The Disc Method

Revolving a region about a line sweeps out a solid of revolution. When the line lies along an edge of the region, each slice perpendicular to the line sweeps out a disc. The volume is a sum of discs.

Revolving about the x-axis or the y-axis

Take the region under

y = f(x) \ge 0

[a, b]

and revolve it about the

x

-axis. A thin vertical strip at

x

sweeps out a disc of radius

f(x)

and thickness

\Delta x

. The cross section at

x

is a circle of area

\pi f(x)^2

. So this is a volume with known cross sections, as on the page on cross sections.

The slices are always perpendicular to the axis. So a horizontal axis means an integral in

x

, and a vertical axis means one in

y

Example 1

The region under

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

for

0 \le x \le 2

is revolved about the

x

-axis. Find the volume of the solid.

The curve is on or above the axis on

[0, 2]

and meets it at both ends. The radius at

x

R(x) = (2 - x)\sqrt{x}

, and squaring it removes the root.

The region, its mirror image across the x-axis, and the disc at x = 0.8 seen edge-on. The solid is a teardrop.

\begin{aligned} V &= \pi\int_{0}^{2} x(2 - x)^2\,dx \\[4pt] &= \pi\int_{0}^{2} \left(4x - 4x^2 + x^3\right)dx \\[4pt] &= \pi\left(2x^2 - \frac{4x^3}{3} + \frac{x^4}{4}\right)\Big|_{0}^{2} \\[4pt] &= \pi\left(8 - \frac{32}{3} + 4\right) = \frac{4\pi}{3}. \end{aligned}

Revolving about other lines

When the axis is the line

y = k

x = h

, the radius is still the distance from the axis to the curve. For a horizontal axis

y = k

, that's

|f(x) - k|

, the top minus the bottom.

Example 3

The region between the line

y = 4

and the curve

y = 4 - 3x^2 + x^3

, for

0 \le x \le 3

, is revolved about the line

y = 4

. Find the volume of the solid.

The curve meets the line where

x^2(3 - x) = 0

, at

x = 0

and

x = 3

, and it's below the line between them. The axis is the top of the region, so the radius is the line minus the curve:

\begin{aligned} R(x) &= 4 - \left(4 - 3x^2 + x^3\right) \\ &= x^2(3 - x). \end{aligned}

The region between y = 4 and the solid curve y = 4 − 3x² + x³, its mirror image above y = 4 (dashed), and the disc at x = 1.5.

\begin{aligned} V &= \pi\int_{0}^{3} x^4(3 - x)^2\,dx \\[4pt] &= \pi\int_{0}^{3} \left(9x^4 - 6x^5 + x^6\right)dx \\[4pt] &= \pi\left(\frac{9x^5}{5} - x^6 + \frac{x^7}{7}\right)\Big|_{0}^{3} \\[4pt] &= \pi\left(\frac{2187}{5} - 729 + \frac{2187}{7}\right) \\[4pt] &= \frac{729\pi}{35}. \end{aligned}

The volume is

\frac{729\pi}{35} \approx 65.435

Example 4

The region between the line

x = 1

and the curve

x = 1 + y^2(2 - y)^2

, for

0 \le y \le 2

, is revolved about the line

x = 1

. Find the volume of the solid.

The curve touches the line at

y = 0

and

y = 2

and lies to its right in between. The axis is vertical, so slice horizontally. The radius at height

y

is the curve's

x

minus 1.

R(y) = y^2(2 - y)^2

The region right of x = 1, its mirror image to the left, and the disc at y = 1.4.

Expand the square of the radius:

\begin{aligned} R(y)^2 &= y^4(2 - y)^4 \\ &= 16y^4 - 32y^5 + 24y^6 \\ &\qquad - 8y^7 + y^8. \end{aligned}

\begin{aligned} V &= \pi\int_{0}^{2} R(y)^2\,dy \\[4pt] &= \pi\Big(\frac{512}{5} - \frac{1024}{3} + \frac{3072}{7} \\ &\qquad - 256 + \frac{512}{9}\Big) \\[4pt] &= \frac{256\pi}{315} \approx 2.553. \end{aligned}