Slicing parallel to the axis of revolution, a shell's volume as circumference times height times thickness, why the method needs no inversion of the boundary, shells about lines other than the coordinate axes, and choosing between shells and washers.
Subject: Calculus I · 65 slides · symbolic lesson
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Title
Calculus I · Chapter 6 — Applications of Integration
Volumes of Revolution: Cylindrical Shells
Objectives
Five outcomes. The formula is one line and the reason it exists is the last two.
OpenStax Calculus Volume 1, §6.3 Volumes of Revolution: Cylindrical Shells §6.3, pp. 569-581 — the section these objectives are drawn from
Warm-up
Section 6.2 stalled on one problem: revolving the region under a cubic about the y-axis needed the boundary solved for x, and it would not solve.
Discussion prompt
What if the strip were swept around the axis instead of sliced across it?
Hint: A vertical strip parallel to a vertical axis sweeps out what shape?
Answer:
A vertical strip parallel to the y-axis, swept about that axis, traces a thin cylindrical shell — a tube. Its radius is the strip's distance from the axis, its height is the strip's height, and its thickness is the strip's width.
\[ dV = 2\pi r h\,dx \]
The strip's height is the function's value, used exactly as given. No inversion is needed, because the strip runs in the direction the function was written in. That single fact is why this method exists alongside Section 6.2's.
Concept
A strip parallel to the axis of revolution sweeps out a thin cylindrical shell. Unrolled, it is a flat sheet of area circumference times height, so its volume is that times its thickness.
cylindrical shell — The thin tube swept out by a strip parallel to the axis of revolution. Its volume is twice pi times its radius, times its height, times its thickness.
\[ V = 2\pi\int_{a}^{b}r(x)h(x)\,dx \]
No radius is ever squared, which makes the integrands different in character from Section 6.2's — and often simpler, since a boundary appears to the first power rather than the second.
Figure (svg): A vertical strip swept about a vertical axis, sweeping out a cylindrical shell
OpenStax Calculus Volume 1, §6.3 Volumes of Revolution: Cylindrical Shells §6.3, pp. 646-656
Section
Section 1
Concept
Cut a thin cylindrical shell down its side and flatten it. The result is a rectangular sheet whose width is the circumference and whose height is the shell's height, with the shell's thickness as its depth.
the shell formula — The volume of a thin cylindrical shell is twice pi times its radius, times its height, times its thickness — the area of the unrolled sheet times its depth.
\[ dV = 2\pi r h\,dx \]
The unrolling is exact in the limit: a thin shell's inner and outer circumferences differ by an amount that vanishes faster than the thickness, so the approximation costs nothing.
Figure (svg): Unrolling a shell: why its volume is circumference times height times thickness
OpenStax Calculus Volume 1, §6.3 Volumes of Revolution: Cylindrical Shells §6.3, pp. 646-656 — the method of cylindrical shells
Picture it
The derivation in one picture.
Figure (svg): Unrolling a shell: why its volume is circumference times height times thickness
Nothing is squared anywhere in the formula. That is the clearest surface difference from Section 6.2, and it changes what the integrands look like.
Worked example
Example 6.13. A strip parallel to the axis.
\[ \text{Revolve the region under } y=x^{2} \text{ on } [0,2] \text{ about the } y\text{-axis, using shells.} \]
Draw a vertical strip
Why: Parallel to the axis.
Identify its radius
Why: Distance from the y-axis.
\[ r = x \]
Identify its height
Why: The function, as given.
\[ h = x ^{2} \]
Write the shell's volume
Why: The formula.
\[ 2 \pi x \times x ^{2} \,dx \]
Integrate from 0 to 2
Why: The power rule.
\[ 2 \pi(\frac{16}{4}) = 8 \pi \]
Figure (svg): A vertical strip swept about a vertical axis, sweeping out a cylindrical shell
\[ V = 2\pi\int_{0}^{2}x^{3}dx = 8\pi \]
Verify: check against Section 6.2's washer computation
Why: That section computed the same solid with washers and also got 8 pi, so the two methods agree as they must. Note what each needed: the washer route required the boundary solved for x as a square root, while the shell route used x squared exactly as given. Both were easy here; the difference matters when the inversion is hard or impossible.
OpenStax Calculus Volume 1, §6.3 Volumes of Revolution: Cylindrical Shells §6.3, pp. 650-652
Fill the middle
A thin cylindrical shell.
