Integrals add up more than area: the length of a curve, the balance point of a shape, and the work to pump out a tank
The recipe behind every integral is the same: chop something into small pieces, approximate each piece by something simple, add, and let the pieces shrink. Applied to a curve it gives length, to a plate its balance point, and to a tank of water the energy needed to empty it.
Press Double and watch the inscribed polyline creep up on the true length. Drag across the curve to magnify one chord, and switch on the surface of revolution.
The true length is computed with adaptive Simpson’s rule on √(1 + f′(x)²), with f′ found symbolically. Where f′ is infinite at an end (the quarter circle at x = 1) the integral is improper, and the demo instead extrapolates polylines of 50 000 and 200 000 chords. The length of sin x on [0, π] has no elementary closed form: it is an elliptic integral.
A chord from x to x + Δx rises Δy, so by Pythagoras its length is √(Δx² + Δy²) = √(1 + (Δy/Δx)²)·Δx. By the mean value theorem, Δy/Δx equals f′ at some point inside the chord, so the polyline length is exactly a Riemann sum for
L = ∫ₐᵇ √(1 + f′(x)²) dx.
The integrand is rarely friendly. Even for the parabola y = x² it needs a logarithm, and for y = sin x the integral is elliptic, with no formula in elementary functions. The happy exceptions are worth knowing: for the catenary y = cosh x, 1 + sinh² x = cosh² x, so the length from −1 to 1 is 2 sinh 1; for y = x^(3/2), 1 + (9/4)x is a perfect target for substitution. The curve y = x^(3/2) (the semicubical parabola) was among the first curves to be measured exactly: William Neile rectified it in 1657.
Spin the curve about the x-axis and each chord sweeps out a slice of a cone, a frustum, with area π(r₁ + r₂)·ds. In the limit, the surface area is S = 2π∫ₐᵇ |f(x)|√(1 + f′(x)²) dx. For the quarter circle, that gives a hemisphere of area 2π, exactly as Archimedes found.
Drag the pin along the plate and let go. It tips towards the heavier side unless the pin is directly at the centre of mass. Make the plate denser at one end and watch the target move.
The pin passes through the plate at the height of ȳ, so the only thing that matters is the horizontal distance between the pin and x̄; the swing is a torque-driven rotation with damping, using the plate’s true moment of inertia about the pin, and it stops on rests at ±24°. The bottom row spins the uniform plate about the x-axis: Pappus’s area times the distance travelled by the centroid matches the washer volume.
Cut the plate into vertical strips. The strip at x has width dx, height f(x) − g(x) and density ρ(x), so its mass is dm = ρ(f − g) dx, sitting at horizontal position x and at mid-height (f + g)/2. Weighting each position by its mass gives the moments
M = ∫ρ(f − g) dx, M_y = ∫xρ(f − g) dx, M_x = ∫ρ(f² − g²)/2 dx,
and the centre of mass is (x̄, ȳ) = (M_y/M, M_x/M). The plate balances on a pin at x̄ because the torque about that point, ∫(x − x̄) dm, is zero; that is exactly what the definition of x̄ says. For uniform density ρ cancels and (x̄, ȳ) is the centroid, a purely geometric point. A centroid need not lie on the plate: the lens between y = x and y = x² has it at (1/2, 2/5), which is inside, but a crescent can have it in the hole.
Pappus of Alexandria noticed, in the fourth century AD, that spinning a region about an axis it does not cross gives a solid of volume V = A · 2πȳ: the area times the distance travelled by the centroid. Under y = √x on [0, 4] the area is 16/3 and the centroid sits at ȳ = 3/4, so V = (16/3)·2π·(3/4) = 8π, the same as the disks in Volumes of Revolution. The demo checks the same identity live for any plate above the axis. Paul Guldin rediscovered the theorem in the seventeenth century, so it is sometimes called Guldin’s rule.
Pick a tank and press Pump. Each layer of water is lifted to the outlet, and the tally adds up its work. Raise the spout, lower the water, and compare the shapes.
SI units throughout: ρ = 1000 kg/m³, g = 9.8 m/s², so a layer of volume V lifted d metres costs 9800·V·d joules. Each layer is lifted from its middle, which makes the layer sum a midpoint Riemann sum. The trough is drawn end-on; it runs 4 m into the page. Pipe friction and the water’s kinetic energy are ignored.
Lifting a weight F through a distance d takes work W = F·d, measured in joules. Water in a tank is not lifted as one object: water near the bottom travels further than water near the top. So slice it. The layer at height y with thickness dy has volume A(y) dy, mass ρA(y) dy and weight ρgA(y) dy, and travels h − y to the outlet. Adding the layers,
W = ∫₀ᵈ ρ g A(y) (h − y) dy.
For the full cylinder of radius 1.5 m and height 3 m, pumped over the rim, A(y) = 2.25π and the integral is 9800 · 2.25π · ∫₀³ (3 − y) dy = 9800 · 2.25π · 4.5 ≈ 311.7 kJ. The answer is also the weight of the water times the distance its centre of mass rises, here 1.5 m. A full cone standing on its point keeps most of its water near the top: its centre of mass is three quarters of the way up, so on average each kilogram rises only a quarter of the depth.
A spring is the other classic case: the force grows as you stretch. Hooke’s law says F = kx, so the work to stretch it by x is ∫₀ˣ ks ds = ½kx², the area of a triangle under the force line. Doubling the stretch quadruples the energy stored.
Spot a mistake or have an idea?