2.2 Determining Volumes by Slicing

Big idea

Area came from slicing a region into thin strips and adding rectangles. Volume comes from slicing a solid into thin slabs and adding prisms. Nothing about the method changes; only the shape of one slice does. That is the whole content of this section, and it is worth stating plainly before any formula appears, because the formulas that follow are all the same formula wearing different clothes.

Cut a solid with planes perpendicular to a chosen axis. A slab between the cuts at $x$ and $x + dx$ is almost a prism: its two faces are nearly the same region, and its thickness is $dx$. A prism has volume equal to the area of its face times its thickness, so the slab contributes about $A(x)\,dx$, where $A(x)$ is the area of the cross-section at position $x$. Add over all slabs and let them thin out, and the sum becomes $\int A(x)\,dx$. The only work left in any particular problem is writing down $A(x)$.

When the solid is made by spinning a plane region about a line, that work is easy, because every cross-section perpendicular to the axis of rotation is a circle or a circle with a circular hole punched out. A circle of radius $R$ has area $\pi R^2$, and an annulus with outer radius $R$ and inner radius $r$ has area $\pi(R^2 - r^2)$. Those two facts, dropped into $\int A(x)\,dx$, are the disk method and the washer method. They are not separate techniques to memorize; they are the slicing formula with the two easiest cross-sections substituted in.

The practical skill is reading a picture and producing $A(x)$: choose the slicing direction, describe one cross-section, express its dimensions in the slicing variable, and find that variable’s range. The integration afterwards is usually the easiest part.

Decoder

If a solid lies between the planes at $x = a$ and $x = b$, and the cross-section cut at position $x$ by a plane perpendicular to that axis has area $A(x)$, then the volume of the solid is the integral of $A(x)$ from $a$ to $b$.

The load-bearing word is “perpendicular.” The cross-sections have to be taken perpendicular to the direction you are integrating in, because only then is the slab’s thickness equal to $dx$. Slice at an angle and the thin piece is a wedge, not a prism, and the product $A(x)\,dx$ no longer measures it.

The second thing to notice is that the statement asks for $A(x)$ as a function, not as a number. Every problem in this section is really the problem of writing the dimensions of one cross-section in terms of the slicing variable. If the cross-section is a square whose side is the vertical extent of some plane region, you need that extent as a formula in $x$ before you can square it.

The classic mistake in the rotation case is squaring after subtracting. The area of a washer is $\pi(R^2 - r^2)$, the difference of two circular areas, and that is not $\pi(R - r)^2$. Write the outer and inner radii separately, square each one, then subtract.

Definitions and results

The slicing formula. For a solid between $x = a$ and $x = b$ with continuous cross-sectional area $A(x)$ perpendicular to the $x$-axis,

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

Slicing perpendicular to the $y$-axis gives the same statement with $A(y)$ and $dy$.

Solids with a known cross-section. A common setup gives a plane region as the base and states the shape of every cross-section standing on it. If the base region has vertical extent $h(x)$ at position $x$, then a square cross-section has area $h(x)^2$, an equilateral triangle has area $\tfrac{\sqrt{3}}{4} h(x)^2$, and a semicircle on that segment as diameter has area $\tfrac{\pi}{8} h(x)^2$.

Solids of revolution. Spinning a plane region about a line in its plane sweeps out a solid of revolution. Cross-sections perpendicular to the axis of rotation are circles or annuli, since rotating a segment about a line produces one of those.

The disk method. If the region between the curve $y = f(x)$ and the $x$-axis on $[a, b]$ is rotated about the $x$-axis, the cross-section at $x$ is a disk of radius $|f(x)|$, so

$$ V = \pi \int_a^b \big[ f(x) \big]^2 \, dx $$

The washer method. If the region between $y = f(x)$ and $y = g(x)$, with $f(x) \ge g(x) \ge 0$, is rotated about the $x$-axis, each cross-section is an annulus with outer radius $f(x)$ and inner radius $g(x)$:

$$ V = \pi \int_a^b \Big( \big[ f(x) \big]^2 - \big[ g(x) \big]^2 \Big) \, dx $$

Rotating about the $y$-axis. Slice perpendicular to the axis of rotation, which now means horizontally. Solve the boundary curves for $x$ as functions of $y$ and integrate in $y$:

$$ V = \pi \int_c^d \Big( \big[ p(y) \big]^2 - \big[ q(y) \big]^2 \Big) \, dy $$

Rotating about another line. The radius is always the distance from the axis of rotation to the boundary curve. Rotating about the horizontal line $y = k$ makes the radii $|f(x) - k|$ and $|g(x) - k|$; rotating about $x = k$ makes them $|p(y) - k|$ and $|q(y) - k|$. Nothing else in the formula changes.

