Build a solid from washers, integrals from easy to advanced lesson

Rotate the region between the line $y = x$ and the parabola $y = x^2$ , for $0 \le x \le 1$ , a full turn about the $x$ -axis. The result is a solid of revolution, the cone swept by the region under the line with the solid swept by the region under the parabola hollowed out of it, and no standard geometry formula gives its volume directly. This page slices the solid perpendicular to the axis, shows that every slice is a washer with an outer and an inner radius, adds the slices with an integral and finds the volume $\tfrac{2\pi}{15}$ .

Key terms

Solid of revolution

Definition Let $R$ be a region in a plane and $L$ a line in the same plane. Rotating $R$ a full turn about $L$ , the axis of rotation, sweeps out a solid of revolution. Each point of $R$ at distance $d$ from $L$ traces a circle of radius $d$ centered on $L$ , in the plane through that point perpendicular to $L$ , and a point on $L$ stays where it is. The volume formulas on this page assume that $R$ lies on one side of $L$ , possibly touching it, because the two sides of a region that crosses the axis sweep overlapping solids.

In plain words Spin a flat shape around a line in its own plane. Every point of the shape moves around a circle centered on the line, and together the circles fill the solid.

Example Rotating the triangle under $y = x$ on $[0, 1]$ about the $x$ -axis gives a cone of height $1$ whose base has radius $1$ , and the cone formula $\tfrac13 \pi r^2 h$ with $r = h = 1$ gives the volume $\tfrac{\pi}{3}$ . Rotating the region between $y = x^2$ and $y = x$ on the same interval gives that cone with the solid swept by the region under the parabola removed from inside it. The point $(\tfrac12, \tfrac14)$ of the region traces a circle of radius $\tfrac14$ in the plane $x = \tfrac12$ .

Outer and inner radius

Definition Let a region lie on one side of a horizontal axis, and suppose that for every $x$ in $[a, b]$ the vertical line through $x$ meets the region in one segment. When that segment rotates about the axis, its far end traces the outer circle and its near end traces the inner circle. The outer radius $R(x)$ and the inner radius $r(x)$ are the distances from the axis to those two ends. For the region $g(x) \le y \le f(x)$ with $0 \le g(x) \le f(x)$ , rotated about the $x$ -axis, $R(x) = f(x)$ and $r(x) = g(x)$ . For the axis $y = k$ with the region above it, $R(x) = f(x) - k$ and $r(x) = g(x) - k$ . With the region below it, $R(x) = k - g(x)$ and $r(x) = k - f(x)$ .

In plain words Both radii are measured from the axis, not from each other. The height of the strip is the gap $R - r$ between the two circles, and it equals the outer radius only when the strip reaches the axis.

Example For the region $x^2 \le y \le x$ on $[0, 1]$ rotated about the $x$ -axis, the line is the far end and the parabola the near end, because $x - x^2 = x(1 - x) \ge 0$ there. So

R(x) = x, \qquad r(x) = x^2.

At $x = \tfrac12$ the strip runs from $y = \tfrac14$ to $y = \tfrac12$ , so the outer radius is $\tfrac12$ , the inner radius is $\tfrac14$ and the gap is $\tfrac14$ . Rotating the same region about the line $y = -1$ instead gives $R(x) = x + 1$ and $r(x) = x^2 + 1$ , because every distance to the axis grows by $1$ , even though the region is unchanged.

Cross-sectional area

Definition The cross-section of a solid at position $x$ is the part of the solid that lies in the plane perpendicular to the $x$ -axis at $x$ , and $A(x)$ denotes its area. When a region as in the card above, with radii $R(x) \ge r(x) \ge 0$ , rotates about a horizontal axis, the cross-section at $x$ is a washer, the ring between two concentric circles, and

A(x) = \pi R(x)^2 - \pi r(x)^2.

When $r(x) = 0$ the washer is a full disk of area $\pi R(x)^2$ .

In plain words A washer is a large disk with a smaller disk removed from its center, so its area is the difference of two disk areas. Using the gap as a radius instead gives $\pi(R - r)^2$ , the area of a different disk.

Example For the region $x^2 \le y \le x$ rotated about the $x$ -axis,

\begin{aligned} A(x) &= \pi x^2 - \pi (x^2)^2 \\[0.4em] &= \pi(x^2 - x^4). \end{aligned}

At $x = \tfrac12$ this is $\pi(\tfrac14 - \tfrac{1}{16}) = \tfrac{3\pi}{16}$ . The tempting expression $\pi(R - r)^2 = \pi(\tfrac14)^2 = \tfrac{\pi}{16}$ is only a third of it. In general

R^2 - r^2 = (R - r)(R + r),

which equals $(R - r)^2$ only when $r = 0$ or $R = r$ .

Volume by slicing

Definition Let a solid lie between the planes $x = a$ and $x = b$ , and let its cross-section at each $x$ in $[a, b]$ have area $A(x)$ , where $A$ is continuous on $[a, b]$ . The volume of the solid is

V = \int_a^b A(x)\,dx.

The slab between the planes at $x$ and $x + \Delta x$ has volume approximately $A(x)\,\Delta x$ , and the integral is the limit of the sums of these slab volumes as the thickest slab shrinks to thickness $0$ .

