Unit 8 · Topic 8.10 Beta

Disc Method About Other Lines

To revolve a region about a line y = k or x = h that it touches all the way across, use discs whose radius is the distance from that line to the curve: V = π∫ from a to b of (k − f(x))² dx (or the mirror image).

Practice 1: Implementing Mathematical Processes

Question set for this topic

Part 1 · Hook

Why this matters

A lathe can spin a piece of wood around any center line the machinist chooses, not just the edge of the table. Revolving a region about a line like

y = 1

x = 3

works the same way as revolving it about an axis. The only change is where you measure the radius from.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. The region under $y = \sqrt{x}$ , $0 \le x \le 4$ , is revolved about the $x$ -axis. Which integral gives the volume?

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

Show the answer

Radius $\sqrt{x}$ , so each disc has area $\pi(\sqrt{x})^2 = \pi x$ , from $x = 0$ to $4$ .

Correct: $\pi\int_0^4 x\,dx$ :
$\pi\int_0^4 \sqrt{x}\,dx$ :
$\int_0^4 x\,dx$ :
$\pi\int_0^2 x\,dx$ :

2. How far is the point $(2, -1)$ from the line $y = 3$ ?

$4$
$2$
$-4$
$1$

Show the answer

A vertical distance is the larger $y$ minus the smaller: $3 - (-1) = 4$ .

Correct: $4$ :
$2$ :
$-4$ :
$1$ :

Part 4 · See it

See it first

Left: the region between the parabola y = x² and the line y = 1, from x = −1 to 1, shaded, with y = 1 dashed as the axis of rotation and one thin vertical rectangle from the parabola up to the axis, of length 1 − x². Right: the solid made by revolving the region about y = 1, shaped like a lemon, with the axis through its center and one disc highlighted in blue whose radius is r = 1 − x². Below: V = the integral from −1 to 1 of π(1 − x²)² dx = 16π/15. — Revolving the region between $y = x^2$ and $y = 1$ about the line $y = 1$ . The radius runs from the axis down to the curve: $r = 1 - x^2$ . LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

The axis of rotation is the line $y = k$ (or $x = h$ )every slice is a disc centered on that line
The region touches that line all the way acrosseach rectangle reaches from the line to the curve and sweeps out a solid disc
The radius is the distance from the line to the curve $r = k - f(x)$ if the line is above the curve, $f(x) - k$ if it is below
Square the whole radius and multiply by $\pi$ $V$ $= \pi\int_a^b \left(f(x) - k\right)^2\,dx$

Part 6 · Key ideas

Key ideas

Revolving about a line: the disc method still applies; only the radius changes.
The radius is a distance from the axis: larger value minus smaller value. With the line $y = k$ above the curve, $r = k - f(x)$ ; below it, $r = f(x) - k$ .
With a negative $k$ , subtract carefully: about $y = -2$ , $r$ $= f(x) - (-2)$ $= f(x) + 2$ .
A vertical line $x = h$ gives horizontal discs: $r = h - x(y)$ or $x(y) - h$ , integrated in $dy$ .
Square $(k - f(x))$ as a whole; $k^2 - f(x)^2$ is a different number.

Part 7 · Misconception

A common mistake

The wrong idea: The radius is always the curve's height $f(x)$ , whatever the axis.

What actually happens: $f(x)$ is the distance from the $x$ -axis. When the axis is $y = k$ , measure from that line: $r = |f(x) - k|$ . For the region between $y = x^2$ and $y = 1$ about $y = 1$ , the radius is $1 - x^2$ , not $x^2$ .

Part 8 · Check yourself

Check yourself

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

Graph

The region $R$ bounded by $y = \sqrt{x}$ , $y = 2$ and the $y$ -axis

1. $R$ is revolved about the line $y = 2$ . Write the radius $r(x)$ of the disc at $x$ , for $0 \le x \le 4$ .

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

Show the answer

The axis $y = 2$ is above the curve, so the radius is (axis) minus (curve): $r(x) = 2 - \sqrt{x}$ .

Answer: $r(x) =$ $2 - \sqrt{x}$

2. Which integral gives the volume of the solid formed when $R$ is revolved about the line $y = 2$ ?

$\pi\int_0^4 \left(2 - \sqrt{x}\right)^2\,dx$
$\pi\int_0^4 \left(2^2 - (\sqrt{x})^2\right)\,dx$
$\pi\int_0^4 \left(\sqrt{x}\right)^2\,dx$
$\pi\int_0^2 \left(2 - \sqrt{x}\right)^2\,dx$

Show the answer

$R$ touches the line $y = 2$ all the way across, so the slices are discs of radius $2 - \sqrt{x}$ , for $0 \le x \le 4$ .