Which variable. Slice perpendicular to the axis of rotation. Rotating about a horizontal line means vertical radii and an integral in $x$; rotating about a vertical line means horizontal radii and an integral in $y$. If solving for the other variable is unpleasant, the method of cylindrical shells is the alternative.

Worked examples

Square cross-sections on a parabolic base

The base of a solid is the region between $y = 1 - x^2$ and the horizontal axis. Every cross-section perpendicular to the $x$-axis is a square. Find the volume.

The base runs from $x = -1$ to $x = 1$. At position $x$, the base region stretches from $y = 0$ up to $y = 1 - x^2$, so the side of the square is $s(x) = 1 - x^2$ and its area is $A(x) = (1 - x^2)^2 = 1 - 2x^2 + x^4$.

$$ V = \int_{-1}^{1} (1 - 2x^2 + x^4)\, dx = 2 \left[ x - \frac{2x^3}{3} + \frac{x^5}{5} \right]_0^1 $$

using evenness. The bracket at $x = 1$ is $1 - \tfrac{2}{3} + \tfrac{1}{5} = \tfrac{15 - 10 + 3}{15} = \tfrac{8}{15}$, so $V = \tfrac{16}{15}$.

Sanity check: the largest square has area $1$ and the squares shrink to nothing at both ends, so the volume must be under the $2 \times 1 \times 1$ box containing the solid. It is about $1.07$.

A disk

Rotate the region under $y = \sqrt{x}$ from $x = 0$ to $x = 4$ about the $x$-axis.

The cross-section at $x$ is a full disk of radius $\sqrt{x}$, so $A(x) = \pi(\sqrt{x})^2 = \pi x$.

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

Check against a shape you know: the solid sits inside the cylinder of radius $2$ and length $4$, whose volume is $\pi \cdot 4 \cdot 4 = 16\pi$. Our solid fills exactly half of it, which is right for a curve that grows like a square root.

A washer

Rotate the region between $y = 2x$ and $y = x^2$ about the $x$-axis.

The curves meet where $2x = x^2$, that is at $x = 0$ and $x = 2$. On that interval $2x \ge x^2$ (test $x = 1$: $2$ against $1$), so the line is the outer boundary and the parabola the inner one.

$$ V = \pi \int_0^2 \big[ (2x)^2 - (x^2)^2 \big] dx = \pi \int_0^2 (4x^2 - x^4)\, dx $$

$$ = \pi \left[ \frac{4x^3}{3} - \frac{x^5}{5} \right]_0^2 = \pi \left( \frac{32}{3} - \frac{32}{5} \right) = 32\pi \cdot \frac{2}{15} = \frac{64\pi}{15} $$

Note what would have gone wrong with $\pi \int (2x - x^2)^2 dx$: that integrand expands to $4x^2 - 4x^3 + x^4$, a different function entirely. Square first, subtract second.

Rotating about a line below the region

Rotate the region bounded by $y = x^2$, $y = 0$ and $x = 1$ about the horizontal line $y = -1$.

The axis of rotation is not a boundary of the region, so a hole appears. The distance from $y = -1$ up to the curve is $x^2 + 1$, and the distance from $y = -1$ up to the lower boundary $y = 0$ is $1$. Those are the outer and inner radii.

$$ V = \pi \int_0^1 \big[ (x^2 + 1)^2 - 1^2 \big] dx = \pi \int_0^1 (x^4 + 2x^2)\, dx $$

since $(x^2+1)^2 = x^4 + 2x^2 + 1$ and the $1$ cancels.

$$ V = \pi \left[ \frac{x^5}{5} + \frac{2x^3}{3} \right]_0^1 = \pi \left( \frac{1}{5} + \frac{2}{3} \right) = \frac{13\pi}{15} $$

The shifted axis changed only the two radii. That is the entire adjustment, and it is why it pays to write “radius equals distance from axis to curve” rather than memorizing a formula per picture.

Practice

First, solids described by their cross-sections. You are given a base region and a shape standing on each slice; the work is expressing that shape’s area as a function of the slicing variable.

Then solids of revolution by disks and washers. Identify the axis, write the outer and inner radii as distances from it, square each, and integrate.

Quiz

Five items on the general slicing formula, known cross-sections, and the disk and washer methods, including rotation about a shifted axis.