Geometric Applications: Planar Area, Arc Length and Volumes of Solids of Revolution

Definite integrals compute planar areas, the length of curves and the volumes of solids of revolution: the area between two curves is ∫ (f − g) dx, the arc length is ∫ √(1 + f′(x)²) dx, and the volume of a solid of revolution about the x-axis is π ∫ f(x)² dx. In brief: the area between y = x and y = x² is 1/6; rotating y = √x on [0, 4] about the x-axis gives the volume π ∫₀⁴ x dx = 8π; the length of y = x^3/2 on [0, 4] is (8/27)(10√10 − 1) ≈ 9.07.

The common idea is “slice, approximate, add up”: cut the figure into thin pieces whose size is easy to compute, sum, and take the limit, which gives a definite integral. This guide presents the formulas for area (including parametric and polar curves), arc length, volume by discs, washers and cylindrical shells, and the surface area of a solid of revolution, with worked examples and a table of results.

How do you compute planar areas?

The area under y = f(x) ≥ 0 from a to b is ∫_a^b f(x) dx. The area between two curves with f(x) ≥ g(x) on [a, b] is ∫_a^b [f(x) − g(x)] dx. When the curves cross, split the interval at the intersection points, or integrate |f − g|.

Example 1

The area between y = x and y = x²: they intersect at x = 0 and 1; x ≥ x² in between. ∫₀¹ (x − x²) dx = 1/2 − 1/3 = 1/6.

Example 2

The area between y = 4 − x² and y = x + 2: intersections: 4 − x² = x + 2, so x² + x − 2 = 0, x = −2 or 1. ∫₋₂¹ (2 − x − x²) dx = [2x − x²/2 − x³/3]₋₂¹ = (2 − 1/2 − 1/3) − (−4 − 2 + 8/3) = 7/6 + 10/3 = 9/2 = 4.5.

Example 3: area under a sine arch

∫₀^π sin x dx = 2. The area enclosed by y = sin x and y = cos x between consecutive intersections (π/4 and 5π/4): ∫_π/4^5π/4 (sin x − cos x) dx = [−cos x − sin x]_π/4^5π/4 = (√2/2 + √2/2) − (−√2/2 − √2/2) = 2√2 ≈ 2.828.

Example 4: integrating with respect to y

The area between x = y² and x = y + 2: y² = y + 2 gives y = −1, 2. Right curve minus left curve: ∫₋₁² (y + 2 − y²) dy = [y²/2 + 2y − y³/3]₋₁² = (2 + 4 − 8/3) − (1/2 − 2 + 1/3) = 10/3 + 7/6 = 9/2.

How do you compute arc length?

The length of y = f(x) on [a, b] is L = ∫_a^b √(1 + f′(x)²) dx. It comes from the Pythagorean theorem for an element: ds = √(dx² + dy²).

Example 5

y = x^3/2 on [0, 4]: y′ = (3/2)x^1/2, 1 + y′² = 1 + 9x/4. L = ∫₀⁴ √(1 + 9x/4) dx = (4/9)(2/3)[(1 + 9x/4)^3/2]₀⁴ = (8/27)(10^3/2 − 1) = (8/27)(31.623 − 1) ≈ 9.073.

Example 6: a circle

The upper half of x² + y² = r²: y = √(r² − x²), y′ = −x/√(r² − x²), √(1 + y′²) = r/√(r² − x²). L = ∫_−r^r r dx/√(r² − x²) = r[arcsin(x/r)]_−r^r = πr ✓, half of the circumference.

Example 7: the catenary

y = cosh x on [0, 1]: y′ = sinh x, and 1 + sinh²x = cosh²x, so L = ∫₀¹ cosh x dx = sinh 1 ≈ 1.1752.

How do you compute the volume of a solid of revolution?

• Discs (about the x-axis): V = π ∫_a^b f(x)² dx. Each slice is a disc of radius f(x).
• Washers (region between f ≥ g ≥ 0): V = π ∫_a^b [f(x)² − g(x)²] dx.
• Cylindrical shells (about the y-axis): V = 2π ∫_a^b x f(x) dx, for 0 ≤ a < b.
• Cross-sections: V = ∫ A(x) dx for a solid with known cross-sectional area A(x).

