Spin a region about an axis and slice the solid into disks, washers or shells; stack cross-sections of any shape
An integral adds up thin slices. Slice a solid and each slice is nearly a flat piece of known volume, so the same idea that measures area measures volume too. The craft is in choosing the slices: coins across the axis, tin cans around it, or squares and triangles standing on a base.
Press Revolve and watch the region sweep out a solid. Then raise N to fill it with thinner washers, and move the axis to see the holes open up.
Drag the 3D view to orbit; drag across the flat plot to move the highlighted slice. Each washer is sampled at the middle of its slice (a midpoint sum), so doubling N cuts the error by about four. The washers are drawn slightly thinner than Δx so you can see between them; the sum uses the full Δx. The dashed outline in the flat plot is the region’s mirror image across the axis, the other half of every cross-section.
Spin the region between y = f(x) and y = g(x), a ≤ x ≤ b, about the x-axis. The thin vertical strip at x, of width dx, sweeps out a washer: a disk of radius R(x) with a hole of radius r(x), where R is the distance from the axis to the far curve and r the distance to the near one. Its volume is π(R² − r²)·dx, and adding the washers gives
V = π ∫ₐᵇ (R(x)² − r(x)²) dx.
When the region touches the axis, r = 0 and the washers are solid disks: V = π∫f(x)² dx. About a line y = k instead, the radii are measured from that line, R = |f(x) − k| and r = |g(x) − k| for a region on one side of it. For y = √x on [0, 4] about the x-axis, π∫₀⁴ x dx = 8π.
The sum of N washers is a midpoint Riemann sum for this integral, so its error shrinks like 1/N². The formula is the slicing method that Cavalieri and Kepler used in the early seventeenth century, made exact by the limit. Kepler’s Nova stereometria doliorum vinariorum (1615) measured the volumes of wine barrels this way.
Spin the region about a vertical axis and nest it in tin cans. Unroll the highlighted shell into a flat slab, then switch to horizontal washers and compare the two sums.
A quarter of the solid is cut away so you can see inside; drag to orbit. Drag across the flat plot to choose a shell. Each shell takes its height at the middle of its strip, so its volume π(r₂² − r₁²)(f − g) is exactly 2π·r·Δr·(f − g) with r the middle radius. Shells and washers are drawn a little thinner than they are so the gaps show.
Revolve the same region about the y-axis. The strip at x, of height f(x) − g(x) and width dx, now sweeps out a thin cylindrical shell of radius x. Cut it down one side and lay it flat: it becomes a slab of length 2πx (the circumference), height f(x) − g(x) and thickness dx. So
V = 2π ∫ₐᵇ x (f(x) − g(x)) dx,
with x replaced by |x − h| for an axis x = h. The flattening is exact in the limit: a shell between radii r₁ and r₂ has volume π(r₂² − r₁²)h = 2π·((r₁ + r₂)/2)·h·(r₂ − r₁), the circumference at its middle radius times its height and thickness.
Washers would cut the same solid horizontally, and then each slice needs the width of the region at height y, which means solving y = f(x) for x. For y = sin x that requires arcsin, and the region at height y runs from arcsin y to π − arcsin y. Shells avoid inverting anything, which is why they are often easier by hand. Both methods compute the same volume, 2π² for sin x on [0, π].
Stand squares, triangles or semicircles on a base and add up the slabs. Then shear a stack of coins, and slice Archimedes’ hemisphere and cylinder at any height.
Drag to orbit. Slabs and coins take the cross-section at the middle of their thickness and are drawn with small gaps so they read as separate pieces; the sums use their full thickness.
Revolution is a special case. Any solid whose cross-section perpendicular to the x-axis has area A(x) has volume V = ∫ₐᵇ A(x) dx. If the base is a disc of radius 1 and every cross-section is a square standing on a chord, the chord at x has length w = 2√(1 − x²), the square has area 4(1 − x²), and the volume is ∫₋₁¹ 4(1 − x²) dx = 16/3.
Bonaventura Cavalieri stated the principle in his Geometria indivisibilibus (1635): two solids that have equal cross-sectional areas at every height have equal volumes. A sheared stack of coins is the picture. In calculus language it says nothing more than that the volume depends only on A(x).
Archimedes had used the idea much earlier. A hemisphere of radius R, cut at height z, gives a disc of area π(R² − z²). A cylinder of radius R and height R with a cone removed (its tip at the bottom, its base the cylinder’s top) gives a ring between radius z and R, of area πR² − πz², the same. So the hemisphere has the volume of the cylinder minus the cone, πR³ − πR³/3 = (2/3)πR³, and the whole sphere (4/3)πR³. In The Method, a text lost for centuries and rediscovered in a palimpsest in 1906, Archimedes describes finding such results by balancing slices, and he asked for a sphere inside a cylinder to be carved on his tomb.
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