Worked examples — Volume of revolution — shell method

Before anything, meet the three pieces of every shell. In the figure below, the radius $r$ (cyan, along the bottom) is the horizontal distance from the axis to the strip; the height $h$ (amber, on the right) is how tall the cyan strip is; and the thickness — which for a vertical strip is exactly the symbol $dx$ (a horizontal axis would make it $dy$ instead) — is how wide the strip is (white, at the top). The amber vertical line marked "axis" is the $y$-axis ($c=0$) — notice the radius arrow starts exactly there. Nothing on this page uses a symbol beyond these three plus $\int$, which just means "add up all the strips."

Figure s01 — one cylindrical shell dissected: cyan radius $r=x$ from the amber axis, amber height $h$, white thickness labelled $dx$.

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The scenario matrix

Every shell-method problem is one (or a blend) of these cells. The whole difficulty is always the radius and the height — get those right and the integral is routine. (Recall $c$ = the number naming the axis, defined just above; $f(x)$ = the top curve's height.)

Cell	What changes	Radius $r$	Height $h$	Thickness	Example
A	Axis = $y$-axis ($c=0$), region to its right	$x$	$f(x)$	$dx$	Ex 1
B	Axis = vertical line $x=c$, region left of it	$c-x$	top − bottom	$dx$	Ex 2
C	Axis = vertical line $x=c$, region right of it	$x-c$	top − bottom	$dx$	Ex 3
D	Region straddles the axis $x=c$ (radius sign flips)	$	x-c	$	$f(x)$
E	Axis = horizontal line $y=c$ → must use $dy$	$	y-c	$	right − left
F	Curve given as $x=g(y)$ (no inversion needed)	$y$ or $	y-c	$	$g(y)$
G	Degenerate: axis touches the region (radius $=0$ at an edge)	$x$	$f(x)$	$dx$	Ex 7
H	Word problem with units	$x$ (distance in cm)	$f(x)$ (length in cm)	$dx$	Ex 8
I	Exam twist: region between two curves + shifted axis	$	x-c	$	top − bottom

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Cell A — axis is the $y$-axis ($c=0$)

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Cells B & C — a vertical line $x=c$ as axis, region on either side

The single subtle thing here is the sign inside the radius. Look at the figure: the cyan strip sits at some $x$ between $0$ and $2$, and the same strip is measured to two different axes. To the right axis ($c=5$) the distance is $5-x$; to the left axis ($c=-1$) it is $x+1$. Both cyan arrows come out positive — that is the whole lesson of this pair.

Figure s02 — the same cyan region measured to two amber axes; cyan arrows show $r=5-x$ (right axis) and $r=x+1$ (left axis), both positive.

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Cell D — the region straddles a vertical axis $x=c$

The danger here is a radius that changes sign as you cross the axis. The figure shows a region running from $x=1$ to $x=4$ split by the axis $x=2$: strips to the left of the axis have radius $2-x$, strips to the right have $x-2$, and the single formula $|x-2|$ captures both. Watch the cyan radius shrink to zero exactly at the axis and grow again on the far side.

Figure s04 — rectangle straddling the amber axis $x=2$; cyan arrows show radius $2-x$ on the left, $x-2$ on the right, both vanishing at the axis.

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Cell E — horizontal axis $y=c$ forces $dy$

When the axis is horizontal, strips must be horizontal too (parallel to the axis), so their thickness is $dy$ and their length is measured left-to-right. In the figure, the amber length is now a width ($x_{\text{right}}-x_{\text{left}}$) and the cyan radius is measured vertically from the axis $y=c$ up to the strip. Here the axis is drawn as the line $y=1$ (so $c=1$), not the $x$-axis — watch the radius arrow start at $y=1$, giving $r=|y-1|$.

Figure s03 — horizontal strip (thickness $dy$) with cyan vertical radius $r=|y-1|$ from the amber axis $y=1$ and amber width $h=$ right $-$ left.

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Cell F — curve already in the form $x=g(y)$

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Cell G — the degenerate case (radius zero at an edge)

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Cell H — a word problem with units

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Cell I — the exam twist (two curves + shifted axis)

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Recall checkpoint

Recall Which cell is each situation?

Axis is $x=4$, region lies in $0\le x\le 3$ — which radius? ::: Region is left of the axis (Cell B): $r=4-x$; here $c=4$. Rotating about the horizontal line $y=2$ — what is the strip thickness? ::: $dy$ (Cell E); strips are horizontal, radius $|y-2|$. Region symmetric across the $y$-axis, rotated about it — what safeguard? ::: Use $r=|x|$ or exploit symmetry (Cell D); a signed $x$ would wrongly cancel. Region crosses the axis $x=2$ — how do you set up the integral? ::: Use $r=|x-2|$ and split at $x=2$ into $2-x$ (left) and $x-2$ (right). A strip sits exactly on the axis — is the integral broken? ::: No (Cell G); that shell has $r=0$, contributing $0$, integral stays finite. Curve given as $x=g(y)$, axis horizontal — why is shell ideal? ::: No inversion needed (Cell F); integrate in $y$ directly.

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Connections

• shell method — the parent formula this page stress-tests.
• Volume of revolution — disk and washer method — the perpendicular-slice alternative; Cell D's check used solid cylinders (the disk method).
• Definite integral as a limit of Riemann sums — why summing infinitely many shells is exact.
• Area between two curves — the "top − bottom" height reused in Cells D, E, I.
• Choosing dx vs dy strips — decides between Cells A–D ($dx$) and Cells E–F ($dy$).
• Arc length and surfaces of revolution — revolve the boundary instead of the region.

Case map