Volume as the integral of a cross-sectional area, solids with known cross-sections, the disk method for a region touching the axis, the washer method for one separated from it, revolving about lines other than the coordinate axes, and why the axis decides the variable.
Subject: Calculus I · 65 slides · symbolic lesson
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Title
Calculus I · Chapter 6 — Applications of Integration
Determining Volumes by Slicing
Objectives
Five outcomes. The idea is Section 6.1's, with a slice's area in place of a strip's height.
OpenStax Calculus Volume 1, §6.2 Determining Volumes by Slicing §6.2, pp. 551-568 — the section these objectives are drawn from
Warm-up
Section 6.1 found an area by adding thin strips: each strip's area was its height times its width, and integrating summed them.
Discussion prompt
How would you find the volume of a solid whose cross-sections perpendicular to an axis are all squares of known size?
Hint: What is a thin slab's volume?
Answer:
A thin slab has volume equal to its cross-sectional area times its thickness — exactly as a thin strip had area equal to its height times its width. So if the cross-section at position x is a square of side s of x, the slab's volume is that side squared times dx.
\[ V = \int_{a}^{b}A(x)\,dx \]
Integrating adds the slabs. That is the entire content of this section, and the disk and washer methods that occupy most of it are simply the two cases where the cross-section happens to be a circle or an annulus.
Concept
Slice the solid perpendicular to an axis. Each slice is a thin slab whose volume is its cross-sectional area times its thickness, and integrating that area over the solid's extent gives the volume.
volume by slicing — The volume of a solid equals the integral of its cross-sectional area, taken perpendicular to an axis, over the interval the solid occupies.
\[ V = \int_{a}^{b}A(x)\,dx \]
The formula is general: any cross-section whose area can be written as a function of position will do. Disks and washers are simply the shapes that arise when a plane region is revolved.
Figure (svg): A solid sliced perpendicular to an axis, with one representative slice
OpenStax Calculus Volume 1, §6.2 Determining Volumes by Slicing §6.2, pp. 620-630
Section
Section 1
Concept
The cross-section need not be a circle. Squares, semicircles and triangles all work, and the method is unchanged: express the area as a function of position and integrate.
cross-sectional area function — The area of the slice at each position along the axis, written as a function of that position. It is the integrand in every volume-by-slicing problem.
\[ A(x) = s(x)^{2} \text{ for squares}, \quad \tfrac{\sqrt3}{4}s(x)^{2} \text{ for equilateral triangles} \]
Starting from the general case makes the disk and washer methods look like instances rather than separate techniques, which is how they are best remembered.
Figure (svg): Solids with known cross-sections that are not disks
OpenStax Calculus Volume 1, §6.2 Determining Volumes by Slicing §6.2, pp. 620-630 — volume by slicing
Picture it
The method does not care which.
Figure (svg): Solids with known cross-sections that are not disks
Every case reduces to writing the area in terms of position. Once that is done the integration is Chapter 5 routine, exactly as it was for areas.
Worked example
Example 6.6. A solid built on a circular base.
\[ \text{A solid has a base bounded by } x^{2}+y^{2}=4 \text{ with square cross-sections perpendicular to the } x\text{-axis. Find its volume.} \]
Find the base's width at position x
Why: Between the two halves of the circle.
\[ 2 \sqrt{4 - x ^{2}} \]
Identify the square's side
Why: The width is the side.
\[ s = 2 \sqrt{4 - x ^{2}} \]
Write the area function
Why: Side squared.
\[ A(x) = 4(4 - x ^{2}) \]
Set the limits
Why: The base's extent.
\[ \text{from } -2\text{ to } 2 \]
Integrate
Why: A polynomial.
\[ 4(16 - \frac{16}{3}) = \frac{128}{3} \]
Figure (svg): Solids with known cross-sections that are not disks
\[ V = \int_{-2}^{2}4\left(4-x^{2}\right)dx = \frac{128}{3} \]
Verify: sanity-check against a bounding box
Why: The solid fits inside a box 4 long, 4 wide and 4 tall, of volume 64 — and it fills about two thirds of that, which is plausible for a shape tapering to nothing at both ends. Note that the integrand was found geometrically, from the base's width, and only then integrated: the geometry is the work and the calculus is the routine part, exactly as in Section 6.1.
