6.5 Volumes of Solids of Revolution: Disk, Washer & Shell Methods

6.5 Volumes of Solids of Revolution: Disk, Washer & Shell Methods

Of every topic in single-variable calculus, this is the one the National Board names outright. The published description of Component 1, Exercise 2 (Geometry) states that candidates will "express the volume of a solid generated by the rotation of a two-dimensional object about an axis," and the released sample prompt asks for exactly that: rotate a right trapezoid about the $y$-axis and write the volume in terms of its parameters, showing all work. The exercise sits in the Geometry exercise rather than a calculus one, which tells you what the National Board is really testing — whether an accomplished teacher can move fluently between a plane figure, the solid it sweeps out, and the integral that measures it.

---

1. The Underlying Idea: Slice, Measure, Accumulate

Section 5.3 established geometric measurement as fixing a unit and fitting copies of it into a figure, and Cavalieri's Principle as the statement that two solids of equal height with equal cross-sectional areas at every level have equal volume. Solids of revolution are the calculus completion of that idea. Partition the axis of rotation, take a thin slab of thickness $dx$ or $dy$, treat its cross-section as a circle or annulus, and accumulate:

$$V = \int_a^b A(x)\, dx \qquad \text{where } A(x) \text{ is the cross-sectional area perpendicular to the axis.}$$

Everything that follows is a special case of this one formula.

---

2. The Disk Method

When the region is bounded by a curve and the axis of rotation itself, each cross-section is a disk of radius $R$ equal to the distance from the curve to the axis.

Worked derivation — the cone. Rotate the region under $y = \frac{r}{h}x$ for $0 \le x \le h$ about the $x$-axis:

$$V = \pi \int_0^h \left(\frac{r}{h}x\right)^{\!2} dx = \frac{\pi r^2}{h^2}\left[\frac{x^3}{3}\right]_0^h = \frac{1}{3}\pi r^2 h$$

Worked derivation — the sphere. Rotate $y = \sqrt{r^2 - x^2}$ on $[-r, r]$ about the $x$-axis:

$$V = \pi \int_{-r}^{r} (r^2 - x^2)\, dx = \pi\left[r^2 x - \frac{x^3}{3}\right]_{-r}^{r} = \pi\left(\frac{4r^3}{3}\right) = \frac{4}{3}\pi r^3$$

Being able to derive the formulas tabulated in Section 5.3, rather than recite them, is precisely the depth Standard II describes when it says accomplished teachers "can explain the volume formula of a cylinder with an arbitrary base … and how it relates to the volume of the cone on the same base."

---

3. The Washer Method

When the region does not touch the axis, the cross-section is an annulus (washer) with outer radius $R$ and inner radius $r$:

$$V = \pi \int_a^b \left( [R(x)]^2 - [r(x)]^2 \right) dx$$

Example. Rotate the region between $y = x$ and $y = x^2$ on $[0,1]$ about the $x$-axis. On this interval $x \ge x^2$, so $R = x$ and $r = x^2$:

$$V = \pi \int_0^1 (x^2 - x^4)\, dx = \pi\left[\frac{x^3}{3} - \frac{x^5}{5}\right]_0^1 = \pi\left(\frac13 - \frac15\right) = \frac{2\pi}{15}$$

---

4. The Shell Method

Slice parallel to the axis instead of perpendicular to it. Each slice sweeps a thin cylindrical shell of radius $x$, height $h(x)$, and thickness $dx$; unrolled, it is a rectangular sheet of area $2\pi x \cdot h(x)$:

$$V = 2\pi \int_a^b (\text{radius})(\text{height})\, dx$$

For a region bounded above by $f(x)$ and below by $g(x)$ on $[a,b]$, rotated about the $y$-axis, the radius is $x$ and the height is $f(x) - g(x)$.

Choosing between washers and shells

Both methods must agree. Verifying a volume by computing it a second way is the single best self-check in a timed exercise, and it is exactly the kind of "thorough discussion" the constructed-response rubrics reward.

