Double integrals, accumulation and density

Try

A rectangular sheet occupies 0≤x≤3 and 0≤y≤2, with density ρ(x,y)=1+x in mass units per square length unit. The left edge has density 1 and the right edge density 4. Approximate the mass by dividing the sheet into small rectangles, taking a representative density in each and multiplying by that rectangle’s area. Adding those pieces explains why both a function value and an area element belong in a double integral.

Predict

Predict whether the mass is 6, 12 or 15. The area is 6, but density is not uniformly one. The average density in this linear example is (1+4)/2=5/2, so mass should be 15. This shortcut relies on the particular linear dependence and uniform rectangular geometry. For a general density, averaging a few corner values is not an integration rule with automatic exactness.

Experiment

Compute directly: m=∫₀³∫₀²(1+x)dy dx. The inner integral is 2(1+x); the outer integral is 2[x+x²/2]₀³=15. Reverse the order and verify the same result. Then replace density by x²+1. The mass becomes 2∫₀³(x²+1)dx=24, whereas averaging endpoint densities would predict 33. Record why the earlier shortcut failed.

A partition table can display each strip’s interval, representative density and area. Refine the strips and compare approximate sums with the exact integral, keeping approximation and equality visibly distinct.

Understand

The double integral ∫∫R f dA is the limiting signed accumulation of function values times small areas, when that limit exists. For continuous f on a compact rectangle, Fubini’s theorem permits either order of iterated integration. Each inner integral treats the outer variable as fixed.

Physical mass uses nonnegative density. A signed function can instead represent gain minus loss or signed height. The integral of f is not necessarily the geometric volume between its graph and the plane z=0: that unsigned volume uses |f|. Average value over a region of positive finite area is (1/area(R))∫∫R f dA. The denominator matters; an integral alone carries accumulation units, whereas an average carries the original function’s units.

Apply

For the worked sheet, find the center of mass in the x-direction. Its first moment is ∫∫xρ dA=2∫₀³(x+x²)dx=27. Dividing by mass 15 gives x̄=9/5, to the right of the geometric midpoint 3/2. Symmetry in y gives ȳ=1. The density is larger on the right, so the direction of the shift is a useful independent check.

Repair “the center of mass is obtained by averaging x and y coordinates of boundary points.” Boundary samples do not represent the sheet’s area or density. A center of mass is a density-weighted average over the entire material. Even uniform density requires accounting for the shape’s area distribution.

Continue

As optional depth, compare Riemann sums with numerical quadrature and ask what information would bound the approximation error. A finite grid can miss narrow features, so refinement and regularity assumptions matter. The present exact examples keep the integration mechanism visible without turning a grid into proof.

Write an accumulation setup with region, integrand, differential element and units on separate lines. Explain what each factor contributes. Then calculate a corresponding average and identify its normalization. In the next lesson, the region will no longer be rectangular, so the geometry will determine variable-dependent inner bounds and sometimes require splitting the integral.