OpenStax Calculus Volume 1, §6.2 Determining Volumes by Slicing §6.2, pp. 623-625
Fill the middle
Square cross-sections whose side is the base's width.
Fill in the blanks
A(x) = \left[2\sqrt2}\right]^___}
Why: The side is squared to give the square's area. Integrating the side itself would give the base's area rather than the solid's volume, and the units reveal the difference.
Worked example
Checkpoint 6.6. Deriving a known formula.
\[ \text{Derive the volume of a pyramid of height } h \text{ and square base of side } b. \]
Set up coordinates
Why: Measure from the apex downward.
\[ x\text{ from } 0\text{ to } h \]
Find the side at depth x
Why: Similar triangles.
\[ s = b x / h \]
Write the area function
Why: A square.
\[ A(x) = b ^{2} x ^{2} / h ^{2} \]
Integrate from 0 to h
Why: The power rule.
\[ (b ^{2} / h ^{2}) (h ^{3} / 3) \]
Simplify
Why: Collect.
\[ b ^{2} h / 3 \]
Figure (svg): The solution to Worked example a pyramid from first principles shown as a ladder of expressions, one row per legal move
\[ V = \frac{b^{2}h}{3} \]
Verify: compare with the formula from geometry
Why: The classical formula is one third of the base area times the height, and that is exactly what came out. The factor of one third, memorised in school geometry without justification, appears here as the integral of a square — it is the three in the power rule's denominator. Deriving a known result is a useful check that the method is being applied correctly before trusting it on shapes with no formula.
OpenStax Calculus Volume 1, §6.2 Determining Volumes by Slicing §6.2, pp. 625-627
Trap
\[ V = \int_{-2}^{2}2\sqrt{4-x^{2}}\,dx \]
Integrate the width rather than the cross-sectional area
Why: The student integrates the base's chord.
That integral gives the AREA of the circular base, not the volume of the solid built on it.
\[ V = \int_{-2}^{2}\left[2\sqrt{4-x^{2}}\right]^{2}dx \]
Square the side to get the square's area, then integrate
Why: The integrand is an area, so its units must be a length squared.
Checking units catches this immediately: integrating a length gives an area, and a volume needs an area integrand. That check applies to every problem in this chapter.
Matching
Each in terms of a side or diameter s.
Match the pairs
Why: Every one is proportional to the square of a length, which is why the units check works so reliably. The constant of proportionality is the only thing that changes between cross-section shapes.
Ranking
Volume by slicing.
Put in order
Why: Steps b and c are the geometry and contain all the difficulty; step e is Chapter 5 routine. That division of labour is the same one Section 6.1 established for areas.
Prediction
Commit before reasoning.
Predict first
In a volume-by-slicing integral, what kind of quantity is the integrand?
Correct: An area.
\[ [\text{area}]\times[\text{length}] = [\text{volume}] \]
Why: The differential supplies a length, so the integrand must be an area for the product to be a volume. That is a complete units check and it catches the commonest error in this section — integrating a width or a radius rather than the area it determines. Any volume integrand that is not visibly a length squared is wrong.
Section
Section 2
Concept
Revolving a region that reaches the axis of revolution produces slices that are full disks, with radius equal to the distance from the axis to the curve.
the disk method — Revolving a region bounded by a curve and the axis produces circular cross-sections of radius equal to the function's value, so the volume is pi times the integral of the function squared.
\[ V = \pi\int_{a}^{b}\left[f(x)\right]^{2}dx \]
The squaring is what makes these integrals different in character from Section 6.1's. A square root revolved gives a polynomial integrand, and a polynomial revolved gives a higher-degree one.
Figure (svg): A region revolved about the x-axis, generating a solid of disks
OpenStax Calculus Volume 1, §6.2 Determining Volumes by Slicing §6.2, pp. 626-636 — the disk method
Picture it
The region and the solid it generates.
Figure (svg): A region revolved about the x-axis, generating a solid of disks
Because the region touches the axis, every slice is a complete disk with no hole. The radius at each position is simply the function's value there.
Worked example
Example 6.8. A square root swept around.
\[ \text{Revolve the region under } y=\sqrt{x} \text{ on } [0,4] \text{ about the } x\text{-axis.} \]
Identify the radius
Why: The distance from the axis to the curve.
\[ R = \sqrt{x} \]
Write the area function
Why: A disk.
\[ A(x) = \pi x \]
Note the simplification
Why: The square undoes the root.