Fill in the blanks
dV = 2\pi\,r\,h\,dx
Why: The unrolled sheet's width is the circumference, twice pi times the radius. Its height is the shell's height and its depth is the thickness, giving the product.
Worked example
Checkpoint 6.13. Section 6.2's failure, resolved.
\[ \text{Revolve the region under } y=x^{3}+x \text{ on } [0,1] \text{ about the } y\text{-axis.} \]
Note why washers fail
Why: The boundary will not invert.
Draw a vertical strip
Why: Parallel to the axis.
Identify its height
Why: The function as given.
\[ h = x ^{3} + x \]
Write the integrand
Why: Radius times height, times 2 pi.
\[ 2 \pi x(x ^{3} + x) \]
Integrate from 0 to 1
Why: Expand and use the power rule.
\[ 2 \pi(\frac{1}{5} + \frac{1}{3}) \]
Figure (svg): The advantage: the boundary is used as given
\[ V = 2\pi\int_{0}^{1}\left(x^{4}+x^{2}\right)dx = \frac{16\pi}{15} \]
Verify: identify exactly what shells avoided
Why: The cubic's inversion was the obstacle and shells never required it: the strip runs vertically, so its height is read directly off the function as written. Multiplying by the radius x turned a cubic into a quartic, which the power rule handles without difficulty. A problem that stopped Section 6.2's method entirely became a two-line integral.
OpenStax Calculus Volume 1, §6.3 Volumes of Revolution: Cylindrical Shells §6.3, pp. 652-654
Trap
\[ V = 2\pi\int_{0}^{2}x^{2}\cdot x^{2}dx \]
Square the radius, as in the disk method
Why: The student carries over Section 6.2's habit.
A shell's volume has no square in it. The radius appears once, in the circumference, and the height appears once.
\[ V = 2\pi\int_{0}^{2}x\cdot x^{2}dx = 8\pi \]
Radius times height, each to the first power
Why: The unrolled sheet's area is circumference times height.
The units check confirms it: two pi times a radius is a length, times a height is an area, times a thickness is a volume. Squaring the radius would give a length cubed before the differential and a four-dimensional quantity after.
Matching
Each factor in the formula.
Match the pairs
Why: Every factor corresponds to a dimension of the flattened sheet, which is why the derivation is worth doing once. Reconstructing the picture is quicker and safer than recalling the formula.
Two truths and a lie
All three are about the shell formula.
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 — that is the volume of a solid CYLINDER, not a thin shell. A shell is a tube with negligible wall thickness, and its volume is the wall's surface area times that thickness.
Prediction
Commit before reasoning.
Predict first
The inner and outer circumferences of a shell differ. Why does treating it as a flat sheet give the exact answer?
Correct: Because the difference vanishes faster than the thickness.
\[ 2\pi(r+dx)h\,dx - 2\pi rh\,dx = 2\pi h(dx)^{2} \to 0 \text{ faster} \]
Why: The two circumferences differ by twice pi times the thickness, so the error in the sheet's area is proportional to the thickness — and the volume error is then proportional to its square, which vanishes relative to the thickness in the limit. That is the same second-order argument that made Section 5.1's rectangles exact in the limit despite each being wrong.
Section
Section 2
Concept
A vertical strip's height is the function's value read directly. Because the strip is parallel to a vertical axis, the boundary is used as written and never has to be solved for the other variable.
the inversion advantage — Shells use the boundary in the form it is given, because the strip's extent runs along the direction the function is defined in. Washers about the same axis would need the inverse.
\[ \text{shells about } y\text{-axis need } y=f(x); \quad \text{washers need } x=g(y) \]
This is the whole reason the method exists. Section 6.2's failure was an inversion failure, not a conceptual one, and shells remove it entirely.
Figure (svg): The advantage: the boundary is used as given
OpenStax Calculus Volume 1, §6.3 Volumes of Revolution: Cylindrical Shells §6.3, pp. 650-660 — advantages of the shell method
Picture it
One stalls, one does not.
Figure (svg): The advantage: the boundary is used as given
The left column is Section 6.2's method meeting an obstacle it cannot pass. The right column never encounters it, because the strip runs the other way.
Worked example
Example 6.15. Shells where washers cannot go.
\[ \text{Revolve the region under } y=e^{x^{2}} \text{ on } [0,1] \text{ about the } y\text{-axis.} \]
Note washers would need the inverse
Why: Solve for x.
Note the deeper problem
Why: The washer integrand would be the inverse squared.
