Volumes by slicing: disks and washers · Calculus II: Integrals & Series

§5.0 sliced a region into thin strips and summed their areas. Slice a solid into thin cross-sections instead, and the same Riemann-sum-to- integral argument from §4.1–§4.2 sums their volumes. The special case where every cross-section is a circle — because the solid was formed by spinning a flat region around an axis — is a solid of revolution, and it's the most common volume integral you'll meet.

The general slicing principle

For any solid, if A(x) is the cross-sectional area at position x (the area of the slice perpendicular to the x-axis), then

$V=\int_a^bA(x)\,dx$

This is exactly §4.2's definite-integral-as-a-limit argument, run one dimension up: chop [a,b] into n pieces, approximate the solid on each piece by a thin slab of area A(x_i) and thickness \Delta x, sum \sum A(x_i)\Delta x, and take n\to\infty. Nothing about this requires circular cross-sections — square, triangular, or arbitrary cross-sections all use the identical formula. Disks and washers are simply the case where A(x) happens to be the area of a circle or an annulus, because revolving a region always produces circular slices.

The disk method

Revolve the region under y=f(x), above the x-axis, on [a,b] around the x-axis. Each cross-section is a disk of radius f(x), so A(x)=\pi\big[f(x)\big]^2:

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

This is §5.0's area-under-a-curve formula, squared and multiplied by \pi — the direct fingerprint of a circle's area formula A=\pi r^2 from §0.1, with r=f(x).

The washer method

If the region being revolved is bounded between two curves, g(x)\le y\le f(x), the solid has a hole down the middle. Each cross-section is an annulus — a washer — with outer radius f(x) and inner radius g(x):

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

Do not square the difference. [f(x)-g(x)]^2\ne[f(x)]^2-[g(x)]^2 — the washer's area is (outer disk) minus (inner disk), two separate circles, not one annulus-shaped algebra shortcut. This is the single most common volumes-by-slicing error.

Disks are the washer method with g(x)=0: a hole of radius zero is no hole at all.

Revolving around a different axis

Revolving around a horizontal line y=k instead of the x-axis shifts every radius by k: replace f(x) with f(x)-k (and g(x) with g(x)-k) before squaring. Revolving around a vertical line, or slicing horizontally around the y-axis, swaps the roles of x and y exactly as §5.0's "integrate against y" option did — solve the boundary curves for x in terms of y, and integrate \pi\int[\,]^2dy instead.

Doing it in Python

The disk method: revolve y=\sqrt x around the x-axis on [0,4] — a paraboloid:

import sympy as sp

x = sp.Symbol('x')
f = sp.sqrt(x)

V = sp.pi * sp.integrate(f**2, (x, 0, 4))
print(f"disk volume, sqrt(x) revolved on [0,4] = {V}")

The washer method, and the common squaring mistake made explicit as a contrast:

import sympy as sp

x = sp.Symbol('x')
outer, inner = x, x**2   # y = x is on top of y = x^2 on [0,1]

correct = sp.pi * sp.integrate(outer**2 - inner**2, (x, 0, 1))
wrong = sp.pi * sp.integrate((outer - inner)**2, (x, 0, 1))
print(f"washer, correct  (outer^2 - inner^2) = {correct}")
print(f"washer, WRONG    (outer - inner)^2   = {wrong}")
print("these differ -- squaring the difference is not the same as")
print("the difference of squares")

General slicing with a non-circular cross-section — a solid whose base is the region under y=\sqrt{1-x^2/4} on [-2,2] and whose cross-sections perpendicular to the x-axis are squares (not disks) of side 2f(x):

import sympy as sp

x = sp.Symbol('x')
f = sp.sqrt(1 - x**2 / 4)

side = 2 * f
A = side**2                       # square cross-section, not pi*f^2
V = sp.integrate(A, (x, -2, 2))
print(f"square cross-sections over an elliptical base: V = {sp.simplify(V)}")
print("same slicing formula V = int A(x) dx -- just a different A(x)")

Worked example

Find the volume of the solid formed by revolving the region between y=x and y=x^2 around the x-axis, for x in [0,1].

First confirm which curve is on top: on (0,1), e.g. at x=0.5, x=0.5>x^2=0.25, so y=x is the outer boundary and y=x^2 is the inner one — the region (and hence the hole) sits between them.

