Volumes of revolution: the disc method

Three quick questions from earlier lessons.

Spin the region under $y = x$ , $0 \le x \le 3$ , about the $x$ -axis: a cone. The slice at $x$ is a disc. What is its radius?

Revolving a region about an axis: slices are discs.

Disc method: revolving the region under

y = f(x) \ge 0

,

a \le x \le b

, about the

x

-axis gives

V = \pi\displaystyle\int_a^b f(x)^2\,dx

.

Revolve the region under $y = x$ , $0 \le x \le 3$ .

V = \pi\displaystyle\int_0^3 x^2\,dx

Radius x.

= 9\pi

x³/3.

Matches

\tfrac{1}{3}\pi r^2 h

Check: a cone of radius 3 and height 3 has (1/3)π·9·3.

Your turn: the region under $y = \sqrt{x}$ , $0 \le x \le 4$ .

V = 8\pi

π∫₀⁴ x dx.

The region under $y = \sqrt{x}$ , $0 \le x \le 1$ , is revolved about the $x$ -axis. Find the volume.
The region under $y = \sqrt{x}$ , $0 \le x \le 3$ , is revolved about the $x$ -axis. Find the volume.

The region under $y = x$ , $0 \le x \le 4$ , is revolved about the $x$ -axis (a cone). Find the volume.
The region under $y = x$ , $0 \le x \le 7$ , is revolved about the $x$ -axis (a cone). Find the volume.

The region under $y = e^x$ , $0 \le x \le 1$ , is revolved about the $x$ -axis. Find the volume.
The region under $y = e^x$ , $0 \le x \le 3$ , is revolved about the $x$ -axis. Find the volume.

Three questions to check you have got it.

The region under $y = \sqrt{x}$ , $0 \le x \le 1$ , is revolved about the $x$ -axis. Find the volume.
The region under $y = \sqrt{x}$ , $0 \le x \le 3$ , is revolved about the $x$ -axis. Find the volume.
The region under $y = \sqrt{x}$ , $0 \le x \le 4$ , is revolved about the $x$ -axis. Find the volume.

Disc method

$V = \pi\displaystyle\int_a^b (\text{radius})^2\,dx$ .

Watch out: square the radius; keep the $\pi$ .

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