Integrate from 0 to 4
Why: The power rule.
\[ \pi(\frac{16}{2}) \]
State
Why: The volume.
\[ 8 \pi \]
Figure (svg): A region revolved about the x-axis, generating a solid of disks
\[ V = \pi\int_{0}^{4}x\,dx = 8\pi \]
Verify: sanity-check against a cylinder
Why: The solid fits inside a cylinder of radius 2 and length 4, of volume 16 pi — and it fills exactly half of that, which is plausible for a shape tapering to a point at one end. Note that squaring turned an awkward square root into a linear function: revolution problems are often computationally easier than the corresponding area problems, which is a pleasant reversal.
OpenStax Calculus Volume 1, §6.2 Determining Volumes by Slicing §6.2, pp. 629-631
Fill the middle
A circular cross-section of radius equal to the function's value.
Fill in the blanks
A(x) = \pi\left[f(x)\right]^___
Why: The cross-section is a disk, so its area is pi times the radius squared. The constant factors outside the integral, which is exactly why it is easy to forget.
Worked example
Checkpoint 6.8. The variable is forced by the axis.
\[ \text{Revolve the region bounded by } y=x^{2}, \; y=4 \text{ and the } y\text{-axis about the } y\text{-axis.} \]
Note the axis of revolution
Why: Vertical.
Solve the boundary for x
Why: The radius must be a function of y.
\[ x = \sqrt{y} \]
Write the area function
Why: A disk.
\[ A(y) = \pi y \]
Find the y limits
Why: From the region.
\[ 0\text{ to } 4 \]
Integrate with respect to y
Why: The power rule.
\[ 8 \pi \]
Figure (svg): The solution to Worked example revolving about the y-axis shown as a ladder of expressions, one row per legal move
\[ V = \pi\int_{0}^{4}y\,dy = 8\pi \]
Verify: notice the variable was not a choice
Why: Unlike Section 6.1, where either variable could be used and one was merely more convenient, here the axis of revolution forces the slicing direction — slices must be perpendicular to it or they are not disks at all. That removes a decision and imposes a constraint: the boundary had to be solved for x, and if that had been impossible a different method would have been needed. Section 6.3's shells exist for exactly that situation.
OpenStax Calculus Volume 1, §6.2 Determining Volumes by Slicing §6.2, pp. 631-633
Error analysis
A student computes a volume of revolution.
Annotate
On: \( V = \int_{0}^{4}\left(\sqrt{x}\right)^{2}dx = 8 \)
The constant is easy to lose because it factors outside the integral and plays no part in the integration. Writing it down at the moment the area function is written, rather than adding it afterwards, is the habit that prevents this.
Sorting
Slice perpendicular to the axis of revolution.
Sort into buckets
Sort each situation.
A horizontal axis always means integrating in x, whatever its height, and a vertical one always means y. The position of the axis affects the radii but never the variable.
Two truths and a lie
All three are about disks.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. For areas either direction worked and one was merely tidier; here the slicing direction is forced by the axis, because only perpendicular slices are disks. That constraint is what makes Section 6.3's shell method necessary for some regions.
Prediction
Commit before reasoning.
Predict first
Revolving the region under a square root gave a linear integrand. Why does this happen so often?
Correct: Because the area squares the radius.
\[ \left(\sqrt x\right)^{2} = x \quad \text{inside the integral} \]
Why: A disk's area is proportional to the radius squared, so a square-root boundary becomes linear and a boundary containing any even root simplifies. Volume-of-revolution integrals are frequently easier than the area integrals for the same region, which is an unexpected reversal — the extra dimension makes the algebra simpler rather than harder.
Section
Section 3
Concept
When the region does not reach the axis, each slice is an annulus. Its area is the outer disk's minus the inner one's — a difference of squares, not the square of a difference.
the washer method — For a region separated from the axis of revolution, the cross-section is an annulus whose area is pi times the outer radius squared minus the inner radius squared.
\[ V = \pi\int_{a}^{b}\left[R(x)^{2}-r(x)^{2}\right]dx \]
The single commonest error in this section is writing the square of the difference of radii instead. Those are different quantities and the mistake is invisible in the arithmetic that follows.