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

$=\pi\left[\frac{x^3}3-\frac{x^5}5\right]_0^1=\pi\left(\frac13-\frac15\right)=\pi\cdot\frac{5-3}{15}=\boxed{\frac{2\pi}{15}}$

Sanity check. The solid sits inside the disk swept by y=x alone (radius up to 1, giving \pi\int_0^1x^2dx=\frac\pi3\approx1.047) with a smaller solid (y=x^2's disk, \frac\pi5\approx0.628) hollowed out — so the answer must be less than \frac\pi3. \frac{2\pi}{15}\approx0.419 is comfortably below that. ✓ It also equals \frac\pi3-\frac\pi5 exactly, as it must: the washer volume is literally outer-disk-volume minus inner-disk-volume, since \int(f^2-g^2)=\int f^2-\int g^2 by linearity.

Your turn

1. Find the volume of the solid formed by revolving y=x^2 on [0,2] around the x-axis (a disk problem — no hole).

2. Find the volume when the region under y=\sin x on [0,\pi] is revolved around the x-axis.

3. Set up (don't necessarily finish) the washer integral for revolving the region between y=\sqrt x and y=x/2 around the x-axis. Find the bounds first.

Solutions

1. Disk method, f(x)=x^2:

$V=\pi\int_0^2(x^2)^2dx=\pi\int_0^2x^4dx=\pi\left[\frac{x^5}5\right]_0^2=\boxed{\frac{32\pi}5}$

2. Disk method, f(x)=\sin x, using the half-angle identity from §4.7:

$V=\pi\int_0^\pi\sin^2x\,dx=\pi\int_0^\pi\frac{1-\cos2x}2dx=\pi\left[\frac x2-\frac{\sin2x}4\right]_0^\pi=\pi\cdot\frac\pi2=\boxed{\frac{\pi^2}2}$

The trig-integral technique from §4.7 feeds directly into a volume once the disk method sets up the square.

3. Bounds: $\sqrt x=\frac x2\Rightarrow x=\frac{x^2}4\Rightarrow 4x=x^2\Rightarrow x(x-4)=0\Rightarrow x=0,4$. Check which is on top: at x=1, \sqrt1=1>\frac12, so y=\sqrt x is outer.

$V=\pi\int_0^4\left[(\sqrt x)^2-\left(\frac x2\right)^2\right]dx=\pi\int_0^4\left(x-\frac{x^2}4\right)dx$

Check yourself in code

Compute two volumes with SymPy: the disk volume of y=\sqrt x revolved on [0,4], and the washer volume between y=x (outer) and y=x^2 (inner) revolved on [0,1].

Print exactly this:

disk volume, sqrt(x) on [0,4]       = 8*pi
washer volume, x and x^2 on [0,1]   = 2*pi/15

import sympy as sp

x = sp.Symbol('x')

f = sp.sqrt(x)
V_disk = sp.pi * sp.integrate(f**2, (x, 0, 4))
print("disk volume, sqrt(x) on [0,4]       = ...")

outer, inner = x, x**2
V_washer = sp.pi * sp.integrate(outer**2 - inner**2, (x, 0, 1))
print("washer volume, x and x^2 on [0,1]   = ...")

import sympy as sp

x = sp.Symbol('x')

f = sp.sqrt(x)
V_disk = sp.pi * sp.integrate(f**2, (x, 0, 4))
print(f"disk volume, sqrt(x) on [0,4]       = {V_disk}")

outer, inner = x, x**2
V_washer = sp.pi * sp.integrate(outer**2 - inner**2, (x, 0, 1))
print(f"washer volume, x and x^2 on [0,1]   = {V_washer}")

Every volume-by-slicing problem starts from V=\int A(x)\,dx, summing cross-sectional areas exactly the way §4.2 summed heights into an area. When the cross-sections are full disks, A(x)=\pi[f(x)]^2; when the region has a hole, A(x)=\pi\big([f(x)]^2-[g(x)]^2\big) — outer disk minus inner disk, never the squared difference. The slicing idea itself doesn't care about circles at all, which is why it generalizes cleanly to any cross-sectional shape, and why it reappears, essentially unchanged, when §11 slices solids in two variables at once.

Next: a second way to build the same solids — by wrapping cylindrical shells around the axis instead of stacking disks perpendicular to it — which turns out to be far less painful whenever solving for x in terms of y is the hard part.

15. Volumes by slicing: disks and washers