Math Core

Spin a flat region around a line and it sweeps out a solid, the way a potter's wheel turns a profile into a vase. These solids of revolution have circular cross sections, so they are a special case of the last lesson: you already know how to integrate cross-sectional area, and now every cross section is a disc.

Why the slices are discs

Take the region under $y = f(x) \ge 0$ from $x = a$ to $x = b$ and revolve it around the $x$ -axis. A vertical slice at position $x$ has one end on the axis and the other on the curve. As it spins, it traces out a circle of radius $r = f(x)$ . So the cross section perpendicular to the axis is a disc with area

A(x) = \pi r^2 = \pi\big[f(x)\big]^2.

Revolving the shaded region under y = √x around the x-axis. The slice at x spins into a disc of radius √x; the dashed curve is the mirror image the solid reaches below the axis.Open in grapher →

The disc method

When a region with one boundary on the axis of rotation is revolved around that axis,

V = \pi\int_a^b \big[r(x)\big]^2\,dx \quad \text{(horizontal axis)} \qquad V = \pi\int_c^d \big[r(y)\big]^2\,dy \quad \text{(vertical axis)},

where the radius $r$ is the distance from the axis to the far boundary of the slice.

Choosing dx or dy

The slice must be perpendicular to the axis of rotation, so the variable of integration matches the axis:

Revolving around a horizontal line (the $x$ -axis, $y = 1$ , …): slices are vertical, integrate with respect to $x$ .
Revolving around a vertical line (the $y$ -axis, $x = 2$ , …): slices are horizontal, integrate with respect to $y$ , and write the curves as $x$ in terms of $y$ .

Axes other than the coordinate axes

The radius is always a distance from the axis, so it is always "far minus near":

Around a horizontal line $y = k$ : $r = |f(x) - k|$ , usually written as (larger $y$ ) $-$ (smaller $y$ ).
Around a vertical line $x = h$ : $r = |g(y) - h|$ , written as (larger $x$ ) $-$ (smaller $x$ ).

For example, if the curve $y = x^2$ is below the axis $y = 1$ , the radius is $1 - x^2$ , not $x^2$ and not $x^2 - 1$ .

Common mistake

Don't forget to square the radius, and don't square the pieces separately. $\big(1 - x^2\big)^2 = 1 - 2x^2 + x^4$ , which is not $1 - x^4$ . Also keep the $\pi$ : dropping it is a lost point on the AP exam even when the integral is right.

Worked examples

Worked example: Around the x-axis

Find the volume when the region under $y = \sqrt{x}$ for $0 \le x \le 4$ is revolved around the $x$ -axis.

Solution. The radius is $r = \sqrt{x}$ , so

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

Worked example: Around the y-axis

The region bounded by $y = x^3$ , the line $y = 8$ and the $y$ -axis is revolved around the $y$ -axis. Find the volume.

Solution. The axis is vertical, so slice horizontally and use $y$ . The curve is $x = y^{1/3}$ , and a horizontal slice runs from the $y$ -axis to the curve, so $r = y^{1/3}$ for $0 \le y \le 8$ .

V = \pi\int_0^8 \left(y^{1/3}\right)^2 dy = \pi\int_0^8 y^{2/3}\,dy = \pi\left[\frac{3}{5}y^{5/3}\right]_0^8 = \pi\cdot\frac{3}{5}\cdot 32 = \frac{96\pi}{5}.

Worked example: Around the line y = 1

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

Solution. The curves meet at $x = \pm 1$ . The axis $y = 1$ is a boundary of the region, so each vertical slice runs from the axis down to the parabola. The radius is $r = 1 - x^2$ .

\begin{aligned} V &= \pi\int_{-1}^{1} (1 - x^2)^2\,dx = \pi\int_{-1}^{1} (1 - 2x^2 + x^4)\,dx \\ &= \pi\left[x - \frac{2x^3}{3} + \frac{x^5}{5}\right]_{-1}^{1} = \pi\cdot 2\left(1 - \frac{2}{3} + \frac{1}{5}\right) = \frac{16\pi}{15}. \end{aligned}

Tip

Before integrating, check the radius at one test point against your sketch. In the last example, at $x = 0$ the region reaches from $y = 0$ to the axis $y = 1$ , so the radius should be 1, and indeed $1 - 0^2 = 1$ .

Practice

Practice 1

The region under $y = x$ for $0 \le x \le 3$ is revolved around the $x$ -axis. Find the exact volume. (It is a cone.)