Volume of revolution — shell method: Question bank

True or false — justify

TF1. "The shell method always uses thickness

d x

False. Thickness matches the strip, and the strip must run parallel to the axis. About a horizontal axis the strips are horizontal, so thickness is

d y

TF2. "Rotating a region about a farther axis always gives a larger volume."

False. Radius grows on the far side but the height profile and where the region sits also matter. The parent's own example gives

8 π

about both

x = 0

and

x = 3

— same volume despite the shift.

TF3. "The radius in a shell integral is always the variable of integration (

x

y

)."

False. Radius is the distance from strip to axis. It equals

x

only when the axis is the

y

-axis at

x = 0

; about

x = c

it is

∣ x - c ∣

TF4. "For a region between two curves

y = f (x)

and

y = g (x)

, the shell height is

f (x) + g (x)

False. Height is top minus bottom,

f (x) - g (x)

(the strip's actual length). Adding would double-count, giving nonsense.

TF5. "You can freely swap between the disk and shell methods; they give the same number."

True on the value, false on the effort. Both compute the same volume, but see Volume of revolution — disk and washer method — one often needs an ugly inverse the other avoids. Pick the easy one.

TF6. "If the axis of rotation passes through the region, the shell setup is unchanged."

False. The radius

∣ x - c ∣

changes sign of

(x - c)

across the axis, so you must split the integral at

x = c

and use the absolute distance on each side. Otherwise negative-radius terms wrongly cancel real volume.

TF7. "A shell whose height is zero contributes zero volume."

True.

d V = 2 π r h d x

; if

h = 0

the strip has no length, so the tube is empty. This is why endpoints where the curves meet contribute nothing.

TF8. "The factor

2 π

comes from the area of a circle."

False. It comes from the circumference

2 π r

— the length you get when you slit and unroll the tube into a flat sheet. Disks use area

π r^{2}

; shells use perimeter.

TF9. "Shells and disks integrate over the same variable for the same problem."

False. They usually use opposite variables: for the same axis, disk strips are perpendicular (one variable) while shell strips are parallel (the other). That is exactly why one avoids inverting

f

Spot the error

SE1. A student rotates

y = x^{2}

0 \leq x \leq 2

about

x = 3

and writes

V = \int_{0}^{2} 2 π x \cdot x^{2} d x

The radius is wrong: about

x = 3

it is

∣ x - 3∣ = 3 - x

, not

x

. They used the

y

-axis radius by habit; correct is

\int_{0}^{2} 2 π (3 - x) x^{2} d x

SE2. Rotating about the

x

-axis, a student writes

V = \int_{a}^{b} 2 π x f (x) d x

with

d x

The axis is horizontal, so strips must be horizontal with thickness

d y

; radius is

∣ y ∣

and height is

x_{right} - x_{left}

. Mixing a horizontal axis with

d x

and

x

-radius is the classic mismatch.

SE3. For the region between

y = x

and

y = x^{2}

[0, 1]

about the

y

-axis, a student sets

h = x^{2} - x

The sign is flipped. On

(0, 1)

x > x^{2}

, so top is

y = x

: height must be

x - x^{2}

. As written,

h

is negative and the volume comes out negative.

SE4. A student writes

d V = π r h d x

, "just like the disk uses

π

Missing the factor of

2

. Unrolling the tube gives a sheet of width

2 π r

(a full loop), so

d V = 2 π r h d x

. Half-remembering the disk's

π

is the trap.

SE5. Thickness is taken as

d x

but the height is written as

h = y = x

... then the student "integrates in

y

Variable inconsistency. If thickness is

d x

, then radius, height, and limits must all be functions of

x

. You cannot integrate a

d x

element with respect to

y

SE6. Rotating about

x = 3

with the region on

0 \leq x \leq 2

, a student writes

r = x - 3

(not the absolute value).

x - 3

is negative on the whole region, so this radius is negative. Correct radius is the distance

3 - x > 0

. Here the sign error would flip the whole volume negative.

SE7. A student sets the limits of a shell integral (about the

y

-axis) using

y

-values because "the region goes from

y = 0

y = 4

The limits must match the thickness variable. Thickness is

d x

, so limits are the

x

-range of the strips, not the

y

-range.