---

5. The Released Exercise 2 Prompt, Worked

A right trapezoid is placed on the coordinate plane with vertices $(0,0)$, $(0,2a)$, $(b,a)$, and $(b,0)$. Write an expression, in terms of $a$ and $b$, for the volume of the solid obtained by rotating the region about the $y$-axis. Show all work.

Step 1 — identify the boundaries. The left edge lies on the $y$-axis, the bottom edge on the $x$-axis from $(0,0)$ to $(b,0)$, the right edge is the vertical segment from $(b,0)$ to $(b,a)$, and the slanted top runs from $(b,a)$ to $(0,2a)$. That slant has slope $\dfrac{2a - a}{0 - b} = -\dfrac{a}{b}$, so its equation is

$$y = 2a - \frac{a}{b}x, \qquad 0 \le x \le b.$$

Step 2 — shell method (cleanest here). Radius $= x$; height $=$ top $-$ bottom $= \left(2a - \frac{a}{b}x\right) - 0$:

$$V = 2\pi \int_0^b x\left(2a - \frac{a}{b}x\right) dx = 2\pi \int_0^b \left(2ax - \frac{a}{b}x^2\right) dx = 2\pi\left[a x^2 - \frac{a}{3b}x^3\right]_0^b$$ $$V = 2\pi\left(ab^2 - \frac{ab^2}{3}\right) = 2\pi \cdot \frac{2ab^2}{3} = \boxed{\dfrac{4\pi a b^2}{3}}$$

Step 3 — verify with disks and washers in $y$. For $0 \le y \le a$ the horizontal cross-section runs from $x = 0$ to $x = b$, a full disk of radius $b$. For $a \le y \le 2a$ the right boundary is the slant, so $x = \dfrac{b(2a - y)}{a}$:

$$V = \pi b^2 a \;+\; \pi\int_a^{2a} \frac{b^2 (2a-y)^2}{a^2}\, dy = \pi a b^2 + \frac{\pi b^2}{a^2}\left[-\frac{(2a-y)^3}{3}\right]_a^{2a} = \pi a b^2 + \frac{\pi a b^2}{3} = \frac{4\pi a b^2}{3}$$

The two methods agree, and a dimensional check confirms plausibility: the expression has dimensions of length$^3$ and is linear in $a$, quadratic in $b$, as any volume swept at radius $b$ must be.

Step 4 — sanity-check against known solids. The solid is a cylinder of radius $b$ and height $a$ ($\pi a b^2$) capped by a cone-like frustum-complement of volume $\frac{1}{3}\pi a b^2$ — precisely the cone on the same base and height, which is the cylinder–cone relationship Standard II asks teachers to explain.

---

6. Pappus's Centroid Theorem — a Fast Cross-Check

If a plane region of area $A$ whose centroid lies a distance $\bar{d}$ from an axis (and which does not cross that axis) is revolved about it, the resulting volume is

$$V = 2\pi \bar{d} A.$$

For the trapezoid above, $A = \frac{1}{2}(2a + a)(b) = \frac{3ab}{2}$, and the centroid's distance from the $y$-axis is $\bar{x} = \frac{4b}{9}$, giving $V = 2\pi \cdot \frac{4b}{9} \cdot \frac{3ab}{2} = \frac{4\pi a b^2}{3}$ — a third independent confirmation. The torus is the classic application: revolving a circle of radius $r$ whose centre sits $D$ from the axis gives $V = 2\pi D \cdot \pi r^2 = 2\pi^2 D r^2$.

---

7. Traps and Teaching Notes

For classroom use, this topic is a natural home for productive struggle (Section 10.2): give students a shape and two methods and let them discover that both agree, rather than prescribing which to use. It also connects directly back to Section 5.3's frustum, cone, and sphere formulas, letting students see memorised formulas re-derived from a single integral — the "whole fabric" view of mathematics that Standard II asks accomplished teachers to model.