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Lesson 89 — Volume by slicing: disks, washers, and cylindrical shells

Solids of revolution and solids with known cross-sections. Disk method, washer method, and cylindrical shell method. Cavalieri's principle.

Used in: Calculus II (BR) · Calc BC AP (USA) · Advanced Math III Japanese · Leistungskurs Class 12 (DE)

V = \int_a^b A(x)\, dx, \qquad A_{\text{disk}} = \pi[f(x)]^2, \qquad A_{\text{washer}} = \pi\bigl([R(x)]^2 - [r(x)]^2\bigr), \qquad V_{\text{shell}} = 2\pi\int_a^b x\,f(x)\,dx

Choose your door

Rigorous notation, full derivation, hypotheses

Rigorous definition and the three methods

Cavalieri's principle and slicing

"If two solids have the same height and equal cross-sections at each level, then the two solids have the same volume." — Cavalieri's Principle (17th century), formalized in Active Calculus §6.2

Cylindrical shell method

"The shell method can be thought of as integrating along the axis parallel to the axis of rotation." — APEX Calculus §7.3

For revolution around $y = c$ (instead of the $x$ -axis): substitute $f(x) \mapsto f(x) - c$ (or $|f(x) - c|$ ). For revolution around $x = c$ with shells: substitute $x \mapsto |x - c|$ in the radius role.

Method selection

Decision rule between disk/washer and shell. Always sketch the region first.

Worked examples

Example— 1· Volume of sphere by disks (application)

Problem: Derive the volume formula of a sphere of radius $r$ using the disk method.

Strategy: A sphere of radius $r$ is generated by revolving the semicircle $y = \sqrt{r^2 - x^2}$ , for $x \in [-r, r]$ , around the $x$ -axis. Each cross-section is a disk of radius $\sqrt{r^2 - x^2}$ .

Solution:

Applying the disk formula:

$V = \pi \int_{-r}^{r} \bigl(\sqrt{r^2 - x^2}\bigr)^2\, dx = \pi \int_{-r}^{r} (r^2 - x^2)\, dx$

$= \pi \left[r^2 x - \frac{x^3}{3}\right]_{-r}^{r} = \pi \left[\left(r^3 - \frac{r^3}{3}\right) - \left(-r^3 + \frac{r^3}{3}\right)\right]$

$= \pi \cdot 2\left(r^3 - \frac{r^3}{3}\right) = \pi \cdot 2 \cdot \frac{2r^3}{3} = \frac{4\pi r^3}{3}$

Verification: For $r = 1$ : $V = 4\pi/3 \approx 4.19$ u³. Classical result confirmed.

Source. OpenStax Calculus Volume 2 §2.2, Example 2.5 — CC-BY-NC-SA license.

Example— 2· Volume by washers — region between two parabolas (intermediate)

Problem: Calculate the volume of the solid generated by revolving, around the $x$ -axis, the region bounded by $y = x$ and $y = x^2$ , with $x \in [0, 1]$ .

Strategy: The two curves intersect at $x = 0$ and $x = 1$ . On $[0, 1]$ , we have $x \geq x^2$ , so $R(x) = x$ (outer radius) and $r(x) = x^2$ (inner radius). Washer method.

Solution:

$V = \pi \int_0^1 \bigl([x]^2 - [x^2]^2\bigr)\, dx = \pi \int_0^1 (x^2 - x^4)\, dx$

$= \pi \left[\frac{x^3}{3} - \frac{x^5}{5}\right]_0^1 = \pi\!\left(\frac{1}{3} - \frac{1}{5}\right) = \pi \cdot \frac{2}{15} = \frac{2\pi}{15}$

Verification: The volume should be less than the cone of radius 1 and height 1 ( $V_{\text{cone}} = \pi/3 \approx 1.047$ ). Indeed, $2\pi/15 \approx 0.419 < \pi/3$ . Plausible — the region between $y=x$ and $y=x^2$ is smaller than the complete triangle.

Source. Active Calculus §6.2, Activity 6.2.2 — CC-BY-NC-SA license.

Example— 3· Volume by shells — comparison with washers (advanced)

Problem: Use cylindrical shells to calculate the volume generated by revolving the region between $y = x^2$ and $y = x$ , on $[0, 1]$ , around the y-axis. Compare with the washer result if you integrated in $y$ .

Strategy: The axis of revolution is $y$ (vertical) and the natural variable is $x$ : shell method. Each vertical strip of width $dx$ generates a shell of radius $x$ , height $f(x) - g(x) = x - x^2$ .