Why questions

WHY1. Why does the

d x^{2}

term in

π h (2 r d x + d x^{2})

get dropped?

It is second order: as

d x \to 0

d x^{2}

shrinks far faster than

d x

, so its contribution to the total sum vanishes in the limit. See Definite integral as a limit of Riemann sums.

WHY2. Why does the shell method often beat disks for

y = f (x)

about the

y

-axis?

Shells keep the natural variable $x$ , so you never have to solve

y = f (x)

for

x = f^{- 1} (y)

, which can be ugly or impossible.

WHY3. Why must shell strips run parallel to the axis, not perpendicular?

A strip parallel to the axis sweeps out a tube (shell) when rotated; a perpendicular strip sweeps a disk/washer instead. The geometry of the swept solid decides the method.

WHY4. Why is the shell height "top − bottom" the same idea as area between curves?

The strip's length is exactly the vertical gap between the two curves — the same

y_{top} - y_{bot}

integrand from Area between two curves. Shells just weight each strip by

2 π r

and revolve it.

WHY5. Why is the circumference

2 π r

and not, say,

π r

π d /2

Going once all the way around a circle of radius

r

has length

2 π r

by definition of

π

. Unrolling the shell lays out that full loop as the width of the flat sheet.

WHY6. Why can the same volume come from integrating in

x

(shells) or

y

(disks)?

Both partition the same solid into thin pieces; the total is independent of how you slice it. Only the algebra differs — this is the "

d x

d y

" choice in Choosing dx vs dy strips.

WHY7. Why must we split the integral when the axis passes through the region?

The distance

∣ x - c ∣

has a kink at

x = c

(it is

c - x

on one side,

x - c

on the other). An integrand with an absolute value is handled by splitting at that point so each piece is smooth and positive.

Edge cases

EC1. What is the volume when the region has zero width (say

a = b

Zero. The integral

\int_{a}^{a} = 0

; there are no strips to revolve, so no solid forms.

EC2. What happens to a shell whose radius

r = 0

(the strip sits exactly on the axis)?

Its circumference

2 π r = 0

, so

d V = 0

. A degenerate tube collapsed to a line has no volume — this single strip contributes nothing.

EC3. If a region straddles the

y

-axis (has both positive and negative

x

) and we rotate about the

y

-axis, what is the radius?

r = ∣ x ∣

, and typically you split at $x = 0$ since the two sides overlap in the swept solid. Using

r = x

alone would give negative contributions on the left.

EC4. A region lies entirely to the left of the axis

x = c

(so

x < c

). What is the radius, and is anything different?

r = c - x > 0

, still just the positive distance. The setup is identical to the "right of axis" case — only the sign inside the absolute value flips, and

∣ x - c ∣

handles it automatically.

EC5. Rotating a region about a horizontal line

y = c

that lies below the whole region — how is the radius set?

r = y - c > 0

for every strip, with horizontal strips of thickness

d y

and height

x_{right} - x_{left}

. No splitting needed since the region never crosses

y = c

EC6. The curves meet at the integration endpoints (like

y = x

and

y = x^{2}

meeting at

x = 0, 1

). What does the height do there?

The height

\to 0

at both meeting points, so the boundary shells vanish. The solid tapers smoothly to nothing at the ends — no discontinuity, no leftover cap.

EC7. Can the shell height ever be a horizontal length while thickness is

d x

No. With thickness

d x

the strips are vertical, so their length is vertical (a

y

-difference). Horizontal heights (

x_{R} - x_{L}

) go with thickness

d y

. Matching these is the orientation rule.

Recall One-line self-test

Cover every answer above and re-derive the reason, not just the verdict. If you can explain why a false statement is false, you own the concept.

Connections

4.2.15 · Shell Method — the parent derivation these traps guard.
Choosing dx vs dy strips — the orientation decision behind half these misconceptions.
Volume of revolution — disk and washer method — the "same value, different effort" comparison.
Area between two curves — where "top − bottom" comes from.
Definite integral as a limit of Riemann sums — why the $d x^{2}$ term vanishes.