Volume of revolution — disk method, washer method: Question bank

Before you start, hold three pictures in your head:

A disk is a solid coin — the region touches the axis of revolution with no gap.
A washer is a coin with a hole — the region floats away from the axis, so a ring appears.
The radius of a slice is always the distance from the axis to the curve, never just "the $y$ -value."

True or false — justify

TF1. "Disk and washer are two different formulas you must memorise separately."

False — a disk is just a washer whose inner radius

r = 0

, so

π (R^{2} - 0^{2}) = π R^{2}

; there is really one idea: outer disk minus hole.

TF2. "If the region touches the axis of revolution, you never get a hole."

True — a hole (washer) appears only when the nearest edge of the region is held away from the axis; touching the axis means the near radius is

0

TF3. "Rotating the region between

y = x

and

y = x^{2}

about the x-axis gives the same volume as rotating it about the y-axis."

False — the axis changes both the radius (distance) and the slicing direction, so the swept solids differ in shape and volume.

TF4. "

\int_{a}^{b} (R - r) d x

gives the washer volume if you multiply by

π

False — that would be

π \int (R - r) d x

, which mixes an area-between-curves integrand into a volume problem; the ring's area is

π R^{2} - π r^{2}

, not

π (R - r)

TF5. "Because volume can't be negative, the radius must always be positive, so you can drop absolute-value signs."

Partly — the radius is a distance so it is

\geq 0

, but we still write

∣ f (x) - k ∣

to be safe; luckily we square it, and squaring already erases the sign, so the answer stays positive automatically.

TF6. "When rotating about

y = 2

, you can use

f (x)

as the radius as long as

f (x)

stays below

2

False — the radius is the distance to the axis

∣ f (x) - 2∣

, not

f (x)

; using

f (x)

measures from

y = 0

, which is the wrong line.

TF7. "The disk method requires the axis to be the x-axis or y-axis."

False — you can revolve about any horizontal line

y = k

or vertical line

x = k

; you just re-measure every radius from that line.

TF8. "Doubling the thickness of every slice doubles the true volume."

False conceptually — thickness is

d x \to 0

, an infinitesimal we sum, not a quantity you can scale; "doubling

d x

" just makes a coarser approximation, not a different solid.

Spot the error

SE1. "Region between

y = x

and

y = x^{2}

[0, 1]

: volume

= π \int_{0}^{1} (x - x^{2})^{2} d x

Error — they squared the difference; correct is

π \int_{0}^{1} (x^{2} - (x^{2})^{2}) d x

, i.e. square each radius then subtract, since

R^{2} - r^{2} \neq = (R - r)^{2}

SE2. "Rotating

y = x

[0, 4]

about the x-axis:

V = π \int_{0}^{4} x d x

Error — the radius

x

must be squared:

π \int_{0}^{4} (x)^{2} d x = π \int_{0}^{4} x d x

; they forgot the

r^{2}

π r^{2}

SE3. "Rotating a region about the x-axis, I'll slice with thickness

d y

and use y-limits."

Error — slices are ⊥ to the axis, so an x-axis rotation gives slices of thickness

d x

and x-limits;

d y

belongs to a y-axis rotation.

SE4. "Region

1 \leq y \leq 3

(a horizontal strip) rotated about the x-axis is a disk of radius

3

Error — the strip does not touch the x-axis (nearest edge is

y = 1

), so it's a washer:

R = 3

r = 1

, giving

π \int (3^{2} - 1^{2}) d x

SE5. "About

y = 2

, region under

y = x

[0, 2]

: radius

= x - 2

, so

V = π \int_{0}^{2} (x - 2) d x

Two errors — radius should be squared

(x - 2)^{2}

, and it need not be forced positive since squaring handles the sign; correct is

π \int_{0}^{2} (x - 2)^{2} d x

SE6. "The washer's outer radius is always the top curve

f (x)

Error — "outer" means farthest from the axis, not "topmost"; if the axis lies above the region, the lower curve can be the outer one, so identify

R

by distance, not by which is higher.

SE7. "To rotate about the y-axis I keep

y = f (x)

and just change the limits to

c

and

d

Error — you must invert to

x = g (y)

so the radius is expressed in the slicing variable

y

V = π \int_{c}^{d} [g (y)]^{2} d y

Why questions

WHY1. Why is a thin slice modelled as a cylinder and not, say, a cone?

Over an infinitesimal thickness

d x

the curve barely changes, so the slice has near-constant radius — a flat coin (cylinder), whose

π r^{2} h

we already know; the cone-like taper vanishes in the limit

d x \to 0

WHY2. Why does

π

appear outside the integral in every formula?