Use shells instead
Why: A vertical strip.
\[ r = x, h = e ^{x ^{2}} \]
Write the integrand
Why: Radius times height.
\[ 2 \pi x e ^{x ^{2}} \]
Recognise the substitution
Why: Section 5.6's form.
\[ u = x ^{2}\text{ gives } \pi(e - 1) \]
Figure (svg): The advantage: the boundary is used as given
\[ V = 2\pi\int_{0}^{1}xe^{x^{2}}dx = \pi(e-1) \]
Verify: notice what the shell's radius contributed
Why: The factor of x from the radius is exactly what makes the exponential integrable — without it the integrand would be the exponential of x squared, which Section 5.6 showed has no elementary antiderivative. So the shell method did not merely avoid an inversion here: the radius supplied the missing factor that made the integral elementary at all. That is a recurring and slightly surprising benefit.
OpenStax Calculus Volume 1, §6.3 Volumes of Revolution: Cylindrical Shells §6.3, pp. 655-657
Fill the middle
A vertical strip under a parabola.
Fill in the blanks
r = x, \quad h = x^2
Why: The radius is the strip's distance from the axis and the height is how tall the strip is — the function's value. They are different quantities and confusing them is the commonest slip here.
Worked example
Checkpoint 6.15. Same solid, two setups.
\[ \text{Set up } y=x^{2} \text{ on } [0,2] \text{ about the } y\text{-axis both ways.} \]
Shells: strip vertical
Why: Radius x, height x squared.
\[ 2 \pi \int\text{ of } x ^{3}\text{ from } 0\text{ to } 2 \]
Washers: slice horizontal
Why: Solve for x.
\[ x = \sqrt{y} \]
Identify the radii
Why: Outer 2, inner the root.
\[ R = 2, r = \sqrt{y} \]
Write the washer integrand
Why: Difference of squares.
\[ \pi(4 - y) \]
Both integrate to
Why: The same volume.
\[ 8 \pi \]
Figure (svg): The solution to Worked example comparing the two integrands shown as a ladder of expressions, one row per legal move
\[ 2\pi\int_{0}^{2}x^{3}dx = \pi\int_{0}^{4}(4-y)\,dy = 8\pi \]
Verify: compare the two setups' difficulty
Why: Both integrands are easy here — a cubic and a linear function — so neither method is clearly better for this region. What differs is that the washer route required an extra step, solving for x, and required noticing the outer radius is the constant 2 rather than a function. Where the region is simple either works; the methods separate on the hard cases, which is why knowing both matters.
OpenStax Calculus Volume 1, §6.3 Volumes of Revolution: Cylindrical Shells §6.3, pp. 657-659
Error analysis
A student sets up a shell integral.
Annotate
On: \( V = 2\pi\int_{0}^{2}x\cdot x\,dx \quad (y=x^{2} \text{ about the } y\text{-axis}) \)
Radius and height are different quantities measured in different directions: the radius runs across to the axis, the height runs along the strip. Labelling both on a sketch keeps them apart.
Sorting
Which form of the boundary does each need?
Sort into buckets
Sort each situation, with the boundary given as y = f(x).
The pattern is exactly complementary: shells about a vertical axis and washers about a horizontal one both use y as a function of x. Which method avoids the inversion depends on the axis, and each method wins on the cases the other loses.
Two truths and a lie
All three are about the advantage.
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. Shells about a HORIZONTAL axis need horizontal strips, whose extent runs in x — so a boundary given as y equal to f of x would have to be inverted. Each method avoids inversion for one axis direction and requires it for the other.
Prediction
Commit before reasoning.
Predict first
Revolving the exponential of x squared gave an elementary integral by shells. Why?
Correct: Because the radius contributes a factor of x.
\[ \int e^{x^{2}}dx \text{ non-elementary}; \quad \int xe^{x^{2}}dx = \tfrac12 e^{x^{2}}+C \]
Why: Section 5.6 showed the exponential of x squared has no elementary antiderivative, but x times that exponential does — the extra factor is precisely what the substitution needs. The shell method multiplies by the radius, which for a vertical axis is x, so it supplies that factor automatically. Shells therefore make elementary a whole class of revolution problems that would otherwise not be.