Figure (svg): The disk and washer cross-sections compared
OpenStax Calculus Volume 1, §6.2 Determining Volumes by Slicing §6.2, pp. 632-642 — the washer method
Picture it
One radius, or two.
Figure (svg): The disk and washer cross-sections compared
The washer's area subtracts the hole's area from the outer disk's. Subtracting the radii first and then squaring gives a different and wrong number.
Worked example
Example 6.10. Two radii.
\[ \text{Revolve the region between } y=x \text{ and } y=x^{2} \text{ about the } x\text{-axis.} \]
Find the intersections
Why: Set equal.
\[ x = 0\text{ and } 1 \]
Identify the outer radius
Why: The farther boundary from the axis.
\[ R = x \]
Identify the inner radius
Why: The nearer one.
\[ r = x ^{2} \]
Write the area function
Why: Difference of squares.
\[ \pi(x ^{2} - x ^{4}) \]
Integrate from 0 to 1
Why: Term by term.
\[ \pi(\frac{1}{3} - \frac{1}{5}) = 2 \pi / 15 \]
Figure (svg): The disk and washer cross-sections compared
\[ V = \pi\int_{0}^{1}\left(x^{2}-x^{4}\right)dx = \frac{2\pi}{15} \]
Verify: check what the wrong form would have given
Why: Squaring the difference instead gives the integral of x minus x squared, all squared — which comes to pi over 30, half the correct answer. Both are small positive numbers and neither looks obviously wrong, which is exactly why the error persists. Expanding the square of the difference shows it contains a cross term that has no business being there.
OpenStax Calculus Volume 1, §6.2 Determining Volumes by Slicing §6.2, pp. 635-637
Fill the middle
An annular cross-section with outer and inner radii.
Fill in the blanks
A(x) = \pi\left[R^r^2-___\right]
Why: The areas subtract, not the radii. Writing the square of the difference instead describes a solid disk of the gap's size, which is a different shape entirely.
Worked example
Checkpoint 6.10. Expanding both.
\[ \text{Compare } R^{2}-r^{2} \text{ with } (R-r)^{2}. \]
Expand the first
Why: A standard factorisation.
\[ (R - r) (R + r) \]
Expand the second
Why: The binomial square.
\[ R ^{2} - 2 R r + r ^{2} \]
Compare
Why: They differ by a cross term.
Interpret geometrically
Why: The second is a disk of radius R minus r.
Note when they agree
Why: Only if r is zero.
Figure (svg): The solution to Worked example why the two forms differ shown as a ladder of expressions, one row per legal move
\[ R^{2}-r^{2} \ne (R-r)^{2} \text{ unless } r=0 \]
Verify: see the geometric meaning of the wrong form
Why: The square of the difference is the area of a disk whose radius is the gap between the two boundaries — a solid disk of that size, which is not the shape being cut at all. The washer is an annulus, and an annulus's area is genuinely a difference of two disk areas. Seeing what the wrong formula describes makes it much harder to write down by accident.
OpenStax Calculus Volume 1, §6.2 Determining Volumes by Slicing §6.2, pp. 637-639
Trap
\[ V = \pi\int_{0}^{1}\left(x-x^{2}\right)^{2}dx = \frac{\pi}{30} \]
Square the difference of the radii
Why: The student carries Section 6.1's habit of subtracting the boundaries.
That describes a disk whose radius is the gap between the curves, not an annulus. The answer is half the correct one.
\[ V = \pi\int_{0}^{1}\left(x^{2}-x^{4}\right)dx = \frac{2\pi}{15} \]
Subtract the AREAS: square each radius, then subtract
Why: An annulus is a disk with a disk removed.
The habit from Section 6.1 is precisely what causes this: there the strip's height WAS a difference of boundaries. Here the boundaries give two radii, and it is their areas that subtract.
Sorting
Does the region reach the axis?
Sort into buckets
Sort each revolution.
The last is the one to notice: the region touches the x-axis but the revolution is about a line below it, so a gap opens up and washers are needed. Whether a hole appears depends on the axis, not on the region alone.
Two truths and a lie
All three are about washers.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. Expanding both forms shows they differ by a cross term, and they agree only when the inner radius is zero — which is exactly the disk case. Section 6.1's habit of subtracting boundaries is what makes this error so persistent.
