluated correctly and still give the wrong answer if its setup does not match the region. Common errors include integrating with respect to the wrong variable, using bounds that do not cover the region, subtracting boundaries in the wrong order, and treating a curve’s coordinate as a radius instead of measuring its distance from the axis.
A reliable setup begins with the slice. Ask which way the slice runs, which variable locates it, and what geometric quantity the slice contributes. For a cross-section, that quantity is its area. For a washer, it is the area between two concentric circles. In either case, checking what the quantities mean before integrating can reveal errors early.
Cross-Section Setups: Direction and Base Length
For cross-sections, the base of a slice is a segment in the region. A vertical base segment runs from a lower curve to an upper curve at a fixed \(x\), so its length is upper \(y\)-value minus lower \(y\)-value. A horizontal base segment runs from a left boundary to a right boundary at a fixed \(y\), so its length is right \(x\)-value minus left \(x\)-value.
The slice direction determines the integration variable. Slices at fixed \(x\) have thickness \(dx\), while slices at fixed \(y\) have thickness \(dy\). The limits must cover the values of that same variable for which the region has slices. If a cross-section is built on a base of length \(b\), its area depends on the specified shape: for example, a square has area \(b^2\), and a semicircle with diameter \(b\) has area \(\frac{\pi b^2}{8}\).
This distinction prevents a common mix-up. A vertical segment is measured by a difference in \(y\)-values, even though the volume integral uses \(dx\) to stack vertical slices. A horizontal segment is measured by a difference in \(x\)-values, even though the volume integral uses \(dy\).
Worked Example: Square Cross-Sections on Vertical Bases
Worked Example: Square Cross-Sections on Vertical Bases
The region lies between \(y=x\) and \(y=x^2\) for \(0\le x\le1\). Each cross-section perpendicular to the \(x\)-axis is a square whose base is the vertical segment in the region. Set up and evaluate the volume.
At a fixed \(x\) between \(0\) and \(1\), compare the curves: \(x-x^2=x(1-x)\ge0\). Thus \(y=x\) is above \(y=x^2\), and the vertical base length is
The slices are located by \(x\), so the bounds are \(0\) and \(1\), and the thickness is \(dx\). The cross-section is a square, so its area is \(A(x)=b(x)^2\). Therefore
Integrate term by term and evaluate at the bounds:
Check the setup: \(x-x^2\ge0\) on \([0,1]\), so the square’s area is nonnegative. The base length is zero at both endpoints, matching the fact that the two boundary curves meet there. The evaluated volume is therefore consistent with the geometry.
Bounds Must Match the Slice Variable
Bounds usually come from the region’s extent in the integration variable. For vertical slices, find the leftmost and rightmost \(x\)-values of the region. For horizontal slices, find its lowest and highest \(y\)-values. When bounds come from curve intersections, solve for the coordinate that describes the slice and use the relevant intersections.
Do not copy \(x\)-bounds into a \(dy\) integral or \(y\)-bounds into a \(dx\) integral without checking what they represent. If a boundary must be written in terms of the other variable, solve for that variable and verify which branch or boundary applies over the interval.
Worked Example: Semicircles on Horizontal Bases
Worked Example: Semicircles on Horizontal Bases
The region is bounded by \(y=x^2\) and \(y=4\). Each cross-section perpendicular to the \(y\)-axis is a semicircle whose diameter is the horizontal segment in the region. Set up and evaluate the volume.
For \(0\le y\le4\), the equation \(y=x^2\) gives \(x=-\sqrt y\) on the left and \(x=\sqrt y\) on the right. The horizontal base length is right minus left:
The slices are horizontal, so integrate with respect to \(y\). Their bounds are \(0\) to \(4\). A semicircle with diameter \(b\) has radius \(b/2\), so its area is \(\frac12\pi(b/2)^2=\frac{\pi b^2}{8}\). Thus
Evaluate:
Check the base length: at \(y=0\), the two \(x\)-boundaries meet and \(b(0)=0\); at \(y=4\), the segment runs from \(-2\) to \(2\), so \(b(4)=4\), as the formula gives. The bounds and the length are consistent with the region.