Section
Section 3
Concept
When the axis is a vertical line other than the y-axis, the shell's radius is the strip's position minus the axis's — the same shift as in Section 6.2, applied to the strip rather than to a boundary.
shifted shell radius — The distance from the axis of revolution to the strip, taken positively. For a vertical axis at position c and a strip at position x, it is the absolute difference.
\[ r(x) = |x-c| \]
Whether the region lies to the left or the right of the axis decides the sign, and a sketch settles it. A radius that comes out negative on part of the interval is the signal that the region straddles the axis.
Figure (svg): Shells about a line other than a coordinate axis
OpenStax Calculus Volume 1, §6.3 Volumes of Revolution: Cylindrical Shells §6.3, pp. 656-666 — shells about other lines
Picture it
A strip's distance to a line beside the region.
Figure (svg): Shells about a line other than a coordinate axis
The radius is the strip's position minus the axis's, which here adds one to every strip. The height is unchanged, because moving the axis does not move the region.
Worked example
Example 6.17. The shifted radius.
\[ \text{Revolve the region under } y=x^{2} \text{ on } [0,2] \text{ about } x=-1. \]
Sketch the axis
Why: One unit left of the origin.
Find the strip's radius
Why: Position minus the axis's.
\[ r = x + 1 \]
Find its height
Why: Unchanged by the axis.
\[ h = x ^{2} \]
Write the integrand
Why: Radius times height.
\[ 2 \pi(x + 1) x ^{2} \]
Expand and integrate
Why: From 0 to 2.
\[ 2 \pi(4 + \frac{8}{3}) \]
Figure (svg): Shells about a line other than a coordinate axis
\[ V = 2\pi\int_{0}^{2}(x+1)x^{2}dx = \frac{40\pi}{3} \]
Verify: check against the unshifted case
Why: About the y-axis the same region gives 8 pi, and moving the axis one unit further away gives 40 pi over 3, about 41.9 — considerably larger, because every strip is now swept around a bigger circle. The increase is not proportional to the shift, since the shift adds a term involving the integral of the height alone; that extra term is exactly 2 pi times the region's area, which is a useful check. The region's area is 8 over 3, and 2 pi times that is about 16.8, which is the difference.
OpenStax Calculus Volume 1, §6.3 Volumes of Revolution: Cylindrical Shells §6.3, pp. 660-662
Fill the middle
A strip at position x, revolved about a line one unit to the left of the origin.
Fill in the blanks
r = x - (-1) = x + 1
Why: The radius is the strip's position minus the axis's. The height is unaffected, since moving the axis does not move the region.
Worked example
Checkpoint 6.17. Everything rotated.
\[ \text{Revolve the region between } x=y^{2} \text{ and } x=4 \text{ about the } x\text{-axis, using shells.} \]
Note the axis is horizontal
Why: So strips must be horizontal.
Find the strip's radius
Why: Its distance from the x-axis.
\[ r = y \]
Find its length
Why: Right boundary minus left.
\[ h = 4 - y ^{2} \]
Set the limits
Why: Where the region lies.
\[ y\text{ from } 0\text{ to } 2,\text{ doubled by symmetry} \]
Integrate
Why: Expand and use the power rule.
\[ 2 \pi(8 - 4) \times 2 = 16 \pi \]
Figure (svg): The solution to Worked example shells about a horizontal axis shown as a ladder of expressions, one row per legal move
\[ V = 2\cdot 2\pi\int_{0}^{2}y\left(4-y^{2}\right)dy = 16\pi \]
Verify: explain why the interval was halved and doubled
Why: The region is symmetric about the x-axis, and both halves sweep the same solid — so integrating over the upper half and doubling is correct, and it avoids a negative radius for y below zero. Integrating from negative 2 to 2 directly would give zero, since the integrand is odd, which is a misleading answer produced by a sign the geometry does not have. Radii are distances and must be taken positively, which is what the symmetry argument enforces here.
OpenStax Calculus Volume 1, §6.3 Volumes of Revolution: Cylindrical Shells §6.3, pp. 662-664
Trap
\[ V = 2\pi\int_{-2}^{2}y\left(4-y^{2}\right)dy = 0 \]
Integrate across the axis without regard to sign
Why: The student treats the radius as a signed quantity.
For y below zero the radius came out negative, so the lower half subtracted the upper. The solid plainly has volume.
\[ V = 2\cdot 2\pi\int_{0}^{2}y\left(4-y^{2}\right)dy = 16\pi \]
A radius is a distance and is never negative
Why: Integrate over one side and use symmetry, or use the absolute value.
This is Chapter 5's signed-against-total distinction once more. An answer of zero for a solid of obvious size is the warning, and it comes from letting a distance carry a sign.
