Volume of Revolution

Four steps. First take the boundary function f and the limits a, b from the task. Second square f(x) as a term: √x becomes x, 2x becomes 4x². Third evaluate the definite integral ∫ₐᵇ f(x)² dx with an antiderivative. Fourth multiply by π and attach the unit volume units (VU). Example: f(x) = √x over [0; 4]: V = π·∫₀⁴ x dx = π·[x²/2]₀⁴ = π·8 = 8π ≈ 25.1 VU. Checking against known solids pays off: f(x) = x over [0; 3] must yield the cone ⅓·π·3²·3 = 9π, and it does.

Because of the circular discs. If you cut the solid of revolution at position x perpendicular to the axis, a circle appears whose radius is exactly the function value f(x), since that is how far the graph is from the rotation axis. The area of this circle is π·r² = π·f(x)². A wafer-thin disc of thickness dx therefore has volume π·f(x)²·dx, and the integral sums all discs from a to b. So the square comes from the circle area formula, not from some integration rule. Whoever integrates f unsquared computes the area under the curve instead, a completely different quantity with a different unit.

Swapping the order of squaring and integrating: π·(∫f(x) dx)² is WRONG, correct is π·∫f(x)² dx. A numerical example shows they differ: for f(x) = x over [0; 2], ∫x² dx = 8/3, so V = 8π/3 ≈ 8.38; the wrong (∫x dx)² = 2² = 4 would give 4π ≈ 12.57. Further classics: forgetting π (the factor 3.14 is missing, which a plausibility check catches), computing (f − g)² instead of f² − g² for hollow solids, and using limits that do not belong to the rotation region. Antidote: write the formula down cleanly, simplify the integrand f² first, finish with a unit and magnitude check.

Mirror-image of the x-axis formula, just thought in y: V = π·∫꜀ᵈ g(y)² dy, where g is the boundary curve solved for x (the inverse function of f) and c, d are the limits on the y-axis. Procedure: solve y = f(x) for x, x = g(y); compute the y-limits from the x-limits (c = f(a), d = f(b)); then integrate g(y)² over [c; d]. Example: f(x) = x² over [0; 2] around the y-axis: g(y) = √y, limits y = 0 to 4, V = π·∫₀⁴ y dy = π·8 = 8π ≈ 25.1 VU. Most common mistake: carrying the x-limits over unchanged as y-limits.

With the annular disc formula V = π·∫ₐᵇ (f(x)² − g(x)²) dx, where f is the outer and g the inner boundary (f ≥ g ≥ 0 on [a; b]). Every cross-section is an annulus: large circle minus small hole, i.e. π·f² − π·g². Example pipe: f(x) = 2 and g(x) = 1 over [0; 3]: V = π·∫₀³ (4 − 1) dx = 9π ≈ 28.3 VU. The crucial warning: f² − g² is NOT (f − g)². For the numbers 2 and 1: 4 − 1 = 3, but (2 − 1)² = 1; the binomial formula shows the difference 2fg − 2g². Before computing, clarify which graph lies outside, otherwise the volume comes out negative.