A single integral ∫ₐᵇ f(x)dx adds up infinitely thin strips over an interval. A double integral ∫∫_R f(x,y)dA does the same thing over a two-dimensional region — it sums up infinitely thin columns of height f(x,y) over every point (x,y) in R. The result is the volume under the surface z=f(x,y) above the region R.
The key theorem that makes them computable is Fubini's Theorem: for well-behaved functions, you can evaluate the double integral by doing two nested single integrals, one after the other.
What is a Double Integral?
∫∫_R f(x,y) dA gives the volume under the surface z=f(x,y) above the region R. Like a single integral sums infinitely thin strips, the double integral sums infinitely thin prisms of volume f(x,y)·dA.
Iterated Integrals — Fubini's Theorem
- For a rectangular region R=[a,b]×[c,d], Fubini's Theorem says you can compute the double integral as two nested single integrals in either order:
- ∫∫_R f dA = ∫ₐᵇ (∫ᶜᵈ f(x,y)dy) dx = ∫ᶜᵈ (∫ₐᵇ f(x,y)dx) dy.
Non-Rectangular Regions
For regions bounded by curves, set up the inner limits as functions of the outer variable. Type I (vertical strips): x from a to b, y from g₁(x) to g₂(x). Type II (horizontal strips): y from c to d, x from h₁(y) to h₂(y).
Changing to Polar Coordinates
When the region R is a circle or sector, polar coordinates simplify the integral. x=r·cos θ, y=r·sin θ, dA=r·dr·dθ. The extra r factor is the Jacobian of the transformation.
The double integral ∬_R f(x,y) dA gives the volume under the surface z = f(x,y) above the region R. It is computed as an iterated integral — two successive single integrals — via Fubini's Theorem.
Building the Double Integral from Riemann Sums
Partition the region R into small rectangles of area ΔA = ΔxΔy. At each rectangle, f(xᵢ, yⱼ) gives an approximate height. The volume of that thin rectangular prism is f(xᵢ, yⱼ)·ΔA. Summing over all rectangles and taking the limit as ΔA → 0 gives the double integral ∬_R f(x,y)dA. This is the direct 2D analogue of the Riemann sum construction for single integrals.
Fubini's Theorem — Reducing to Two Single Integrals
For a rectangular region R = [a,b]×[c,d]: ∬_R f(x,y)dA = ∫ₐᵇ[∫ᶜᵈ f(x,y)dy]dx = ∫ᶜᵈ[∫ₐᵇ f(x,y)dx]dy. You can integrate in either order, as long as the limits are adjusted correctly. The inner integral is computed first (treating the outer variable as constant), then the outer integral evaluates the result.
Non-Rectangular Regions — Setting Up Limits
For a region bounded by curves, the inner limits become functions of the outer variable. Type I (vertical strips): x runs from a to b (constants), y runs from g₁(x) to g₂(x) (functions of x). ∬_R f dA = ∫ₐᵇ∫_{g₁(x)}^{g₂(x)} f(x,y)dy dx. Type II (horizontal strips): y runs from c to d, x from h₁(y) to h₂(y). Choosing the right type depends on which gives simpler limits — always sketch the region first.
Polar Coordinates for Double Integrals
When the region R is a circle, disk, or annulus, polar coordinates dramatically simplify the computation. Substituting x = r cos θ, y = r sin θ, and — critically — dA = r dr dθ (the Jacobian factor r is essential). ∬_R f(x,y)dA = ∫∫ f(r cos θ, r sin θ)·r dr dθ. Example: ∬_{x²+y²≤1} e^(x²+y²)dA = ∫₀^(2π)∫₀¹ e^(r²)·r dr dθ = 2π·[e^(r²)/2]₀¹ = π(e−1).
Applications
- Volume: V = ∬_R f(x,y)dA gives the volume under z = f(x,y) above region R.
- Mass: If ρ(x,y) is surface density, mass M = ∬_R ρ(x,y)dA.
- Centre of mass: x̄ = (1/M)∬_R x·ρ dA, ȳ = (1/M)∬_R y·ρ dA.
- Probability: For a joint density f(x,y): P((x,y)∈R) = ∬_R f(x,y)dA.
Switching the Order of Integration
Sometimes the inner integral is impossible to evaluate as set up, but switching the order of integration makes it tractable. To switch: draw the region R, rewrite it as a Type II region (horizontal strips) instead of Type I (vertical strips), and re-express the limits accordingly. Classic example: ∫₀¹∫ₓ¹ e^(y²)dy dx. The inner integral ∫e^(y²)dy has no closed form. Switching: the region is 0≤x≤y, 0≤y≤1. New integral: ∫₀¹∫₀ʸ e^(y²) dx dy = ∫₀¹ y·e^(y²) dy = [e^(y²)/2]₀¹ = (e−1)/2. Switching the order converted an impossible integral into a straightforward one.
Computing Double Integrals Over Non-Rectangular Regions
Triple Integrals
Triple integrals extend double integrals to three dimensions: ∭_V f(x,y,z)dV. Evaluated as three nested single integrals. Used to compute: mass of a 3D object with density ρ(x,y,z) — M = ∭ρ dV; volume of a 3D region — V = ∭1 dV; centre of mass; and moment of inertia. Cylindrical coordinates (r,θ,z): dV = r dr dθ dz. Spherical coordinates (ρ,φ,θ): dV = ρ²sin(φ) dρ dφ dθ. Choosing the right coordinate system for the region's geometry is the key to tractable evaluation.
- Stewart, J. (2015). Multivariable Calculus, §15.1–15.4. Cengage.
- Apostol, T. (1969). Calculus, Vol. 2, Ch. 11. Wiley.
- Marsden, J. & Tromba, A. (2011). Vector Calculus, Ch. 5. Freeman.