Sorting
The region stays where it is.
Sort into buckets
Sort each quantity.
Only the radius changes among the setup's ingredients, and it changes the volume through it. That makes shifted-axis shell problems noticeably simpler than the corresponding washer ones, where both radii shift.
Two truths and a lie
All three are about shifted axes.
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. Shifting the axis adds a term equal to the shift times the integral of the height, which is the region's area — so the increase is twice pi times the shift times the area, an additive change rather than a proportional one.
Prediction
Commit before reasoning.
Predict first
Moving a vertical axis one unit further from a region increases the volume by how much?
Correct: By twice pi times the region's area.
\[ 2\pi\int(x+c)h\,dx = 2\pi\int xh\,dx + 2\pi c\int h\,dx \]
Why: The integrand becomes twice pi times the quantity x plus one, times the height — which splits into the original integral plus twice pi times the integral of the height alone. That second integral is precisely the region's area. So the shift's effect is computable in advance and gives a useful check: for the worked example the area is 8 over 3 and the increase is about 16.8, matching the difference between 8 pi and 40 pi over 3.
Section
Section 4
Concept
Shells suit an axis perpendicular to the variable the boundary is written in; washers suit an axis parallel to it. The choice is decided by which avoids an inversion.
choosing the method — Compare the axis's direction with the variable the boundary is expressed in. If they are perpendicular, shells use the boundary as given; if parallel, washers do.
\[ y=f(x) \text{ about } y\text{-axis} \Rightarrow \text{shells}; \quad \text{about } x\text{-axis} \Rightarrow \text{washers} \]
Both methods always give the same answer when both are available. The choice is about the amount of work, and occasionally about whether the problem can be done at all.
Figure (svg): Deciding between shells and washers
OpenStax Calculus Volume 1, §6.3 Volumes of Revolution: Cylindrical Shells §6.3, pp. 658-668 — choosing between shells and washers
Picture it
Which method for which combination.
Figure (svg): Deciding between shells and washers
The rule of thumb is that shells suit an axis at right angles to the given variable. Sketching the region and the axis makes the pairing visible immediately.
Worked example
Example 6.18. The method swaps with the axis.
\[ \text{For the region under } y=\sqrt{x} \text{ on } [0,4], \text{ choose a method for each axis.} \]
About the x-axis
Why: The axis is parallel to the given variable.
Set it up
Why: Disks, since the region touches the axis.
\[ \pi \int\text{ of } x\text{ from } 0\text{ to } 4 \]
About the y-axis
Why: The axis is perpendicular to the given variable.
Set it up
Why: Radius x, height the root.
\[ 2 \pi \int\text{ of } x \sqrt{x} \]
Note both avoid inversion
Why: Each method for its own axis.
Figure (svg): Deciding between shells and washers
\[ V_{x}=8\pi, \qquad V_{y}=\frac{128\pi}{5} \]
Verify: confirm the complementary pattern
Why: Each axis has a method that uses the square root exactly as written, and the other method for that axis would require squaring or inverting. That complementarity is the practical content of the rule: for any boundary written in one variable, one method works cleanly for a parallel axis and the other for a perpendicular one. Knowing both means never being stuck.
OpenStax Calculus Volume 1, §6.3 Volumes of Revolution: Cylindrical Shells §6.3, pp. 663-665
Sorting
The boundary is given as y = f(x).
Sort into buckets
Sort each axis by the method that avoids inversion.
The pattern depends only on the axis's direction, never on its position. A vertical axis at any location calls for shells when the boundary is written in x.
Worked example
Checkpoint 6.18. Choosing on effort alone.
\[ \text{Revolve the region between } y=x \text{ and } y=x^{2} \text{ about the } y\text{-axis, both ways.} \]
Shells: vertical strips
Why: Height is the difference.
\[ 2 \pi \int\text{ of } x(x - x ^{2})\text{ from } 0\text{ to } 1 \]
Evaluate
Why: Expand.
\[ 2 \pi(\frac{1}{3} - \frac{1}{4}) = \frac{\pi}{6} \]
Washers: solve both for x
Why: The line and the parabola.
\[ x = y\text{ and } x = \sqrt{y} \]
Identify the radii
Why: The root is outer.
\[ R = \sqrt{y}, r = y \]
Evaluate
Why: Difference of squares.
\[ \pi(\frac{1}{2} - \frac{1}{3}) = \frac{\pi}{6} \]
Figure (svg): The solution to Worked example when both work shown as a ladder of expressions, one row per legal move
\[ V = \frac{\pi}{6} \]
Verify: compare the effort each required
Why: Both routes were short here, but the shell route needed no inversion at all while the washer route needed both boundaries solved for x and required noticing which becomes the outer radius — the square root, since it is larger on the unit interval. That extra decision is a place to go wrong. When both methods are available, the one needing fewer decisions is usually the better choice.
