Build a solid from cylindrical shells, integrals from easy to advanced lesson

Rotate the region under $y = e^{-x^2}$ and above the $x$ -axis, for $0 \le x \le 1$ , a full turn about the $y$ -axis. Slices perpendicular to that axis would need $x$ as a function of $y$ and two different formulas, while a vertical strip keeps the simple height $e^{-x^2}$ . When it turns, such a strip sweeps out a thin cylindrical shell. This page finds the volume of a shell, builds the volume integral $2\pi\int_0^1 x e^{-x^2}\,dx$ from the radius, height and thickness of the shells, and evaluates it to $\pi(1 - e^{-1})$ .

Key terms

Cylindrical shell

Definition A cylindrical shell is the solid between two circular cylinders with a common axis and a common height $h$ , with inner radius $r_1$ and outer radius $r_2$ , where $0 \le r_1 < r_2$ . Its volume is the difference of two cylinder volumes,

\begin{aligned} V &= \pi r_2^2 h - \pi r_1^2 h \\[0.4em] &= 2\pi\,\bar r\,h\,\Delta r, \end{aligned}

where $\bar r = \tfrac12(r_1 + r_2)$ is the average radius and $\Delta r = r_2 - r_1$ is the thickness. The estimate $2\pi r_1 h\,\Delta r$ , which uses the inner circumference, is too small by exactly $\pi h (\Delta r)^2$ .

In plain words Cut the wall of a thin tube along a seam parallel to its axis and unroll it. The result is nearly a flat slab whose length is the circumference, whose height is $h$ and whose thickness is $\Delta r$ . The factor $2\pi r$ is a circumference, not the area of a disk. The inner and outer circumferences differ by $2\pi\,\Delta r$ , which is small for a thin shell, and measuring the circumference at the average radius makes the slab picture exact.

Example A shell with inner radius $1$ , outer radius $\tfrac65$ and height $3$ has volume

\begin{aligned} V &= \pi\left(\tfrac{36}{25} - 1\right) \cdot 3 \\[0.4em] &= \frac{33\pi}{25}. \end{aligned}

The average radius $\tfrac{11}{10}$ and the thickness $\tfrac15$ give $2\pi \cdot \tfrac{11}{10} \cdot 3 \cdot \tfrac15 = \tfrac{33\pi}{25}$ as well. The estimate with the inner circumference gives $\tfrac{6\pi}{5} = \tfrac{30\pi}{25}$ , which is too small by $\pi \cdot 3 \cdot \tfrac{1}{25} = \tfrac{3\pi}{25}$ .

Strip parallel to the axis

Definition In a region rotated about an axis, a thin strip parallel to the axis stays between two nearby distances from it, so it sweeps out approximately a thin cylindrical shell. A thin strip perpendicular to the axis sweeps out approximately a washer or a disk. For rotation about the $y$ -axis the parallel strips are vertical, their positions are values of $x$ and their thickness is $\Delta x$ . The shell method uses parallel strips and integrates in the direction perpendicular to the axis.

In plain words Pick the direction of the strips first, and the method follows. Strips across the axis give washers, and strips along the axis give shells. Choose the direction in which the length of the strip has one simple formula across the whole region.

Example For the region $0 \le y \le e^{-x^2}$ , $0 \le x \le 1$ , rotated about the $y$ -axis, the vertical strip at $x$ has length $e^{-x^2}$ for every $x$ in $[0, 1]$ . Horizontal slices would need two formulas. Since $e^{-x^2}$ decreases from $1$ to $e^{-1}$ , a slice at a height $y$ with $0 \le y \le e^{-1}$ runs across the whole interval from $x = 0$ to $x = 1$ , while a slice at a height between $e^{-1}$ and $1$ stops at $x = \sqrt{-\ln y}$ . Rotated, these slices are disks, and

\begin{aligned} V &= \pi \int_0^{1/e} 1\,dy \\[0.4em] &\quad + \pi \int_{1/e}^1 (-\ln y)\,dy, \end{aligned}

which needs the inverse function and an integration by parts. The vertical strips avoid both.

Shell radius and height

Definition Let a region $g(x) \le y \le f(x)$ , $a \le x \le b$ , be rotated about a vertical axis $x = k$ that does not pass through the interior of the region. For the vertical strip at $x$ , the shell radius is the distance $\lvert x - k \rvert$ from the strip to the axis, and the shell height is the length of the strip, top minus bottom, $h(x) = f(x) - g(x)$ . The thickness is the width $\Delta x$ of the strip, and the shell has volume approximately

2\pi \lvert x - k \rvert\,h(x)\,\Delta x.

For the $y$ -axis, $k = 0$ , and the radius is $x$ when $x \ge 0$ .

In plain words Three separate measurements go into each shell. The radius says how far the strip is from the axis, the height says how long the strip is, and the thickness says how wide it is. The radius comes from the position of the strip, and the height comes from the curves.

Example For the region under $y = e^{-x^2}$ on $[0, 1]$ rotated about the $y$ -axis, the strip at $x$ has radius $x$ , height $e^{-x^2}$ and thickness $\Delta x$ , so its shell has volume approximately $2\pi x e^{-x^2}\,\Delta x$ . At $x = \tfrac12$ the radius is $\tfrac12$ and the height is $e^{-1/4} \approx 0.779$ . Rotating the same region about the line $x = -1$ would keep the height $e^{-x^2}$ but change the radius to $x + 1$ .

Rotation about the y-axis

Definition Rotating a region of the half-plane $x \ge 0$ a full turn about the $y$ -axis moves each point $(x, y)$ of the region around a circle of radius $x$ that is centered on the axis at height $y$ and lies in the plane perpendicular to the axis at that height. The shell formula with radius $x$ assumes the region lies in $x \ge 0$ . If a region crosses the axis, its parts on the two sides sweep overlapping solids, and the solid of revolution is the one swept by the right part together with the mirror image of the left part.

