Double and multiple integrals

In Cartesian coordinates dA=dx dy is an infinitesimal rectangle. The integral sums f(x,y)·Δx·Δy over rectangles and takes the limit to obtain the exact total. In polar coordinates the area element is r dr dθ because the area of a small sector grows with radius. The element dA carries area units, so an areal density integrates to a total quantity. In polar coordinates, dA=r dr dθ; the factor r is the Jacobian and must not be omitted. A quick unit check shows whether, for example, mass per area really becomes mass.

For continuous functions on a bounded region, Fubini’s theorem permits switching the order. Improper or singular integrals require additional conditions. In practice, recompute the bounds after switching: the region is unchanged, but its inner/outer description changes. Switching the order is useful when it makes the inner bounds simpler. Sketch the region, mark its intersections and write both descriptions carefully. For improper integrals, switching is not automatic; convergence and integrability conditions must be checked. Fubini’s theorem safely applies to continuous functions on bounded rectangles. For complicated regions, a sketch helps read reversed bounds without gaps.

If f(x,y) is nonnegative, ∬_D f dA gives the volume between z=f(x,y) and the base region D. If f is an areal density, the same integral gives total mass. Its meaning depends on the units and interpretation of the integrand. For a density ρ(x,y), total mass is M=∬ρ dA. For a probability density, integrating over a subregion gives its probability; over the full support the integral must equal one. This interpretation provides a useful plausibility check. A quick special case is f(x,y)=1: the integral then equals the area of the region and also checks the chosen bounds.

Sketch the region and describe it using vertical or horizontal strips. For 0≤x≤1 and 0≤y≤x, write ∫₀¹∫₀ˣ f(x,y)dy dx. Reversing the order gives 0≤y≤1 and y≤x≤1. A sketch helps avoid missing or double-counting regions. To reverse the bounds, describe every point in the region again, using the other coordinate as the outer variable. A triangular region may then require several integrals. If so, split the sketch where the upper or lower boundary curve changes. For the triangular region in the example, integrating f=1 gives area 1/2; this checks both descriptions of the bounds.

Compare the mean integrand value times area. For x+y on [0,2]×[0,1], the mean is 1.5 and the area is 2, so the integral should be 3. Also check dimensions: a density in kg/m² integrated over m² must yield mass in kg. For nonnegative f, the integral lies between the minimum times area and the maximum times area. This rough bound catches many missing factors or incorrect limits. If f changes sign, use absolute values or split the region into positive and negative contributions.