OpenStax Calculus Volume 1, §6.3 Volumes of Revolution: Cylindrical Shells §6.3, pp. 665-667
Error analysis
A student uses shells about the x-axis with a boundary given in x.
Annotate
On: \( V = 2\pi\int_{0}^{4}x\cdot\sqrt{x}\,dx \quad \text{(about the } x\text{-axis)} \)
Strips must be parallel to the axis of revolution, or they do not sweep shells at all. Drawing the axis and one strip on a sketch makes the mismatch obvious before any algebra.
Fill the middle
Shells require a particular strip direction.
Fill in the blanks
\textparallel \; ___ \; \text___
Why: Only strips parallel to the axis sweep out cylindrical shells. Perpendicular ones sweep disks or washers, which is Section 6.2's method.
Ranking
A volume by cylindrical shells.
Put in order
Why: Steps c and d are the two quantities most easily confused, and they are measured in perpendicular directions. Labelling both on the sketch keeps them straight, exactly as labelling top and bottom did in Section 6.1.
Prediction
Commit before reasoning.
Predict first
When both shells and washers can be applied to the same solid, can they give different answers?
Correct: No: the choice is about effort.
\[ 2\pi\int_{0}^{1}x(x-x^{2})dx = \pi\int_{0}^{1}\left(y-y^{2}\right)dy = \tfrac{\pi}{6} \]
Why: Both are ways of decomposing the same solid, so both must give its volume — and when both were applied to the region between a line and a parabola, both gave pi over 6. What differs is the work: the number of inversions needed, whether radii must be identified as outer and inner, and how pleasant the integrand is. Disagreement always means an error in one of the setups, which makes computing both a strong check.
Section
Section 5
Concept
A shell's radius is measured across to the axis and its height along the strip. Confusing them is the section's characteristic error, and a labelled sketch prevents it.
the two lengths — The radius runs perpendicular to the strip, from the axis to it; the height runs along the strip, between the region's boundaries. They are measured in different directions and are generally different functions.
\[ r \perp \text{ strip}, \qquad h \parallel \text{ strip} \]
This is the same discipline Section 6.1 required for a strip's height and Section 6.2 for a slice's radii. Every method in this chapter reduces to describing one representative piece correctly.
Figure (svg): Shells against washers: which slicing direction for which axis
OpenStax Calculus Volume 1, §6.3 Volumes of Revolution: Cylindrical Shells §6.3, pp. 650-668 — setting up shell integrals
Picture it
The same region, sliced both ways.
Figure (svg): Shells against washers: which slicing direction for which axis
The left picture's slices give washers and the right picture's give shells. Both revolve about the same axis and both are correct; only the direction of slicing differs.
Worked example
Example 6.16. The height is a difference.
\[ \text{Revolve the region between } y=x \text{ and } y=x^{2} \text{ about the } y\text{-axis, using shells.} \]
Find the intersections
Why: Set equal.
\[ x = 0\text{ and } 1 \]
Draw a vertical strip
Why: Between the curves.
Find the height
Why: Upper minus lower.
\[ h = x - x ^{2} \]
Write the integrand
Why: Radius times height.
\[ 2 \pi x(x - x ^{2}) \]
Integrate from 0 to 1
Why: Expand.
\[ 2 \pi(\frac{1}{3} - \frac{1}{4}) = \frac{\pi}{6} \]
Figure (svg): Shells against washers: which slicing direction for which axis
\[ V = 2\pi\int_{0}^{1}\left(x^{2}-x^{3}\right)dx = \frac{\pi}{6} \]
Verify: note where Section 6.1's skill reappeared
Why: The height was found exactly as in Section 6.1: identify which curve is on top and subtract. So the area work of that section is embedded in this one, and the shell method adds only the radius factor and the constant. Recognising that the height of a shell is precisely the height of an area strip makes these setups routine once Section 6.1 is secure.
