Unit 8 · Topic 8.9 Beta

Disc Method About the Axes

A region that touches the axis of rotation along its whole length makes a solid whose slices are discs.

Practice 1: Implementing Mathematical Processes

Part 1 · Hook

Why this matters

A potter spins clay on a wheel and shapes its outline with one hand. Every horizontal slice of the finished pot is a circle. Spin a flat region about an axis and you get the same kind of solid: every slice perpendicular to the axis is a disc, and you already know a disc's area.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. A solid has cross sections perpendicular to the $x$ -axis with area $A(x)$ for $0 \le x \le 2$ . Its volume is

$\int_0^2 A(x)\,dx$
$A(2) - A(0)$
$\pi\int_0^2 A(x)\,dx$
$\int_0^2 A(x)^2\,dx$

Show the answer

Volume is the integral of the cross-sectional area along the axis.

Correct: $\int_0^2 A(x)\,dx$ :
$A(2) - A(0)$ :
$\pi\int_0^2 A(x)\,dx$ :
$\int_0^2 A(x)^2\,dx$ :

2. Solve $y = x^2$ , $x \ge 0$ , for $x$ .

$x = \sqrt{y}$
$x = y^2$
$x = \frac{y}{2}$
$x = -\sqrt{y}$

Show the answer

Take the square root of both sides, keeping the nonnegative root.

Correct: $x = \sqrt{y}$ :
$x = y^2$ :
$x = \frac{y}{2}$ :
$x = -\sqrt{y}$ :

Part 4 · See it

See it first

Left: the region under y = √x from x = 0 to 4, shaded, with one thin vertical rectangle whose height is √x. Right: the solid made by revolving the region about the x-axis, a rounded horn drawn in 3D with the x-axis through its center. The rectangle sweeps out one thin disc, highlighted in blue, with radius r = √x. Below: V = the integral from 0 to 4 of π(√x)² dx = 8π. — Revolving the region under $y = \sqrt{x}$ about the $x$ -axis. Each vertical rectangle sweeps out a disc of radius $r = \sqrt{x}$ . LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Revolve a region that touches the axis all along its lengthevery slice perpendicular to the axis is a solid disc
The rectangle at $x$ reaches from the axis to the curveit sweeps out a disc whose radius is that distance, $r(x)$
A disc of radius $r$ and thickness $\Delta x$ has volume $\pi r^2\,\Delta x$
Add the discs and let them get thin $V$ $= \pi\int_a^b r(x)^2\,dx$
About the $y$ -axis the discs are horizontalwrite the radius as $x$ in terms of $y$ and integrate in $dy$

Part 6 · Key ideas

Key ideas

Disc method: $V$ $= \pi\int_a^b r(x)^2\,dx$ , the cross-section method with $A(x) = \pi r^2$ .
The radius runs from the axis of rotation to the curve. About the $x$ -axis, $r = f(x)$ .
About the $y$ -axis, slices are horizontal: $r$ is the curve's $x$ written in $y$ , and the limits are $y$ -values.
Do not forget $\pi$ , and square the radius inside the integral.
Checks: a cylinder gives $\pi r^2 h$ , a cone $\frac{1}{3}\pi r^2 h$ , a ball $\frac{4}{3}\pi r^3$ .

Part 7 · Misconception

A common mistake

The wrong idea: The volume of a solid of revolution is $\int_a^b \pi r(x)\,dx$ , or $\int_a^b r(x)^2\,dx$ .

What actually happens: Each slice is a disc, and a disc's area is $\pi r^2$ : both the $\pi$ and the square are needed. Without the square you are adding up lengths; without $\pi$ you are stacking squares, not discs.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

The region $R$ bounded by $y = x^2$ , $y = 4$ and the $y$ -axis

1. $R$ is revolved about the $y$ -axis. Which integral gives the volume of the solid?

$\pi \int_0^4 y\,dy$
$\pi \int_0^2 x^4\,dx$
$\pi \int_0^4 \sqrt{y}\,dy$
$\pi \int_0^4 y^2\,dy$

Show the answer

Revolving about the $y$ -axis makes horizontal discs. At height $y$ the radius reaches from the axis to the curve, $x = \sqrt{y}$ , so the disc's area is $\pi(\sqrt{y})^2 = \pi y$ .

Correct: $\pi \int_0^4 y\,dy$ : $\pi r^2$ with $r = \sqrt{y}$ , stacked from $y = 0$ to $y = 4$ .
$\pi \int_0^2 x^4\,dx$ : That revolves the region under $y = x^2$ about the $x$ -axis: a different region and a different axis.
$\pi \int_0^4 \sqrt{y}\,dy$ : The radius $\sqrt{y}$ was not squared. A disc's area is $\pi r^2$ .
$\pi \int_0^4 y^2\,dy$ : That uses $y$ as the radius. The radius is the horizontal distance $x = \sqrt{y}$ .