Prediction
Commit before reasoning.
Predict first
Why is a washer's area a difference of squares rather than the square of a difference?
Correct: Because a disk's area is removed from a disk's area.
\[ \pi R^{2}-\pi r^{2} = \pi(R-r)(R+r) \ne \pi(R-r)^{2} \]
Why: The annulus is literally what remains when a smaller disk is cut from a larger one, so the quantity subtracted is the smaller disk's area — pi times its radius squared. Subtracting the radii first describes a disk whose radius is the gap, which is a solid shape of a completely different size. Picturing the physical washer settles it in a second.
Section
Section 4
Concept
When the axis is a line other than a coordinate axis, each radius is the distance from that line to the boundary — the boundary's position minus the axis's, taken positively.
radius as a distance — The distance from the axis of revolution to a boundary curve. For a horizontal axis at height c, a boundary at height f gives a radius equal to the absolute difference of the two.
\[ R(x) = |f(x)-c| \]
Sketching the axis and one radius before writing anything is the reliable way to get these right. The shift is easy to omit and the resulting error is a plausible-looking number.
Figure (svg): Revolving about a line other than a coordinate axis: the radii shift
OpenStax Calculus Volume 1, §6.2 Determining Volumes by Slicing §6.2, pp. 638-646 — revolving about other lines
Picture it
A radius measured to a line below the origin.
Figure (svg): Revolving about a line other than a coordinate axis: the radii shift
The radius is the boundary's height minus the axis's, which here adds one rather than subtracting. Reading the sign off a sketch is safer than reasoning about it abstractly.
Worked example
Example 6.12. The shift changes both radii.
\[ \text{Revolve the region under } y=x^{2} \text{ on } [0,2] \text{ about } y=-1. \]
Sketch the axis
Why: One unit below the origin.
Find the outer radius
Why: To the curve.
\[ R = x ^{2} - (-1) = x ^{2} + 1 \]
Find the inner radius
Why: To the region's lower boundary, the x-axis.
\[ r = 0 - (-1) = 1 \]
Write the area function
Why: A washer.
\[ \pi [(x ^{2} + 1) ^{2} - 1] \]
Expand and integrate
Why: From 0 to 2.
\[ \pi(\frac{32}{5} + \frac{8}{3}) \]
Figure (svg): Revolving about a line other than a coordinate axis: the radii shift
\[ V = \pi\int_{0}^{2}\left[(x^{2}+1)^{2}-1\right]dx = \frac{136\pi}{15} \]
Verify: check what omitting the shift would have given
Why: Using x squared as the outer radius and zero as the inner gives 32 pi over 5, about 20.1 — noticeably smaller and entirely plausible-looking. The shift matters because moving the axis away from the region enlarges every radius, and the volume depends on their squares. Note also that a hole appeared: the region touches the x-axis but not the line y equals negative one, so washers were needed where disks would have sufficed before.
OpenStax Calculus Volume 1, §6.2 Determining Volumes by Slicing §6.2, pp. 641-643
Fill the middle
A boundary at height f, revolved about a line one unit below the origin.
Fill in the blanks
R = f(x) - (-1) = f(x) + 1
Why: The radius is the distance from the axis to the boundary, so the axis's position is subtracted. Here that adds one, which is easy to get backwards without a sketch.
Worked example
Checkpoint 6.12. The same idea sideways.
\[ \text{Revolve the region between } x=y^{2} \text{ and } x=4 \text{ about } x=4. \]
Note the axis
Why: Vertical.
Find the outer radius
Why: From the axis to the parabola.
\[ R = 4 - y ^{2} \]
Check for an inner radius
Why: The region touches the axis at x = 4.
\[ r = 0:\text{ disks} \]
Write the area function
Why: A disk.
\[ \pi(4 - y ^{2}) ^{2} \]
Integrate from -2 to 2
Why: Expand first.
\[ 512 \pi / 15 \]
Figure (svg): The solution to Worked example revolving about a vertical line shown as a ladder of expressions, one row per legal move
\[ V = \pi\int_{-2}^{2}\left(4-y^{2}\right)^{2}dy = \frac{512\pi}{15} \]
Verify: note why the radius subtracts this way round
Why: The axis is at x equal to 4 and the parabola is to its left, so the distance is 4 minus the parabola's x value — the larger position minus the smaller. Writing it the other way round would give a negative radius, which squares to the same thing here but signals confused thinking that causes real errors on washer problems. A sketch fixes the order immediately, and the even integrand also lets Section 5.4's symmetry shortcut halve the work.