In plain words Cut the solid into thin slabs perpendicular to an axis, like a sliced loaf. Each slab is nearly a flat piece with face area $A(x)$ and thickness $\Delta x$ , and the integral adds them. Because the slices are perpendicular to the $x$ -axis, the variable $x$ gives their positions and $\Delta x$ their thickness.

Example Slicing works for solids that are not solids of revolution. Take the region $x^2 \le y \le x$ on $[0, 1]$ as the base of a solid whose cross-sections perpendicular to the $x$ -axis are squares with one side in the base. The side of the square at $x$ is the gap $x - x^2$ , so

\begin{aligned} V &= \int_0^1 (x - x^2)^2\,dx \\[0.4em] &= \int_0^1 (x^2 - 2x^3 + x^4)\,dx \\[0.4em] &= \frac13 - \frac12 + \frac15 = \frac{1}{30}. \end{aligned}

Squaring the gap is right here, because the gap really is the side of each square.

Washer method

Definition Let $f$ and $g$ be continuous on $[a, b]$ with $0 \le g(x) \le f(x)$ for every $x$ in $[a, b]$ . Rotating the region of points with $a \le x \le b$ and $g(x) \le y \le f(x)$ about the $x$ -axis gives a solid whose cross-sections are washers with $R(x) = f(x)$ and $r(x) = g(x)$ , so slicing gives the volume

\begin{aligned} V &= \pi \int_a^b f(x)^2\,dx \\[0.4em] &\quad - \pi \int_a^b g(x)^2\,dx. \end{aligned}

For a horizontal axis $y = k$ with the region on one side, use the outer and inner radii $R(x)$ and $r(x)$ measured from that axis in place of $f(x)$ and $g(x)$ . For a vertical axis the slices are horizontal, the radii are functions of $y$ and the integral is in $y$ .

In plain words Slice perpendicular to the axis, read the outer and inner radius of each slice as distances from the axis, and integrate $\pi$ times outer squared minus inner squared. The geometry chooses both the variable of integration, which locates the slices, and the integrand, which is their area.

Example For the region $x^2 \le y \le x$ on $[0, 1]$ rotated about the $x$ -axis, the radii are $R(x) = x$ and $r(x) = x^2$ , so

\begin{aligned} V &= \pi\int_0^1 (x^2 - x^4)\,dx \\[0.4em] &= \pi\left[\frac{x^3}{3} - \frac{x^5}{5}\right]_0^1 \\[0.4em] &= \pi\left(\frac13 - \frac15\right) = \frac{2\pi}{15}. \end{aligned}

The volume is about $0.419$ cubic units. It also equals the volume $\pi\int_0^1 x^2\,dx = \tfrac{\pi}{3}$ of the cone minus the volume $\pi\int_0^1 x^4\,dx = \tfrac{\pi}{5}$ swept by the region under the parabola, the same subtraction done for the whole solid at once instead of slice by slice. As it should be, the answer is smaller than the volume of the cone.

Check yourself

Rotate the region under $y = \sqrt{x}$ for $0 \le x \le 4$ about the $x$ -axis. Find the volume.

Show the answer

The region touches the axis, so every slice is a disk with outer radius $\sqrt{x}$ and inner radius $0$ . Then

\begin{aligned} V &= \pi\int_0^4 (\sqrt{x})^2\,dx \\[0.4em] &= \pi\int_0^4 x\,dx = 8\pi. \end{aligned}

The solid fits inside a cylinder of radius $2$ and length $4$ , whose volume is $16\pi$ , and it fills exactly half of it.

Rotate the region enclosed by $y = 2x$ and $y = x^2$ about the $x$ -axis. Find the volume.

Show the answer

The curves meet at $x = 0$ and $x = 2$ , and $2x - x^2 = x(2 - x) \ge 0$ on $[0, 2]$ , as in the lesson on the region between curves. Both curves are nonnegative there, so the outer radius is $2x$ , the inner radius is $x^2$ and

\begin{aligned} V &= \pi\int_0^2 (4x^2 - x^4)\,dx \\[0.4em] &= \pi\left(\frac{32}{3} - \frac{32}{5}\right) = \frac{64\pi}{15}. \end{aligned}

Rotate the region $x^2 \le y \le x$ , $0 \le x \le 1$ , about the line $y = 1$ . A classmate takes the upper curve as the outer boundary, with outer radius $1 - x$ and inner radius $1 - x^2$ . What goes wrong, and what is the volume?

Show the answer

The region lies below the axis, since $y \le x \le 1$ , so a point at height $y$ is at distance $1 - y$ from the axis. The far end of the strip at $x$ is therefore the parabola, at distance $1 - x^2$ , and the near end is the line, at distance $1 - x$ . The classmate has swapped the radii, and that integral comes out as $-\tfrac{\pi}{5}$ , which cannot be a volume. With the correct radii,

\begin{aligned} V &= \pi\int_0^1 (1 - x^2)^2\,dx \\[0.4em] &\quad - \pi\int_0^1 (1 - x)^2\,dx \\[0.4em] &= \pi\left(\frac{8}{15} - \frac13\right) = \frac{\pi}{5}. \end{aligned}

At $x = \tfrac12$ , for example, the outer radius is $\tfrac34$ and the inner radius is $\tfrac12$ .