2. Find the volume of the solid formed when $R$ is revolved about the $y$ -axis.

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$V$ $= \pi\int_0^4 (\sqrt{y})^2\,dy$ $= \pi\int_0^4 y\,dy$ $= \pi \cdot \frac{y^2}{2}\Big|_0^4$ $= 8\pi$ .

Answer: $8\,\pi$ ≈ 25.133

3. When $R$ is revolved about the $y$ -axis, write the radius $r(y)$ of the disc at height $y$ .

Type an expression for $r(y) =$ .

Show the answer

The radius is the distance from the axis ( $x = 0$ ) to the curve at height $y$ : $x = \sqrt{y}$ .

Answer: $r(y) =$ $\sqrt{y}$

4. The solid from revolving $R$ about the $y$ -axis is cut by the plane $y = 3$ . Find the area of that cross section.

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

At $y = 3$ the cross section is a disc of radius $\sqrt{3}$ , with area $\pi(\sqrt{3})^2 = 3\pi$ .

Answer: $3\,\pi$ ≈ 9.425

5. When $R$ is revolved about the $y$ -axis, each thin slice of the solid is

a horizontal disc with radius $\sqrt{y}$ and thickness $\Delta y$
a vertical disc with radius $x^2$ and thickness $\Delta x$
a horizontal disc with radius $y$ and thickness $\Delta y$
a vertical disc with radius $4 - x^2$ and thickness $\Delta x$

Show the answer

Slices are perpendicular to the axis of rotation. For the $y$ -axis they are horizontal, stacked in $y$ , with radius from the axis to the curve, $\sqrt{y}$ .

Correct: a horizontal disc with radius $\sqrt{y}$ and thickness $\Delta y$ : Perpendicular to the $y$ -axis, radius $\sqrt{y}$ .
a vertical disc with radius $x^2$ and thickness $\Delta x$ : Vertical slices belong to a horizontal axis. About the $y$ -axis the slices are horizontal.
a horizontal disc with radius $y$ and thickness $\Delta y$ : $y$ is the height of the slice, not its radius.
a vertical disc with radius $4 - x^2$ and thickness $\Delta x$ : Vertical slices belong to a horizontal axis, and $4 - x^2$ is a height, not a distance from the $y$ -axis.

6. The region under $y = e^x$ for $0 \le x \le 1$ is revolved about the $x$ -axis. Find the volume.

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$V$ $= \pi\int_0^1 e^{2x}\,dx$ $= \pi \cdot \frac{e^{2x}}{2}\Big|_0^1$ $= \frac{\pi(e^2 - 1)}{2}$ .

Answer: $\frac{\pi\,(e^{2} - 1)}{2}$ ≈ 10.036

7. The triangle under $y = \frac{x}{2}$ for $0 \le x \le 6$ is revolved about the $x$ -axis. The solid is a cone with radius 3 and height 6. What volume does the disc method give?

$18\pi$
$9\pi$
$54\pi$
$36\pi$

Show the answer

$V$ $= \pi\int_0^6 \frac{x^2}{4}\,dx$ $= \pi \cdot \frac{216}{12}$ $= 18\pi$ , which matches the cone formula $\frac{1}{3}\pi(3)^2(6) = 18\pi$ .

Correct: $18\pi$ : It agrees with $\frac{1}{3}\pi r^2 h$ .
$9\pi$ : $\pi\int_0^6 \frac{x}{2}\,dx$ $= 9\pi$ : the radius was not squared.
$54\pi$ : $54\pi$ is the cylinder $\pi(3)^2(6)$ , three times the cone: the antiderivative of $x^2$ needs the division by 3.
$36\pi$ : $36\pi$ comes from writing $\left(\frac{x}{2}\right)^2$ as $\frac{x^2}{2}$ : square the 2 as well, $\frac{x^2}{4}$ .

Part 9 · Summary

Summary

A region that touches the axis of rotation along its whole length makes a solid whose slices are discs. With radius

r

measured from the axis to the curve,

V

= \pi\int_a^b r(x)^2\,dx

about the

x

-axis, or

\pi\int_c^d r(y)^2\,dy

about the

y

-axis. Keep the

\pi

and square the radius.

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

The region RR bounded by y=x2y = x^2, y=4y = 4 and the yy-axis

Summary

What comes next

Connections

The region $R$ bounded by $y = x^2$ , $y = 4$ and the $y$ -axis