Single-Variable-Calculus · Unit 18 · Video 2 · Interactive Practice

The Shell Method: Volume of a Parabolic Cauldron

IKey Formulas

Formula	Meaning	What you need
$dV = (2\pi x)\,(y_{\text{top}} - y_{\text{bottom}})\,dx$	Shell method: a vertical slice revolved about the $y$ -axis	Radius $x$ , height from the diagram, thickness $dx$
$dV = 2\pi x\,(a - x^2)\,dx$	The cauldron: $y = x^2$ below, $y = a$ on top	Substitute $y = x^2$ into the height $a - y$
$V = \displaystyle\int_0^{\sqrt{a}} 2\pi(ax - x^3)\,dx = \frac{\pi a^2}{2}$	Volume of the cauldron	Limits on ONE side of the axis: $x = 0$ to the rim $x = \sqrt{a}$

Key Insight: A shell unrolls into a slab of length $2\pi x$ , height $a - x^2$ and thickness $dx$ . One full turn of the right half already fills the solid, so the limits start at $x = 0$ , not at $x = -\sqrt{a}$ .

IIVisualization 1 — The Cauldron as Nested Shells

For $a = 4$ , the shells' volumes $2\pi x_i (a - x_i^2)\,\Delta x$ add up to $\tfrac{\pi a^2}{2} = 8\pi$ .

Number of shells n 4

IIIVisualization 2 — One Shell, Three Lengths

The shell at radius $x$ has volume $2\pi x\,(4 - x^2)\,dx$ ; the area under that integrand is the volume.

Slice at x 1

IVVisualization 3 — Which Setup Fills the Cauldron Once?

Every correct setup sweeps each point of the cauldron exactly once and gives $8\pi$ for $a = 4$ .

Setup

VQuiz Questions

Problem 1 · Shells Under a Downward Parabola

Given: the region under $y = 4 - x^2$ and above the $x$ -axis, for $0 \le x \le 2$ , is revolved about the $y$ -axis — find the volume.

✅ Correct!

V = \int_0^2 2\pi x(4 - x^2)\,dx = 2\pi\left[2x^2 - \tfrac{x^4}{4}\right]_0^2 = 2\pi(8 - 4) = 8\pi

❌ That counts the solid twice. Revolving the right half alone already fills the solid; adding the mirror half

-2 \le x \le 0

doubles the answer.

❌ That revolves the region about the $x$ -axis.

\int_0^2 \pi(4 - x^2)^2\,dx = \tfrac{256\pi}{15}

is the disk method; about the

y

-axis each slice becomes a shell of radius

x

❌ The circumference is $2\pi x$ , not $\pi x$ .

\pi x

is the half-turn factor, used only when both halves are turned through half a circle.

❌ Not quite. The slice at

x

has radius

x

, height

4 - x^2

and thickness

dx

; integrate

2\pi x(4 - x^2)

from

0

2

Show solution

A vertical slice at $x$ runs from $y = 0$ up to $y = 4 - x^2$ , so its height is $4 - x^2$ . Revolved about the $y$ -axis it is a shell of radius $x$ :

dV = (2\pi x)(4 - x^2)\,dx.

The slices run from the axis, $x = 0$ , to where the curve meets the $x$ -axis, $x = 2$ :

V = 2\pi\int_0^2 (4x - x^3)\,dx = 2\pi\left[2x^2 - \frac{x^4}{4}\right]_0^2 = 2\pi(8 - 4) = 8\pi.

Problem 2 · Limits for a Deeper Cauldron

Given: the cauldron made by revolving the region between $y = x^2$ and $y = 9$ about the $y$ -axis — which integral gives its volume?

✅ Correct! The rim is where

x^2 = 9

, at

x = 3

, and the integral equals

\tfrac{81\pi}{2} = \tfrac{\pi a^2}{2}

with

a = 9

❌ $9$ is the rim's $y$ -value. The integral is in

x

, and the rim is at

x = \sqrt{9} = 3

; past it the height

9 - x^2

turns negative.

❌ The left half is already covered. A full turn of the right half fills the cauldron. Over

-3 \le x \le 3

the integrand is odd, and this integral is

0

❌ That height is $y_{\text{bottom}}$ alone.

x^2

measures the region under the parabola. The cauldron's slice runs from the curve up to

y = 9

, so its height is

9 - x^2

❌ Not quite. The height is

y_{\text{top}} - y_{\text{bottom}}

, and the limits run from the axis to the rim, in

x

Show solution

Height: top $y = 9$ , bottom $y = x^2$ , so $9 - x^2$ . Radius: $x$ . Limits: from the axis $x = 0$ to the rim, where $9 = x^2$ , $x = 3$ .

V = \int_0^3 2\pi x(9 - x^2)\,dx = 2\pi\left[\frac{9x^2}{2} - \frac{x^4}{4}\right]_0^3 = 2\pi\left(\frac{81}{2} - \frac{81}{4}\right) = \frac{81\pi}{2}.

Problem 3 · Between a Line and a Parabola

Given: the region between $y = x$ (above) and $y = x^2$ (below), for $0 \le x \le 1$ , is revolved about the $y$ -axis — find the volume.

✅ Correct!

V = \int_0^1 2\pi x(x - x^2)\,dx = 2\pi\left(\tfrac13 - \tfrac14\right) = \tfrac{\pi}{6}

❌ That uses $\pi x$ for the length around. A full turn of a slice at radius

x

traces a circle of circumference

2\pi x

❌ The height is upside down. On

(0, 1)

x > x^2

, so the line is on top and the height is

x - x^2

; a volume is never negative.

❌ That adds the two curves. The height of the slice is the gap

y_{\text{top}} - y_{\text{bottom}} = x - x^2

, not

x + x^2

❌ Not quite. Integrate

2\pi x(x - x^2)

from

0

1

Show solution

The curves meet where $x = x^2$ , at $x = 0$ and $x = 1$ , and $y = x$ is on top between them. A slice at $x$ has height $x - x^2$ and radius $x$ :

V = \int_0^1 2\pi 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 \cdot \frac{1}{12} = \frac{\pi}{6}.

Problem 4 · Shells Under a Hyperbola

Given: the region under $y = \tfrac{1}{x}$ and above the $x$ -axis, for $1 \le x \le 2$ , is revolved about the $y$ -axis — find the volume.

✅ Correct! The radius

x

cancels the height

\tfrac1x

V = \int_1^2 2\pi x \cdot \tfrac1x\,dx = \int_1^2 2\pi\,dx = 2\pi

❌ The radius factor $x$ is missing.

\int_1^2 2\pi \cdot \tfrac1x\,dx = 2\pi\ln 2

drops the

x

in the circumference

2\pi x

❌ That is the disk method about the $x$ -axis.

\int_1^2 \pi \cdot \tfrac{1}{x^2}\,dx = \tfrac{\pi}{2}

revolves the region about the wrong axis.

❌ That is half a turn. The region lies on one side of the axis and turns a full circle, so the length around is

2\pi x

❌ Not quite. Multiply the circumference

2\pi x

, the height

\tfrac1x

and

dx

, then integrate from

1

2

Show solution

The slice at $x$ runs from $y = 0$ to $y = \tfrac1x$ ; revolved about the $y$ -axis it is a shell of radius $x$ :

dV = (2\pi x)\left(\frac1x\right)dx = 2\pi\,dx.

V = \int_1^2 2\pi\,dx = 2\pi(2 - 1) = 2\pi.

Every shell has the same volume per unit thickness: the taller shells near $x = 1$ are exactly as much narrower as they are taller.

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