OpenStax Calculus Volume 1, §6.2 Determining Volumes by Slicing §6.2, pp. 643-645
Error analysis
A student revolves a region about a line below the axis.
Annotate
On: \( V = \pi\int_{0}^{2}\left(x^{2}\right)^{2}dx \quad \text{(about } y=-1\text{)} \)
The function's value is the radius only when the axis is the x-axis itself. Sketching the axis and drawing one radius takes seconds and makes the correct expression obvious.
Sorting
Does the region still touch the axis?
Sort into buckets
Sort each revolution of the region under y = x^2 on [0,2].
Only the first gives disks, and moving the axis by any amount opens a hole. That the same region gives disks about one line and washers about another is worth internalising — the method depends on the pair, not on the region alone.
Ranking
Revolution about a line other than a coordinate axis.
Put in order
Why: Step c is where the shift is caught and step d is where the switch from disks to washers is noticed. Both are visible in a sketch and neither is reliably visible in the algebra.
Prediction
Commit before reasoning.
Predict first
The same region gives disks about the x-axis and washers about y = -1. Why?
Correct: Because a gap opens between the axis and the region.
\[ \text{region touches axis} \Rightarrow \text{disks}; \quad \text{gap} \Rightarrow \text{washers} \]
Why: Disks occur when the region reaches the axis, so every slice is solid. Move the axis away and the nearest boundary is now at a positive distance, which sweeps out a cylindrical hole through the middle. The region is unchanged; what changed is its relationship to the axis, and the method depends on the pair rather than on the region alone.
Section
Section 5
Concept
Only slices perpendicular to the axis of revolution are circular. That forces the variable of integration, and when the boundaries cannot be written in that variable the method fails.
the forced variable — The slicing direction is determined by the axis of revolution: perpendicular slices are circular and others are not. A horizontal axis means integrating in x, a vertical one in y.
\[ \text{horizontal axis} \Rightarrow dx, \qquad \text{vertical axis} \Rightarrow dy \]
This is a real constraint. If the boundary cannot be solved for the required variable, disks and washers are unavailable and Section 6.3's cylindrical shells are the alternative.
Figure (svg): Slicing perpendicular to the axis of revolution decides the variable
OpenStax Calculus Volume 1, §6.2 Determining Volumes by Slicing §6.2, pp. 626-646 — choosing the variable
Picture it
The axis's direction decides.
Figure (svg): Slicing perpendicular to the axis of revolution decides the variable
The axis's position affects the radii and never the variable. Only its direction — horizontal or vertical — determines which way the slices run.
Worked example
Example 6.11. A boundary that cannot be inverted.
\[ \text{Revolve the region under } y=x^{3}+x \text{ on } [0,1] \text{ about the } y\text{-axis.} \]
Note the axis
Why: Vertical.
The radius must be x as a function of y
Why: Solve the boundary.
\[ y = x ^{3} + x \]
Attempt to solve
Why: A cubic in x.
Conclude the method stalls
Why: The radius cannot be written.
Note the alternative
Why: Slice parallel to the axis instead.
\[ \text{Section } 6.3' s\text{ shells} \]
Figure (svg): Slicing perpendicular to the axis of revolution decides the variable
\[ y=x^{3}+x \text{ is not invertible in elementary terms} \]
Verify: identify precisely what failed
Why: The function is perfectly well behaved — continuous, increasing, and easy to integrate. What failed is the inversion: the disk method needs the radius as a function of the slicing variable, and solving a general cubic for x is not something to attempt. This is a limitation of the method rather than of the problem, and Section 6.3 removes it by slicing the other way.
OpenStax Calculus Volume 1, §6.2 Determining Volumes by Slicing §6.2, pp. 634-636
Matching
The direction decides.
Match the pairs
Why: Only the axis's direction matters, never its position. A horizontal axis at any height means slicing vertically and integrating in x; the height affects the radii alone.