OpenStax Calculus Volume 1, §6.3 Volumes of Revolution: Cylindrical Shells §6.3, pp. 658-660
Matching
Two lengths, measured differently.
Match the pairs
Why: The first two are the ones confused, and they are perpendicular to each other. Naming each by the direction it is measured in, rather than by which formula it appears in, is what keeps them apart.
Worked example
Checkpoint 6.16. The discipline.
\[ \text{Describe how to read a shell's radius and height from a sketch.} \]
Draw the axis
Why: And the region.
Draw one strip parallel to the axis
Why: Anywhere in the region.
Measure across to the axis
Why: Perpendicular to the strip.
Measure along the strip
Why: Between the boundaries.
Write both as functions
Why: Of the strip's position.
Figure (svg): The solution to Worked example reading both lengths off a sketch shown as a ladder of expressions, one row per legal move
\[ r \text{ across}, \quad h \text{ along} \]
Verify: see why the discipline generalises
Why: This is the same procedure as Section 6.1's — draw one representative piece and describe it — and it is what Sections 6.4 through 6.6 will use for arc length, work and centres of mass. The specific formula changes but the discipline does not, and a student who can reliably describe a representative piece can set up every problem in the rest of this chapter.
OpenStax Calculus Volume 1, §6.3 Volumes of Revolution: Cylindrical Shells §6.3, pp. 660-661
Trap
\[ V = 2\pi\int_{0}^{1}(x-x^{2})\cdot x\,dx \quad \text{with the roles named backwards} \]
Call the difference the radius and the position the height
Why: The student swaps the two lengths.
The arithmetic happens to give the same number here, because the two factors are multiplied — but the reasoning is wrong and will fail as soon as the axis is shifted.
\[ r = x \text{ (across to the axis)}, \quad h = x-x^{2} \text{ (along the strip)} \]
Name each length by the direction it is measured in
Why: The radius is perpendicular to the strip and the height is along it.
Because the two are multiplied, swapping them is invisible on an unshifted axis — which is exactly why it survives. Shift the axis and the radius picks up a constant while the height does not, and the error appears.
Fill the middle
A shell between two curves.
Fill in the blanks
h = x - x^2
Why: The height is found exactly as in Section 6.1: upper boundary minus lower. The shell method adds the radius factor and the constant, and nothing else.
Sorting
Direction decides.
Sort into buckets
Sort each measurement for a vertical strip revolved about a vertical axis.
Two descriptions of each quantity are given deliberately: the radius is both a distance to the axis and a difference of positions, and the height is both a distance between boundaries and a difference of functions. Recognising both phrasings is what makes worded problems tractable.
Prediction
Commit before reasoning.
Predict first
Interchanging radius and height gave the right number for one problem. Why?
Correct: Because the two are multiplied.
\[ (x+1)(x-x^{2}) \ne (x-x^{2}+1)x \]
Why: The integrand is a product, so exchanging the two factors leaves it unchanged — as long as both are what they were. The error is in the reasoning rather than the arithmetic, and it surfaces the moment the axis is shifted: the radius then gains a constant and the height does not, so the shifted product is no longer symmetric in the two. An error that hides until the problem gets harder is worth catching early.
Comparison
Fill the blanks. The two methods are complementary.
Comparison matrix
| Shells | Washers | |
|---|---|---|
| Slice | parallel to the axis | perpendicular to the axis |
| Piece swept | a thin tube | a disk or annulus |
| Formula | 2 pi r h dx | pi(R^2 - r^2) dx |
| Boundary needed as | given, for a perpendicular axis | given, for a parallel axis |
The last row is the practical content. Each method uses the boundary as written for one axis direction and requires an inversion for the other, so together they cover every case.
Pattern
Given a solid of revolution to compute by shells.
Steps three and four are perpendicular measurements and are the pair most often confused. Labelling both on the sketch, before any algebra, is what keeps them apart.
Stewart, Calculus: Early Transcendentals 8e, §6.3 Volumes by Cylindrical Shells §6.3, pp. 449-454
Check
The formula.
Check your understanding
What is a thin cylindrical shell's volume?
Answer: A
Why: Unrolled it is a sheet of width 2 pi r and height h, with thickness dx.
Check
Choosing.
Check your understanding
A boundary is given as y = f(x) and the axis is the y-axis. Which method avoids an inversion?
Answer: A
Why: A vertical strip's height is the function's value, read directly.
Check
Shifted axes.
Check your understanding
Revolving a region about a vertical line one unit further away increases the volume by how much?