Washer Setups: Radii Are Distances
A washer is the solid cross-section formed by rotating a region slice around an axis when the slice leaves a hole. If the outer radius is \(R\) and the inner radius is \(r\), then \(R\ge r\ge0\). The washer’s area is the outer disk’s area minus the inner disk’s area:
This formula follows because the inner disk is precisely the part removed from the disk of radius \(R\). Adding the washer areas across slices gives the volume integral. Importantly, a radius is a distance from the axis of rotation to a boundary of the region. It is not necessarily just the boundary’s \(y\)-value or \(x\)-value. For an axis \(y=c\), a point with coordinate \(y\) is at distance \(|y-c|\) from the axis; for an axis \(x=c\), a point with coordinate \(x\) is at distance \(|x-c|\).
To order the radii, compare those distances at each slice. “Outer” means farther from the axis; “inner” means closer. A curve that is higher on a graph is not automatically the outer boundary when the axis is shifted or when the region lies on the other side of the axis.
Worked Example: Washer Radii About a Shifted Axis
Worked Example: Washer Radii About a Shifted Axis
The region between \(y=x^2\) and \(y=x\), for \(0\le x\le1\), is rotated about \(y=-1\). Set up and evaluate the washer volume.
At each \(x\), the upper curve is \(y=x\) and the lower curve is \(y=x^2\), since \(x-x^2\ge0\). The axis \(y=-1\) lies below both curves. The distance from the axis to the upper curve is \(x-(-1)=x+1\); the distance to the lower curve is \(x^2-(-1)=x^2+1\). Since \(x+1\ge x^2+1\) on \([0,1]\), the outer radius is \(R(x)=x+1\) and the inner radius is \(r(x)=x^2+1\).
The slices are vertical and perpendicular to the horizontal axis of rotation, so use \(dx\) with bounds \(0\) to \(1\):
Expand the squares and simplify:
Now integrate and evaluate:
Check the radius ordering at an interior value such as \(x=\frac12\): \(R=\frac32\) and \(r=\frac54\), so \(R>r\). The integrand there is positive, as a washer area must be. At \(x=0\) and \(x=1\), the two boundary curves meet, the radii agree, and the washer area is zero.
A Common Mistake: Subtracting in the Wrong Order
For the last example, a tempting but incorrect setup is to call the distance to the lower curve the outer radius and write \(\pi\int_0^1[(x^2+1)^2-(x+1)^2]\,dx\). This reverses the radii, making the integrand negative wherever the region has positive thickness. For instance, at \(x=\frac12\), the incorrect difference is \((\frac54)^2-(\frac32)^2=\frac{25}{16}-\frac{36}{16}=-\frac{11}{16}\).
The correction is to compare distances from the axis before assigning inner and outer: \(x+1\) is farther from \(y=-1\) than \(x^2+1\) on the interval. Use outer squared minus inner squared. A negative washer area signals a setup error, not a negative volume.
Why These Checks Matter
Volume setups translate geometry into an integral. Each piece of the integral has a geometric role: the bounds describe where the slices exist, the differential describes how they are stacked, and the area describes each slice. If one piece uses a different direction or measures the wrong segment, the integral may still be easy to evaluate but will represent a different solid.
Before evaluating, check the setup with a simple value from the interval, especially an endpoint or midpoint. Confirm that the slice fits the region, that a base length or radius is nonnegative, and that an area is not negative. These checks do not replace the setup, but they often expose mistakes before they affect the final answer.
State whether it is vertical or horizontal and choose \(dx\) or \(dy\) accordingly.
Use the region’s extent in the integration variable, checking intersections or boundary branches as needed.
For a cross-section, subtract the endpoints of its base segment in the correct order. For a washer, measure from the axis to each boundary.
Verify that lengths and radii are nonnegative and that the outer radius is at least the inner radius.
Check Your Understanding
For each question, focus on the geometric meaning of the setup before doing any lengthy calculation.
- For a vertical slice through a region bounded above by \(y=3\) and below by \(y=x^2\), what expression gives the base length?
- A horizontal slice runs from \(x=1-y\) to \(x=2+y\). What is its base length?
- For a \(dy\) integral, what feature of the region should determine the bounds?
- A region is rotated about \(y=2\). What distance from the axis corresponds to a boundary at \(y=f(x)\)?
- At a slice, a proposed washer setup has \(R=2\) and \(r=3\). What does this tell you to check?