Worked example
Checkpoint 6.11. Different axes, different work.
\[ \text{Revolve the region under } y=x^{2} \text{ on } [0,2] \text{ about each axis.} \]
About the x-axis
Why: Disks of radius x squared.
\[ \pi \int\text{ of } x ^{4}\text{ from } 0\text{ to } 2 \]
Evaluate
Why: The power rule.
\[ 32 \pi / 5 \]
About the y-axis
Why: Solve for x.
\[ x = \sqrt{y} \]
Set up washers
Why: Outer radius 2, inner the square root.
\[ \pi \int\text{ of } (4 - y)\text{ from } 0\text{ to } 4 \]
Evaluate
Why: The power rule.
\[ 8 \pi \]
Figure (svg): The solution to Worked example the same region, both axes shown as a ladder of expressions, one row per legal move
\[ V_{x} = \frac{32\pi}{5}, \qquad V_{y} = 8\pi \]
Verify: explain why the two volumes differ
Why: The same region gives different solids because the material far from the axis contributes most — volume depends on the square of the distance. About the y-axis the region's far edge is at x equal to 2 throughout much of its height, sweeping a wide solid; about the x-axis the tall part is confined to large x. There is no reason to expect the two to agree, and 6.4 pi against 25.1 shows how different they can be.
OpenStax Calculus Volume 1, §6.2 Determining Volumes by Slicing §6.2, pp. 636-638
Trap
\[ \text{revolve about the } y\text{-axis, integrate in } x \text{ with disks} \]
Keep the variable used for the area problem
Why: The student slices in the familiar direction.
Slices parallel to the axis of revolution are not circles at all — they are cylindrical shells, and their area formula is different.
\[ \text{revolve about the } y\text{-axis} \Rightarrow \text{slice perpendicular: integrate in } y \]
Let the axis decide the variable
Why: Only perpendicular slices are disks or washers.
Slicing parallel is not wrong in itself — it is Section 6.3's shell method, which is genuinely useful. What is wrong is slicing parallel and then applying the disk formula.
Fill the middle
What the disk method needs.
Fill in the blanks
\textslicing \; ___ \; \text___
Why: If the boundary cannot be solved for that variable, the method stalls — which is exactly the situation cylindrical shells were developed for.
Sorting
Can the boundary be written in the required variable?
Sort into buckets
Sort each revolution.
Revolving about the x-axis never has this problem, since the boundary is already given as a function of x. The difficulty arises only when the axis forces an inversion, and Section 6.3 exists to remove it.
Prediction
Commit before reasoning.
Predict first
The boundary cannot be solved for the slicing variable. What now?
Correct: Slice parallel to the axis, using shells.
\[ \text{cannot invert} \Rightarrow \text{slice parallel} \Rightarrow \text{shells} \]
Why: The solid and its volume are perfectly well defined; only this method fails. Slicing parallel to the axis produces thin cylindrical shells whose volume can be written using the boundary as originally given, with no inversion needed. That is exactly what Section 6.3 develops, and it is the standard response to this obstacle.
Comparison
Fill the blanks. The difference is whether a gap exists.
Comparison matrix
| Disk | Washer | |
|---|---|---|
| When | the region touches the axis | a gap separates them |
| Cross-section | a full circle | an annulus |
| Area | pi R^2 | pi(R^2 - r^2) |
| Common error | omitting pi | squaring the difference of radii |
The last cell is the one that costs most. Squaring a difference of radii describes a solid disk of the gap's size, which is a different shape and a different number.
Pattern
Given a solid of revolution or a solid with known cross-sections.
Steps one and three are where the shift and the hole are caught, and neither shows up in the arithmetic afterwards. A sketch with one radius drawn on it prevents most of this section's errors.
Stewart, Calculus: Early Transcendentals 8e, §6.2 Volumes §6.2, pp. 438-448
Check
The disk method.
Check your understanding
What is the volume when the region under y = sqrt(x) on [0,4] is revolved about the x-axis?
Answer: A
Why: The integrand is pi times x, giving pi times 8.
Check
Washers.
Check your understanding
For a washer with outer radius R and inner radius r, what is the cross-sectional area?
Answer: A
Why: The smaller disk's area is removed from the larger one's.
Check
The variable.
Check your understanding
Revolving about the line x = 3, which variable do you integrate in?
Answer: A
Why: Slices must be perpendicular to the axis, so a vertical axis means horizontal slices.