Answer: A
Why: The extra term is 2 pi times the integral of the height, which is the area.
Real world
A water reservoir is a circular basin whose depth profile is measured along a radius from the centre outward, giving depth as a function of distance from the centre. The engineers need the stored volume.
Discussion prompt
Explain why shells are the natural method, what each factor represents, and why measuring along a radius makes this so.
Hint: The basin is a solid of revolution about a vertical axis.
Answer:
The basin is generated by revolving its depth profile about the vertical axis through its centre, and the profile is measured as a function of distance from that axis — exactly the form shells use. A thin annular ring at distance r from the centre has circumference twice pi times r, depth given by the profile, and width dr.
\[ V = 2\pi\int_{0}^{R}r\,d(r)\,dr \]
Each factor is physically meaningful: the circumference is the length of the ring around the basin, the depth is what a survey pole measures at that distance, and the width is the spacing between survey rings. The integral literally adds the water in each concentric band.
Washers would require inverting the profile — expressing the distance from the centre as a function of depth — which is possible only if the basin's sides are monotonic and is meaningless if the bed has a rise partway out. The survey data is naturally indexed by radius, so the method that consumes it directly is the right one.
Note the practical consequence of the radius factor: water at the outer edge counts far more heavily than water near the centre, because the outer bands are so much longer. A small depth error at the rim costs more volume than a large one at the middle, which tells the surveyors where to measure carefully.
Commit first
Answer, then rate your confidence honestly.
Predict first
Why does the shell method need no inversion when revolving y = f(x) about the y-axis?
Correct: Because the strip's height is the function's value.
\[ \text{vertical strip} \Rightarrow h=f(x) \text{ directly} \]
Why: A vertical strip runs from the region's lower boundary to its upper one, and both are given as functions of x — so the height is read straight off. Washers about the same axis slice horizontally and need the radius as a function of y, which requires solving the boundary for x. The advantage comes from the strip's direction matching the direction the function is written in, and it disappears if the axis is horizontal instead.
Explain it
They cannot see why a shell's volume has no square in it when a disk's does.
Discussion prompt
In four sentences or fewer, give them the picture.
Hint: Ask them to unroll it.
Answer:
Ask them to imagine cutting the tube down its side and flattening it out: it becomes a rectangular sheet whose width is the circumference and whose height is the tube's height. Its volume is that sheet's area times its thickness, and nothing there gets squared.
A disk is different because it is solid all the way across, so its area genuinely involves the radius squared. A shell is only a thin wall, and a wall's area is a length times a length — the circumference and the height.
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 the formula, unroll the shell into a sheet rather than memorising. For radius and height, name each by the direction it is measured in. For shifted axes, subtract the axis's position from the strip's. For choosing, ask whether the axis is parallel or perpendicular to the variable the boundary is written in. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top, draw a region with a vertical strip and, beside it, the tube that strip sweeps about a vertical axis — then draw the tube cut and flattened into a rectangle, labelling its width, height and thickness. Write the formula beneath. Below, draw the same region twice with the same axis, sliced perpendicular in one and parallel in the other, and write which method each gives. In the middle of the page, write the case Section 6.2 could not do and set it up by shells in four lines. Beside it, write the two-column comparison of what each method needs the boundary to look like. In the lower half, draw a region with a shifted vertical axis, label the radius as a difference, and write the identity showing the shift adds twice pi times the area. At the bottom, list the five setup steps with the sketch first.
If your unrolled rectangle has a squared quantity anywhere on it, look again — every side of that sheet is a plain length, and that is exactly why the shell formula has no square in it.
Recap
Five things, and the third is the reason this method exists at all.
| If you see | Then |
|---|---|
| A strip parallel to the axis | It sweeps a shell |
| A squared radius in a shell integral | Something is wrong: nothing is squared |
| A boundary that will not invert | Shells, not washers |
| y = f(x) about a vertical axis | Shells use it as given |
| y = f(x) about a horizontal axis | Washers use it as given |
| An axis that is not a coordinate axis | Subtract its position from the strip's |
| A negative radius | Split or use symmetry: distances are positive |
Section 6.4 leaves volume behind for length. Arc length uses the same discipline — describe one representative piece and integrate — but the piece is a short segment of curve, and the Pythagorean theorem supplies its length.
OpenStax Calculus Volume 1, §6.3 Volumes of Revolution: Cylindrical Shells §6.3, pp. 569-581 — everything on these slides traces back here
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