In plain words The $y$ -axis is vertical, so strips parallel to it are vertical and the shell radius is simply the $x$ -coordinate of the strip. Changing the axis changes the solid, even when the region stays the same.

Example The region under $y = e^{-x^2}$ on $[0, 1]$ gives two different solids. About the $y$ -axis it gives the solid of this page. About the $x$ -axis its slices are disks of radius $e^{-x^2}$ , and the volume

\pi\int_0^1 e^{-2x^2}\,dx \approx 1.879

has an integrand with no elementary antiderivative, so it is found numerically. The shell integral about the $y$ -axis carries an extra factor $x$ , which lets a substitution finish it. The region under the same curve on $[-1, 1]$ sweeps exactly the same solid about the $y$ -axis as the region on $[0, 1]$ , because its left half is the mirror image of its right half.

Shell method

Definition Let $f$ and $g$ be continuous on $[a, b]$ , where $0 \le a < b$ , with $g(x) \le f(x)$ for every $x$ in $[a, b]$ , and write $h(x) = f(x) - g(x)$ for the shell height. Rotating the region of points with $a \le x \le b$ and $g(x) \le y \le f(x)$ about the $y$ -axis gives a solid of volume

V = 2\pi \int_a^b x\,h(x)\,dx.

It is the limit of the sums of shell volumes $2\pi \bar x_i\,h(\bar x_i)\,\Delta x_i$ , with $\bar x_i$ the midpoint of the $i$ th strip, as the widest strip shrinks to width $0$ . For a vertical axis $x = k$ with the region on one side, the radius $x$ becomes $\lvert x - k \rvert$ . For a horizontal axis the strips are horizontal and the integral is in $y$ .

In plain words Multiply circumference by height by thickness for each strip, and let the integral add the shells from the innermost to the outermost. The method needs the region on one side of the axis, and it is the natural choice when the strip lengths have one formula in the variable that measures distance from the axis.

Example For the region $0 \le y \le e^{-x^2}$ , $0 \le x \le 1$ , rotated about the $y$ -axis, the radius is $x$ and the height is $e^{-x^2}$ , so

V = 2\pi \int_0^1 x e^{-x^2}\,dx.

The substitution $u = x^2$ is continuously differentiable, has $du = 2x\,dx$ and sends the bounds $0$ and $1$ to $0$ and $1$ , so

\begin{aligned} V &= \pi \int_0^1 e^{-u}\,du \\[0.4em] &= \pi\bigl[-e^{-u}\bigr]_0^1 \\[0.4em] &= \pi(1 - e^{-1}) \approx 1.986. \end{aligned}

The volume is in cubic units. Two cylinders of radius $1$ bound it. Since $e^{-1} \le e^{-x^2} \le 1$ on $[0, 1]$ , the solid contains the cylinder of height $e^{-1}$ and fits inside the cylinder of height $1$ , so $\pi e^{-1} \le V \le \pi$ , roughly $1.156 \le V \le 3.142$ . The disk integral from the card on parallel strips agrees, since $\int_{1/e}^1 (-\ln y)\,dy = 1 - 2e^{-1}$ and $\pi e^{-1} + \pi(1 - 2e^{-1}) = \pi(1 - e^{-1})$ .

Check yourself

Rotate the region under $y = 1 - x$ for $0 \le x \le 1$ about the $y$ -axis. Find the volume with shells, and compare it with the volume of a cone.

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The strip at $x$ has radius $x$ and height $1 - x$ , so

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

The region is the triangle with vertices $(0, 0)$ , $(1, 0)$ and $(0, 1)$ , so the solid is a cone with base radius $1$ and height $1$ , and the cone formula $\tfrac13 \pi r^2 h$ gives the same $\tfrac{\pi}{3}$ .

Rotate the region under $y = \sin(x^2)$ for $0 \le x \le \sqrt{\pi}$ about the $y$ -axis. First check that the shell height is nonnegative, then find the volume.

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For $0 \le x \le \sqrt{\pi}$ the inner value $x^2$ lies in $[0, \pi]$ , where the sine is nonnegative, so the height $\sin(x^2)$ is nonnegative. With radius $x$ ,

V = 2\pi\int_0^{\sqrt{\pi}} x \sin(x^2)\,dx.

With $u = x^2$ , $du = 2x\,dx$ and the bounds $0$ and $\pi$ ,

\begin{aligned} V &= \pi\int_0^{\pi} \sin u\,du \\[0.4em] &= \pi\bigl[-\cos u\bigr]_0^{\pi} = 2\pi. \end{aligned}

The solid fits inside a cylinder of radius $\sqrt{\pi}$ and height $1$ , whose volume is $\pi^2 \approx 9.87$ , and $2\pi \approx 6.28$ is smaller, as it must be.

Rotate the region under $y = 4 - x^2$ for $-2 \le x \le 2$ about the $y$ -axis. A classmate writes

V = 2\pi\int_{-2}^{2} x(4 - x^2)\,dx = 0.

What went wrong, and what is the volume?

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The region crosses the axis. On the left half the factor $x$ is negative, so it is not a radius there, and the integrand is odd, so the two halves cancel. The left half is the mirror image of the right half and sweeps the same solid, so only $[0, 2]$ counts, and

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

Replacing $x$ by $\lvert x \rvert$ over $[-2, 2]$ would give $16\pi$ , which counts the same solid twice. Washers agree with $8\pi$ , since the horizontal slices are disks of squared radius $4 - y$ and $\pi\int_0^4 (4 - y)\,dy = 8\pi$ .