Single-Variable-Calculus · Unit 18 · Video 1 · Interactive Practice

The Disk Method: Volume of a Ball and a Spherical Cap

IKey Formulas

Formula	Meaning	What you need
$V = \displaystyle\int_a^b A(x)\,dx$	Volume by slicing: face area times thickness, added up	The area $A(x)$ of the slice at $x$ and where the slices start and end
$dV = \pi y^2\,dx$	Disk method: a vertical slice revolved about the $x$ -axis	$y^2$ from the curve and the limits from the picture
$\displaystyle\int_0^{2a} \pi(2ax - x^2)\,dx = \tfrac{4}{3}\pi a^3$	Volume of a ball of radius $a$	The circle $(x - a)^2 + y^2 = a^2$ , so $y^2 = 2ax - x^2$
$V(x) = \pi\left(a x^2 - \tfrac{x^3}{3}\right)$	Volume of the cap of width $x$ cut off the end of the ball	$0 \le x \le 2a$

Key Insight: The disk method needs $y^2$ , never $y$ , so the circle's equation gives the integrand with no square root. Setting up the integrand and the limits is the real work; the integration is routine.

IIVisualization 1 — A Ball as a Stack of Disks

Each slice of the ball of radius $a = 2$ is a disk of volume $\pi y^2\,\Delta x$ ; their sum approaches $\tfrac{4}{3}\pi a^3$ .

Number of disks n 4

IIIVisualization 2 — The Cap Volume V(x)

Stopping the slices at $x$ leaves a cap of volume $V(x) = \pi\left(ax^2 - \tfrac{x^3}{3}\right)$ , exactly half the ball at $x = a$ .

Cut at x 1

IVQuiz Questions

Problem 1 · Revolving a Square-Root Curve

Given: the region under $y = \sqrt{x}$ for $0 \le x \le 4$ is revolved about the $x$ -axis — find the volume of the solid.

✅ Correct! With

y^2 = x

V = \int_0^4 \pi x\,dx = \pi \cdot \tfrac{16}{2} = 8\pi

❌ That integrates $\pi y$ , not $\pi y^2$ . The face of each disk has area

\pi(\text{radius})^2

and the radius is

y = \sqrt{x}

, so the integrand is

\pi(\sqrt{x})^2 = \pi x

❌ The factor $\pi$ is missing.

\int_0^4 x\,dx = 8

is the area-like part; each disk's face is

\pi y^2

, so the volume is

8\pi

❌ That squares $x$ instead of $y$ .

\int_0^4 \pi x^2\,dx = \tfrac{64\pi}{3}

would be the solid under

y = x

. Here

y^2 = (\sqrt{x})^2 = x

❌ Not quite. Write

dV = \pi y^2\,dx

, substitute

y^2 = x

, and integrate from

0

4

Show solution

A vertical slice at $x$ revolves into a disk of radius $y = \sqrt{x}$ and thickness $dx$ :

dV = \pi y^2\,dx = \pi x\,dx.

The slices run from $x = 0$ to $x = 4$ :

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

Problem 2 · Limits From the Picture

Given: a ball of radius $3$ is formed by revolving the upper half of the circle $(x - 3)^2 + y^2 = 9$ about the $x$ -axis — which integral gives its volume?

✅ Correct! The circle crosses the axis at

x = 0

and

x = 6

, and

y^2 = 9 - (x - 3)^2 = 6x - x^2

. The integral equals

36\pi = \tfrac{4}{3}\pi \cdot 27

❌ Those limits belong to a circle centred at the origin. This circle is centred at

(3, 0)

, so it spans

0 \le x \le 6

. With the wrong limits the integral even comes out negative:

-18\pi

❌ That stops at the centre.

\int_0^3 \pi(6x - x^2)\,dx = 18\pi

is the half-ball; the slices continue to the right end of the circle at

x = 6

❌ That integrates $\pi y$ . A disk's face has area

\pi y^2

, and

y^2 = 6x - x^2

directly, with no square root.

❌ Not quite. Solve the circle for

y^2

, then read the leftmost and rightmost points off the picture.