Correct: $\pi\int_0^4 \left(2 - \sqrt{x}\right)^2\,dx$ : Square the distance from the axis to the curve.
$\pi\int_0^4 \left(2^2 - (\sqrt{x})^2\right)\,dx$ : This squares each height separately. It measures from the $x$ -axis, which is not the axis here.
$\pi\int_0^4 \left(\sqrt{x}\right)^2\,dx$ : That radius is measured from the $x$ -axis.
$\pi\int_0^2 \left(2 - \sqrt{x}\right)^2\,dx$ : The disc's thickness is $dx$ , so the limits are $x$ -values, 0 to 4.

3. Find the volume of the solid formed when $R$ is revolved about the line $y = 2$ .

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

Show the answer

$(2 - \sqrt{x})^2 = 4 - 4\sqrt{x} + x$ , so $V$ $= \pi\left[4x - \frac{8}{3}x^{3/2} + \frac{x^2}{2}\right]_0^4$ $= \pi\left(16 - \frac{64}{3} + 8\right)$ $= \frac{8\pi}{3}$ .

Answer: $\frac{8\,\pi}{3}$ ≈ 8.378

4. Instead, $R$ is revolved about the $y$ -axis. Find the volume.

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

Show the answer

About the $y$ -axis the discs are horizontal. At height $y$ the region reaches from $x = 0$ to $x = y^2$ , so $V$ $= \pi\int_0^2 (y^2)^2\,dy$ $= \pi \cdot \frac{32}{5}$ .

Answer: $\frac{32\,\pi}{5}$ ≈ 20.106

5. When $R$ is revolved about the line $y = 2$ , where is the largest disc?

At $x = 0$ , where the radius is 2
At $x = 4$ , where $\sqrt{x} = 2$
At $x = 1$ , where $\sqrt{x} = 1$
At $x = 2$ , halfway across $R$

Show the answer

The radius $2 - \sqrt{x}$ is largest where $\sqrt{x}$ is smallest, at $x = 0$ : $r = 2$ .

Correct: At $x = 0$ , where the radius is 2: $2 - \sqrt{0} = 2$ , the farthest the curve gets from the axis.
At $x = 4$ , where $\sqrt{x} = 2$ : At $x = 4$ the curve meets the axis, so the radius there is 0.
At $x = 1$ , where $\sqrt{x} = 1$ : There the radius is $2 - 1 = 1$ , smaller than at $x = 0$ .
At $x = 2$ , halfway across $R$ : There the radius is $2 - \sqrt{2} \approx 0.59$ .

6. The region bounded by $y = x^2$ and $y = 1$ is revolved about the line $y = 1$ . Find the volume.

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

Show the answer

Radius $1 - x^2$ for $-1 \le x \le 1$ : $V$ $= \pi\int_{-1}^{1} (1 - 2x^2 + x^4)\,dx$ $= \pi\left(2 - \frac{4}{3} + \frac{2}{5}\right)$ $= \frac{16\pi}{15}$ .

Answer: $\frac{16\,\pi}{15}$ ≈ 3.351

7. The region between $y = 2 - x^2$ and the line $y = -2$ is revolved about the line $y = -2$ . What is the radius of the disc at $x$ ?

$4 - x^2$
$-x^2$
$2 - x^2$
$x^2 - 4$

Show the answer

The curve is above the axis: radius $= (2 - x^2) - (-2)$ $= 4 - x^2$ .

Correct: $4 - x^2$ : Curve minus axis, with the minus sign on $-2$ handled.
$-x^2$ : That is $(2 - x^2) - 2$ : subtracting $-2$ adds 2.
$2 - x^2$ : That is measured from the $x$ -axis, not from $y = -2$ .
$x^2 - 4$ : That is axis minus curve, which is negative here.

Part 9 · Summary

Summary

To revolve a region about a line

y = k

x = h

that it touches all the way across, use discs whose radius is the distance from that line to the curve:

V

= \pi\int_a^b \left(k - f(x)\right)^2\,dx

(or the mirror image). Measure from the axis, subtract in the order that makes the radius positive, and square the whole 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=xy = \sqrt{x}, y=2y = 2 and the yy-axis

Summary

What comes next

Connections

The region $R$ bounded by $y = \sqrt{x}$ , $y = 2$ and the $y$ -axis