Real world
A workshop turns a wooden bowl on a lathe. The outer profile is a curve rotated about the lathe's axis, and the inside is hollowed to a second profile, leaving a wall of varying thickness.
Discussion prompt
Explain which method applies, what each radius represents, and how the wood removed is computed.
Hint: The bowl is a solid of revolution with a hole.
Answer:
The bowl is exactly a washer solid: each cross-section perpendicular to the lathe's axis is an annulus whose outer radius is the outer profile and whose inner radius is the hollowed profile. The volume of wood in the finished bowl is the integral of pi times the difference of their squares.
\[ V_{\text{wood}} = \pi\int_{a}^{b}\left[R(x)^{2}-r(x)^{2}\right]dx \]
The wood removed is the integral of pi times the inner radius squared — the volume of the cavity, computed as a disk solid in its own right. The original blank's volume is the outer disk integral, and the three quantities satisfy the obvious accounting: blank equals bowl plus shavings.
The washer error matters here in a way it does not on paper. Using pi times the square of the wall thickness instead of the difference of squares would underestimate the wood in the bowl badly for a thin-walled piece — the thickness is small and its square is smaller still, while the true expression stays large because both radii are large. A workshop costing timber by volume would order far too little.
Note that the axis is the lathe's spindle and the profiles are given as functions of position along it, so the variable is forced and the boundaries are already in the right form — the geometry of the machine supplies exactly what the method needs.
Commit first
Answer, then rate your confidence honestly.
Predict first
Which quantity does a washer's cross-sectional area use?
Correct: The difference of the squares.
\[ \pi R^{2}-\pi r^{2} \ne \pi(R-r)^{2} \quad \text{unless } r=0 \]
Why: A washer is a disk with a smaller disk removed, so what is subtracted is the smaller disk's area — pi times its radius squared. The square of the difference describes a solid disk whose radius is the gap, a different shape of a different size, and for the worked example it gives half the correct volume. Section 6.1's habit of subtracting boundaries is what makes this error persistent.
Explain it
They wrote a washer's area as pi times the square of the difference of the radii.
Discussion prompt
In four sentences or fewer, show them what that describes.
Hint: Ask them to picture a real washer.
Answer:
Ask them to picture an actual metal washer: it is a big disk with a small disk punched out, so what you remove is the small disk's AREA — pi times its radius squared. Their formula describes a solid disk whose radius is the gap between the two circles, which is a completely different object.
Expanding both shows they differ by a cross term, and they agree only when the inner radius is zero, which is the disk case. Squaring each radius first and then subtracting is the order that matches the picture.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For area functions, check the integrand is a length squared. For disks against washers, ask whether the region touches the axis. For shifted axes, draw one radius on a sketch and read it off as a distance. For the variable, take it from the axis's direction, not the region. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top, draw a solid sliced perpendicular to an axis with one slab highlighted, and write the general volume formula beside it with a note that the integrand must be an area. Below, draw a disk and a washer side by side with their radii labelled, write both area formulas, and write out the expansion showing why the square of the difference is not the difference of squares. In the middle of the page, sketch a region and revolve it about the x-axis, then about a line below it, drawing one radius in each case and noting where the shift and the new hole came from. Beside it, write the four axis-to-variable rules. In the lower half, work one washer problem completely from sketch to number, and beside it write what the wrong formula would have given. At the bottom, write the five steps of the procedure, with the sketch first.
If your two radius drawings look the same, redraw the second — the shifted axis should make every radius visibly longer, and seeing that once is what makes the shift impossible to forget.
Recap
Five things, and all of them are Section 6.1's argument with an area in place of a height.
| If you see | Then |
|---|---|
| A solid with known cross-sections | Integrate the area function |
| An integrand that is not an area | Something is wrong: check the units |
| A region touching the axis | Disks |
| A gap between region and axis | Washers, with a difference of squares |
| An axis that is not a coordinate axis | Every radius is a distance: subtract the axis |
| A horizontal axis | Integrate in x, whatever its height |
| A boundary that will not invert | Shells, in Section 6.3 |
Section 6.3 slices the other way. Cylindrical shells run parallel to the axis rather than perpendicular to it, which needs no inversion and handles exactly the solids this section's method could not.
OpenStax Calculus Volume 1, §6.2 Determining Volumes by Slicing §6.2, pp. 551-568 — everything on these slides traces back here
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