Show solution

Integrand: $y^2 = 9 - (x - 3)^2 = 9 - (x^2 - 6x + 9) = 6x - x^2$ , so $dV = \pi(6x - x^2)\,dx$ .

Limits: the circle has centre $(3, 0)$ and radius $3$ , so it meets the $x$ -axis at $x = 0$ and $x = 6$ .

V = \int_0^6 \pi(6x - x^2)\,dx = \pi\left[3x^2 - \frac{x^3}{3}\right]_0^6 = \pi(108 - 72) = 36\pi,

which matches $\tfrac{4}{3}\pi(3)^3 = 36\pi$ .

Problem 3 · A Thin Cap

Given: a ball of radius $a = 3$ , set up as in the video with its left end at $x = 0$ — find the volume of the cap of width $1$ cut off its left end.

✅ Correct!

V(1) = \pi\left(3 \cdot 1^2 - \tfrac{1^3}{3}\right) = \pi\left(3 - \tfrac13\right) = \tfrac{8\pi}{3}

, about

7\%

of the ball's

36\pi

❌ The $\tfrac13$ is missing. An antiderivative of

x^2

\tfrac{x^3}{3}

, so the cap formula is

\pi\left(ax^2 - \tfrac{x^3}{3}\right)

, not

\pi(ax^2 - x^3)

❌ That is a whole ball of radius $1$ . The cap is a slice of the big ball, flat on one side; its volume comes from

V(x) = \pi\left(ax^2 - \tfrac{x^3}{3}\right)

with

a = 3

x = 1

❌ Check the sign. The integrand is

\pi(2ax - x^2)

, so its antiderivative is

\pi\left(ax^2 - \tfrac{x^3}{3}\right)

with a minus sign.

❌ Not quite. Evaluate

V(x) = \pi\left(ax^2 - \tfrac{x^3}{3}\right)

a = 3

x = 1

Show solution

With $y^2 = 2ax - x^2 = 6x - x^2$ , the slices from $0$ to $1$ give

V(1) = \int_0^1 \pi(6x - x^2)\,dx = \pi\left[3x^2 - \frac{x^3}{3}\right]_0^1 = \pi\left(3 - \frac13\right) = \frac{8\pi}{3}.

Check: $\tfrac{8\pi/3}{36\pi} = \tfrac{2}{27} \approx 0.074$ , a small fraction, as a thin cap should be.

Problem 4 · A Cone by Disks

Given: the triangle under $y = \tfrac{3}{4}x$ for $0 \le x \le 4$ is revolved about the $x$ -axis, giving a cone of radius $3$ and height $4$ — find its volume.

✅ Correct!

V = \int_0^4 \pi \cdot \tfrac{9}{16}x^2\,dx = \tfrac{9\pi}{16} \cdot \tfrac{64}{3} = 12\pi

, one third of the cylinder

\pi r^2 h = 36\pi

❌ That is the cylinder $\pi r^2 h$ . The disks shrink toward the tip: the radius at

x

\tfrac34 x

, not always

3

❌ That integrates $\pi y$ .

\int_0^4 \pi \cdot \tfrac34 x\,dx = 6\pi

; the disk's face is

\pi y^2 = \tfrac{9\pi}{16}x^2

❌ That treats the cone as half a cylinder. The face area grows like

x^2

, not like

x

, so the cone is one third of the cylinder, not one half.

❌ Not quite. Square the radius

\tfrac34 x

, multiply by

\pi

, and integrate from

0

4

Show solution

The slice at $x$ is a disk of radius $y = \tfrac34 x$ :

dV = \pi y^2\,dx = \frac{9\pi}{16}x^2\,dx.

V = \int_0^4 \frac{9\pi}{16}x^2\,dx = \frac{9\pi}{16}\left[\frac{x^3}{3}\right]_0^4 = \frac{9\pi}{16} \cdot \frac{64}{3} = 12\pi.

This is the cone formula $\tfrac13 \pi r^2 h = \tfrac13 \pi \cdot 9 \cdot 4 = 12\pi$ , derived by slicing.

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