Example 8: a cone

Rotate y = rx/h, 0 ≤ x ≤ h, about the x-axis: V = π ∫₀^h (rx/h)² dx = π (r²/h²) [x³/3]₀^h = π r²h/3.

Example 9: a sphere

Rotate y = √(r² − x²) on [−r, r]: V = π ∫_−r^r (r² − x²) dx = π[r²x − x³/3]_−r^r = π(2r³ − 2r³/3) = (4/3)πr³ ✓.

Example 10: a washer

The region between y = x and y = x² rotated about the x-axis: V = π ∫₀¹ (x² − x⁴) dx = π(1/3 − 1/5) = 2π/15 ≈ 0.419.

Example 11: shells

The region under y = x² on [0, 2], rotated about the y-axis: V = 2π ∫₀² x · x² dx = 2π [x⁴/4]₀² = 8π. Check with discs in y: the region for y ∈ [0, 4]: washers with the outer radius 2 and inner radius √y: π ∫₀⁴ (4 − y) dy = π(16 − 8) = 8π ✓.

Example 12: Gabriel’s horn

y = 1/x for x ≥ 1 rotated about the x-axis: V = π ∫₁^∞ x⁻² dx = π, finite, though the surface area 2π ∫ (1/x)√(1 + 1/x⁴) dx ≥ 2π ∫ dx/x diverges.

How do you compute the surface area of revolution?

Rotating y = f(x) ≥ 0 about the x-axis gives the surface area S = 2π ∫_a^b f(x) √(1 + f′(x)²) dx. Each element is a band of radius f(x) and slant length ds.

Example 13

The sphere: f = √(r² − x²), f√(1 + f′²) = √(r² − x²) · r/√(r² − x²) = r. S = 2π ∫_−r^r r dx = 4πr² ✓. And the cone y = x, 0 ≤ x ≤ 1: S = 2π ∫₀¹ x√2 dx = π√2 ≈ 4.443.

What about parametric and polar curves?

• Parametric curve (x(t), y(t)), a ≤ t ≤ b: arc length L = ∫ √(x′² + y′²) dt; area under the curve ∫ y x′ dt.
• Polar curve r = r(θ): area of a sector region (1/2) ∫ r² dθ; arc length ∫ √(r² + r′²) dθ.

Example 14: area of a circle by polar integral

r = R, 0 ≤ θ ≤ 2π: A = (1/2) ∫ R² dθ = πR² ✓.

Example 15: a cycloid

x = t − sin t, y = 1 − cos t, 0 ≤ t ≤ 2π: x′ = 1 − cos t, y′ = sin t, x′² + y′² = 2 − 2 cos t = 4 sin²(t/2). L = ∫₀^2π 2 sin(t/2) dt = [−4 cos(t/2)]₀^2π = 8. The area under one arch: ∫ y x′ dt = ∫₀^2π (1 − cos t)² dt = 3π.

Example 16: a cardioid

r = 1 + cos θ: A = (1/2) ∫₀^2π (1 + cos θ)² dθ = (1/2)(2π + π) = 3π/2.

Table of results

Where do these methods come from?

Archimedes computed the area of a parabolic segment, the volume of a sphere and the surface area of a sphere by methods that are essentially integration. Kepler and Cavalieri developed volume computations in the seventeenth century (Cavalieri’s principle), and Pappus’s theorems on centroids of solids of revolution were known earlier. The arc length formula was found by Hendrik van Heuraet and William Neile in the 1650s for the semicubical parabola, the first rectification of an algebraic curve other than a circle.

Where are they used?

Engineers compute the volume of a tank of given shape, the length of a cable or a road, and the surface area to be painted. Volumes of solids of revolution are used in the design of vases, lenses and turbine parts, and the cross-section method is used in medical imaging to estimate the volume of organs from slices. Arc length integrals describe the distance travelled along a trajectory.

Which mistakes are common?

Subtracting in the wrong order

The upper curve minus the lower curve gives a positive area. If the curves cross, split the interval.

Forgetting to square in the disc formula

V = π ∫ f² dx, with the square of the radius, and the factor π.

Using washers incorrectly

The washer volume is π ∫ (f² − g²) dx, not π ∫ (f − g)² dx.

Forgetting the square root in arc length

L = ∫ √(1 + f′²) dx. Setting L = ∫ (1 + f′) dx is wrong.