Every formula this week follows the same three-step pattern established for area in Week 7: cut the object into thin, simple pieces; write down the volume, length, or area of one typical piece; then let the pieces' thickness shrink to zero and integrate.
10.1 Volumes by Cross-Sections
If a solid's cross-sectional area A(x), taken perpendicular to the x-axis, is known for every x in [a,b], slicing the solid into thin slabs of thickness dx and summing (exactly the Riemann-sum idea of Week 7) gives
This is the master formula behind everything else this week — every other volume formula is a special case obtained by writing down A(x) for a particular shape.
10.2 Solids of Revolution: The Disk Method
Revolving the region under y=f(x) ≥ 0, [a,b], about the x-axis produces a solid whose cross-section at x is a disk of radius f(x), so A(x) = π[f(x)]2:
Example 10.1 — The Disk Method
Find the volume generated by revolving y=√x, [0,4], about the x-axis.
Step 1 — Apply Eq. 10.2
V = π∫04 (√x)2 dx = π∫04 x dx
Step 2 — Evaluate
π[x2/2]04 = π(8−0)
V = 8π
10.3 The Washer Method
When the region revolved is between two curves, f(x) ≥ g(x) ≥ 0, the cross-section is an annulus (a "washer") — a disk with a hole:
The washer's area is π(R2−r2), not π(R−r)2. Square the outer and inner radii separately before subtracting — exactly the same trap as (a+b)2 ≠ a2+b2.
10.4 Volumes by Cylindrical Shells
Instead of slicing perpendicular to the axis of revolution, the shell method sums thin cylindrical shells parallel to it. Revolving the region under y=f(x) ≥ 0 on [a,b] (with 0 ≤ a) about the y-axis, each shell at position x has radius x, height f(x), and thickness dx, giving surface area 2πxf(x) times the thickness:
Example 10.2 — The Shell Method
Find the volume generated by revolving the region under y=x2, [0,2], about the y-axis.
Step 1 — Apply Eq. 10.4
V = 2π∫02 x·x2 dx = 2π∫02 x3 dx
Step 2 — Evaluate
2π[x4/4]02 = 2π(4−0)
V = 8π
10.5 Choosing Disks/Washers vs. Shells
Both methods are the same master formula (Eq. 10.1) in disguise, sliced in perpendicular directions — they always agree when both apply. The practical choice comes down to which slicing direction produces cross-sections describable by a single, simple formula:
- Revolving about the x-axis, with the region naturally described as y=f(x): disks/washers are usually simpler.
- Revolving about the y-axis, with the region still described as y=f(x) (not solved for x): shells avoid rewriting the function.
10.6 Arc Length
Approximate a curve y=f(x) by short line segments between nearby points; by the Pythagorean theorem, each segment has length √(Δx2+Δy2) = √(1+(Δy/Δx)2) Δx. Summing and taking the limit as the segments shrink (with Δy/Δx becoming f′(x), exactly as in Week 2) gives the arc length:
Example 10.3 — Arc length
Find the length of y = (2/3)x3/2 from x=0 to x=3.
Step 1 — Find f' and set up the integrand
f′(x) = x1/2, so 1+[f′(x)]2 = 1+x.
Step 2 — Apply Eq. 10.5
L = ∫03 √(1+x) dx
Step 3 — Substitute u=1+x (Week 9, §9.6) and evaluate
u=1+x, du=dx; limits become 1 to 4.
∫14 u1/2 du = (2/3)[u3/2]14 = (2/3)(8−1)
L = 14/3
10.7 Areas of Surfaces of Revolution
Revolving the curve y=f(x) ≥ 0 about the x-axis sweeps out a surface. Each short arc-length piece of length ds = √(1+[f′(x)]2) dx sweeps out a thin band of a cone with radius f(x) and slant height ds, contributing area 2πf(x) ds:
Note the arc-length integrand from §10.6 reappears inside this formula — surface area is arc length weighted by the distance travelled around the axis (2π times the radius), the same "distance times length" logic as the shell method in §10.4.
10.8 Worked Capstone Example
Example 10.4 — Surface area of a sphere
Use Eq. 10.6 to derive the surface area of a sphere of radius r, by revolving y=√(r2−x2) about the x-axis, [−r,r].
Step 1 — Find f' and the arc-length integrand (§10.6)
f′(x) = −x/√(r2−x2), so 1+[f′(x)]2 = 1 + x2/(r2−x2) = r2/(r2−x2).
Step 2 — Simplify the surface integrand (§10.7)
√(1+[f′]2) = r/√(r2−x2) = r/f(x), so f(x)·√(1+[f′]2) = r — a constant, independent of x.
Step 3 — Apply Eq. 10.6
S = 2π∫−rr r dx = 2πr·(2r)
S = 4πr2 — the familiar sphere surface area formula, derived rather than assumed.
10.9 Nomenclature
| Symbol | Meaning |
|---|---|
| A(x) | Cross-sectional area at position x |
| disk / washer method | Slicing perpendicular to the axis of revolution |
| shell method | Slicing parallel to the axis of revolution, as thin cylinders |
| L | Arc length |
| ds | An infinitesimal element of arc length |
| S | Surface area of a surface of revolution |
10.10 Problems
- Cross-sections. A solid has base the region between y=x2 and y=4, and cross-sections perpendicular to the x-axis are squares. Find its volume.
- Disk method. Find the volume generated by revolving y=x2, [0,2], about the x-axis.
- Washer method. Find the volume generated by revolving the region between y=x and y=x2, [0,1], about the x-axis.
- Shell method. Find the volume generated by revolving the region under y=√x, [0,4], about the y-axis.
- Choosing a method. Set up (but do not evaluate) the integral for the volume generated by revolving y=x3, [0,1], about the y-axis using the shell method, and explain why the shell method is more convenient than washers here.
- Arc length. Find the length of y = (1/3)(x2+2)3/2 from x=0 to x=3.
- Arc length, line check. Use Eq. 10.5 to find the length of the line segment y=2x+1 from x=0 to x=3, and confirm it matches the distance formula.
- Surface of revolution. Find the area of the surface generated by revolving y=√x, [1,4], about the x-axis.
- Combined. Find the volume and the arc length contribution needed to find the surface area of revolving y=x3/3, [0,1], about the x-axis (set up both integrals; evaluating the surface integral is optional).
10.11 References
- Thomas, G. B., Weir, M. D. and Hass, J. Thomas' Calculus, SI Units, 15th ed. Pearson, 2024. — §6.1 (volumes using cross-sections), §6.2 (volumes using cylindrical shells), §6.3 (arc length), §6.4 (areas of surfaces of revolution).
- Pearson MyLab Math — the Week 10 homework set is assigned directly from the sections above; see the Calendar for the due date.