Solution:

$V = 2\pi \int_0^1 x(x - x^2)\, dx = 2\pi \int_0^1 (x^2 - x^3)\, dx$

$= 2\pi \left[\frac{x^3}{3} - \frac{x^4}{4}\right]_0^1 = 2\pi\!\left(\frac{1}{3} - \frac{1}{4}\right) = 2\pi \cdot \frac{1}{12} = \frac{\pi}{6}$

Cross-check by washers in $y$ : The curves in $y$ : from $y = x^2$ we have $x = \sqrt{y}$ ; from $y = x$ we have $x = y$ . Outer radius $R(y) = \sqrt{y}$ , inner radius $r(y) = y$ :

$V = \pi \int_0^1 (y - y^2)\, dy = \pi\!\left[\frac{y^2}{2} - \frac{y^3}{3}\right]_0^1 = \pi\!\left(\frac{1}{2} - \frac{1}{3}\right) = \frac{\pi}{6} \checkmark$

Source. APEX Calculus §7.3, Example 7.3.2 — CC-BY-NC license.

Example— 4· Volume of cone by disks — formula derivation (demonstration)

Problem: Derive the formula $V = \pi R^2 h/3$ of a right circular cone of radius $R$ and height $h$ .

Strategy: The cone is generated by revolving the line $y = (R/h)x$ around the $x$ -axis, for $x \in [0, h]$ . Each cross-section is a disk of radius $(R/h)x$ .

Solution:

$V = \pi \int_0^h \left(\frac{R}{h} x\right)^2 dx = \pi \frac{R^2}{h^2} \int_0^h x^2\, dx = \pi \frac{R^2}{h^2} \cdot \frac{h^3}{3} = \frac{\pi R^2 h}{3}$

Verification: For $R = h = 1$ : $V = \pi/3 \approx 1.047$ u³. Pappus's Theorem confirms: the centroid of the right triangle with base $R$ and height $h$ is at $\bar{x} = h/3$ from the vertex. Volume of the shell: $2\pi \cdot (R/2) \cdot (Rh/2) = \pi R^2 h / 2 \neq \pi R^2 h/3$ — here Pappus applies to the solid, not the triangle. Direct derivation by disks is cleanest.

Source. OpenStax Calculus Volume 2 §2.2, Example 2.4 — CC-BY-NC-SA license.

Example— 5· Pumping work — spherical tank (modeling)

Problem: A spherical tank of radius 3 m is full of water (density $\rho = 1000$ kg/m³, $g = 9.8$ m/s²). Calculate the work required to pump all the water to the top of the tank.

Strategy: Place the sphere's center at the origin. The top is at $y = 3$ . A slice of thickness $dy$ at height $y$ must be lifted $(3 - y)$ meters. The circular cross-section at height $y$ has radius $\sqrt{9 - y^2}$ , hence area $\pi(9 - y^2)$ .

Solution:

$W = \int_{-3}^{3} \rho g \cdot \pi(9 - y^2) \cdot (3 - y)\, dy$

$= 1000 \cdot 9.8 \cdot \pi \int_{-3}^{3} (9 - y^2)(3 - y)\, dy$

Expanding the product:

$(9 - y^2)(3 - y) = 27 - 9y - 3y^2 + y^3$

$\int_{-3}^{3} (27 - 9y - 3y^2 + y^3)\, dy = \left[27y - \frac{9y^2}{2} - y^3 + \frac{y^4}{4}\right]_{-3}^{3}$

The odd terms ( $-9y$ and $y^3$ ) cancel by symmetry. Even terms:

$= 2\left(27 \cdot 3 - 3^3\right) = 2(81 - 27) = 108$

$W = 9800\pi \cdot 108 \approx 3.33 \times 10^6 \text{ J} = 3.33 \text{ MJ}$

Verification: Order of magnitude: moving $\approx 113$ t of water $\approx 1.5$ m (centroid 1.5 m from the top) gives $113,000 \times 9.8 \times 1.5 \approx 1.66$ MJ. Factor 2 expected because the actual centroid (by symmetry, at $y = 0$ , distance 3 from the top) gives 3.32 MJ. Confirms.

Source. OpenStax Calculus Volume 2 §6.5, Example 6.24 — CC-BY-NC-SA license.