Every slice's area carries the same constant factor

π

(from

π r^{2}

), and a constant factor can be pulled out of a sum — hence out of the integral.

WHY3. Why do we square the radius but only take the first power of the thickness?

The circular face is 2D so its area

π r^{2}

scales with radius squared; the thickness is the one remaining dimension, contributing a single factor

d x

— together they make a 3D volume.

WHY4. Why does the definite integral, and not a plain sum, give the exact volume?

Finite slabs only approximate; the integral is the limit of that Riemann sum as

d x \to 0

, which removes the approximation error entirely.

WHY5. Why is

R^{2} - r^{2}

correct but

(R - r)^{2}

wrong for a ring's area?

The ring is outer disk minus inner disk,

π R^{2} - π r^{2}

; expanding

(R - r)^{2} = R^{2} - 2 R r + r^{2}

shows the extra

- 2 R r + 2 r^{2}

terms that don't belong — squaring is not linear.

WHY6. Why does the Area Between Curves integrand use

f - g

while washers use

f^{2} - g^{2}

Area sums heights (a 1D difference

f - g

); volume sums circular areas, and each area depends on the radius squared, so the difference of areas is

f^{2} - g^{2}

WHY7. Why must the radius be re-measured when rotating about

y = k

instead of

y = 0

Radius means distance from the actual axis; shifting the axis to

y = k

shifts every distance to

∣ f (x) - k ∣

, so keeping

f (x)

would measure from the wrong line.

WHY8. Why can u-substitution make a shifted-axis integral like

π \int_{0}^{2} (2 - x)^{2} d x

easier?

Setting

u = 2 - x

turns the shifted radius into a clean

π \int u^{2} d u

, a basic power integral, hiding the algebra of expanding

(2 - x)^{2}

WHY9. Why is the Shell Method sometimes preferred over disks/washers for the same solid?

Shells slice parallel to the axis, so if a region is awkward to invert (to get

x = g (y)

) or would need two separate washer integrals, one shell integral in the natural variable is simpler.

Edge cases

EC1. What is the volume if the region has zero area (a single curve, no enclosed region)?

Zero — with no width there is nothing to sweep; the "solid" is just a 2D surface of revolution with no volume.

EC2. The region touches the axis at exactly one point (e.g.

y = x

meeting

y = 2

x = 2

). Disk or washer there?

At that point the inner radius is

0

, so the slice is a full disk; nothing special happens — the washer smoothly degenerates to a disk at the touch point.

EC3. What happens to the volume if the curve dips below the axis, so

f (x) < 0

on part of

[a, b]

Nothing bad — the radius is the distance

∣ f (x) ∣

, and squaring gives

[f (x)]^{2} \geq 0

regardless of sign, so the negative part still adds positive volume (it sweeps the same coin on the other side).

EC4. If two curves cross inside

[a, b]

, so "outer" and "inner" swap, can you use one washer integral?

No — where they swap,

R

and

r

trade places; you must split at the crossing point into separate integrals, each with the correct

R

and

r

EC5. Rotating a region about a line that passes through its interior — is that still a washer?

No single washer — the axis cuts the region into a piece on each side; you handle the piece with the larger radius (or add both correctly), because a slice can't have an outer radius smaller than its inner.

EC6. What is the disk volume when the radius

f (x) = 0

over the whole interval?

Zero — every slice has area

π \cdot 0^{2} = 0

; a degenerate "solid" collapsed onto the axis.

EC7. If the interval has zero length (

a = b

), what is the volume?

Zero — the integral

π \int_{a}^{a} [\dots] d x = 0

; there is no extent along the axis to sweep, so no solid forms.

EC8. Rotating a region that is symmetric about the axis (equal parts above and below) — do we double or ignore the bottom half?

Ignore it — the top half sweeps the same solid as the bottom half (they overlap when spun), so you use only one side; doubling would count the solid twice.

Recall One-line survival kit

Radius = distance from axis (square it). Thickness matches the axis ( $d x$ for x-axis, $d y$ for y-axis). Ring = big disk − small disk, squares first, subtract second.

Connections

Definite Integral as a Limit of Sums — why the limit of slices is exact (WHY4).
Area Between Curves — the $f - g$ vs $f^{2} - g^{2}$ distinction (WHY6).
Shell Method — alternative slicing when inversion is ugly (WHY9).
Volume by Cross-Sections — disks/washers are the circular special case.
u-substitution — clean-up tool for shifted-axis integrals (WHY8).