Exercise list

45 exercises · 11 with worked solution (25%)

Application 28Understanding 3Modeling 5Challenge 6Proof 3

Ex. 89.1ApplicationAnswer key
Revolution of $y = \sqrt{x}$ on $[0, 4]$ around the $x$ -axis. Calculate the volume.
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Ex. 89.2Application
Revolution of $y = x$ on $[0, 1]$ around the $x$ -axis. Calculate the volume (it is a cone).
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Ex. 89.3ApplicationAnswer key
Revolution of $y = x^2$ on $[0, 2]$ around the $x$ -axis. Calculate the volume.
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Ex. 89.4ApplicationAnswer key
Revolution of $y = \sin x$ on $[0, \pi]$ around the $x$ -axis. (Hint: $\sin^2 x = (1 - \cos 2x)/2$ .)
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Ex. 89.5Application
Revolution of $y = e^x$ on $[0, 1]$ around the $x$ -axis.
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Ex. 89.6Application
Revolution of $y = \cos x$ on $[0, \pi/2]$ around the $x$ -axis.
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Ex. 89.7Proof
Show that the volume of a sphere of radius $r$ is $4\pi r^3/3$ , using $V = \pi\int_{-r}^r (r^2 - x^2)\,dx$ .
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Ex. 89.8ProofAnswer key
Show that the volume of a cone of radius $R$ and height $h$ is $\pi R^2 h/3$ , by revolving $y = (R/h)x$ on $[0, h]$ .
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Ex. 89.9Application
Revolution of $y = 1/x$ on $[1, 2]$ around the $x$ -axis.
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Ex. 89.10Application
Revolution of $y = \sqrt{4 - x^2}$ on $[-2, 2]$ around the $x$ -axis. (Identify the resulting solid.)
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Ex. 89.11ApplicationAnswer key
Washer: region between $y = x$ and $y = x^2$ on $[0, 1]$ , rotated around the $x$ -axis.
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Ex. 89.12Application
Washer: region between $y = \sqrt{x}$ and $y = x$ on $[0, 1]$ , rotated around the $x$ -axis.
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Ex. 89.13Application
Washer: region between $y = 2x$ and $y = x^2$ on $[0, 2]$ , rotated around the $x$ -axis.
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Ex. 89.14Application
Washer: region between $y = x^2 + 1$ and $y = 5 - x^2$ , rotated around the $x$ -axis.
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Ex. 89.15ApplicationAnswer key
Region between $y = x$ and $y = x^2$ on $[0, 1]$ , rotated around $y = -1$ (shifted axis).
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Ex. 89.16Application
What is the volume generated by revolving the region between $y = 1$ and $y = x$ on $[0, 1]$ around the $x$ -axis?
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Ex. 89.17Application
Region between $y = x^2$ and $y = x$ on $[0, 1]$ , rotated around $x = 2$ .
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Ex. 89.18Modeling
A bicycle inner tube can be modeled as a torus with central radius $R = 30$ cm and circular cross-section of radius $r = 2$ cm. Calculate the internal volume of the tube (in cm³) using Pappus's Theorem.
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Ex. 89.19Modeling
The rectangular region $[0, h] \times [0, R]$ is rotated around the $x$ -axis. Identify the resulting solid and calculate the volume.
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Ex. 89.20Understanding
To calculate the volume generated by revolving $y = f(x)$ , $x \in [a,b]$ , around the $y$ -axis, integrating in $x$ , which method is most natural?
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Ex. 89.21Application
Revolution of $y = x^2$ on $[0, 2]$ around the $y$ -axis (shells).
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Ex. 89.22Application
Region between $y = x$ and $y = x^2$ on $[0,1]$ , rotated around the $y$ -axis (shells).
Solve online
Ex. 89.23Application
Revolution of $y = \sqrt{x}$ on $[0, 1]$ around the $y$ -axis (shells).
Solve online
Ex. 89.24ApplicationAnswer key
Region between $y = x$ and $y = x^2$ on $[0,1]$ , rotated around $x = 2$ .
Solve online
Ex. 89.25Application
Revolution of $y = e^{-x^2}$ on $[0, 1]$ around the $y$ -axis (shells). (Hint: substitution $u = x^2$ .)
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Ex. 89.26Understanding
Which integral represents the volume generated by revolving $y = \cos x$ , $x \in [0, 1]$ , around the $y$ -axis (shells)?
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Ex. 89.27ApplicationAnswer key
Revolution of $y = 1/x$ on $[1, 3]$ around the $y$ -axis (shells).
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Ex. 89.28Application
Revolution of $y = \sin x$ on $[0, \pi]$ around the $y$ -axis (shells). (Hint: integration by parts.)
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Ex. 89.29Application
Region between $y = x^2$ and $y = x^3$ on $[0, 1]$ , rotated around $x = 1$ .
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Ex. 89.30Challenge
Calculate the volume of the region bounded by $y = x^2$ and $y = 4$ , rotated around the $y$ -axis, using two methods (disks in $y$ and shells in $x$ ). Confirm that the results match.
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Ex. 89.31ApplicationAnswer key
A solid has a base on the interval $[0, \pi]$ on the $x$ -axis, with square cross-sections perpendicular to the $x$ -axis. The side of each square is $f(x) = \sin x$ . Calculate the volume.
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Ex. 89.32Application
A solid has base $[0, 4]$ on the $x$ -axis, with semicircular cross-sections perpendicular to the $x$ -axis. The diameter of each semicircle is $y = \sqrt{x}$ . Calculate the volume.
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Ex. 89.33Proof
Show, via slicing, that the volume of a pyramid with base $A_0$ and height $h$ is $A_0 h/3$ , independent of the shape of the base.
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Ex. 89.34Modeling
Spherical tank of radius $R = 3$ m, filled with water. Calculate the work to pump all the water to the top (in Joules). Use $\rho = 1000$ kg/m³ and $g = 9.8$ m/s².
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Ex. 89.35UnderstandingAnswer key
Which statement about the choice between disk/washer and cylindrical shells is correct?
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Ex. 89.36Modeling
A decorative vase has a profile generated by revolving $x = \sqrt{y}$ (cm) around the $y$ -axis, for $y \in [0, 20]$ cm. Calculate the capacity of the vase in mL ( $1$ cm³ $= 1$ mL).
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Ex. 89.37Challenge
Derive the formula $V = 2\pi^2 R r^2$ of the torus via the washer method, without using Pappus's Theorem.
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Ex. 89.38Challenge
Gabriel's Horn paradox. Consider the surface generated by revolving $y = 1/x$ , $x \in [1, +\infty)$ , around the $x$ -axis. (a) Calculate the volume of the solid. (b) Show that the lateral area is infinite. (c) Interpret the paradox.
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Ex. 89.39Application
Region between $y = x^2$ and $y = 1$ , rotated around $y = 1$ . Calculate the volume (washer with shifted axis).
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Ex. 89.40Application
Revolution of $y = x^2 + 1$ on $[0, 3]$ around the $y$ -axis (shells).
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Ex. 89.41Application
Region between $y = e^x$ and $y = x$ on $[0, 1]$ , rotated around the $x$ -axis (washers).
Solve online
Ex. 89.42Modeling
A hemispherical tank of radius $R = 2$ m (opening facing up) contains water to a depth of $h = 1$ m. Calculate the volume of water (in m³).
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Ex. 89.43ChallengeAnswer key
Region between $y = \cos x$ and $y = \sin x$ on $[0, \pi/2]$ , rotated around the $x$ -axis. (Warning: the curves cross at $x = \pi/4$ .)
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Ex. 89.44Challenge
Revolution of $y = \sin(x^2)$ on $[0, \pi]$ around the $y$ -axis (shells). (Hint: substitution $u = x^2$ .)
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Ex. 89.45Challenge
The triangle with vertices $(0,0)$ , $(1,0)$ , $(0,1)$ is rotated around the line $x = 2$ . Calculate the volume using Pappus's Theorem. Verify by integrating directly.
Solve online

Sources

Active Calculus — Matt Boelkins, David Austin, Steve Schlicker · Grand Valley State University · 2024 · CC-BY-NC-SA. Sections §6.2 and §6.3. Exercises from activities 6.2.1–6.3.7 used in the list.
APEX Calculus — Hartman, Heinold, Siemers, Chalishajar · Virginia Military Institute · 2023 · CC-BY-NC. Sections §7.2 (disks/washers) and §7.3 (shells). Exercises ex. 7.2.5–7.2.25 and ex. 7.3.5–7.3.9 used in the list.
OpenStax Calculus Volume 2 — OpenStax (Herman, Strang et al.) · Rice University · 2023 · CC-BY-NC-SA. Sections §2.2–2.3 (volumes) and §6.5 (physical applications). Exercises and examples 2.2.50–2.2.92 and 6.5.258–6.5.262 used.