Double Integrals and Accumulation

Weight every sample by its area

Divide a planar region into small cells of area $\Delta A_i$ . A sum $\sum f(x_i,y_i)\Delta A_i$ approximates an accumulated quantity. Under suitable integrability assumptions, its limit is $\iint_R f\,dA$ . For nonnegative height this is volume under a surface; for density it is mass; signed values represent signed accumulation.

For continuous $f$ on a rectangle, Fubini's theorem permits evaluation in either iterated order. In the inner integral, the other variable is a constant. More general Fubini statements require integrability conditions; singular signed integrands cannot be rearranged casually.

Worked example: a rectangular region

On $0\le x\le2$ , $0\le y\le1$ ,

\int_0^2\int_0^1(x+2y)\,dy\,dx=\int_0^2(x+1)dx=4.

Reversing order gives $\int_0^1\int_0^2(x+2y)dxdy=\int_0^1(2+4y)dy=4$ . The area is $2$ , so the average value of the integrand is $4/2=2$ .

Explore

Area cells and accumulated height

Try this. Increase the cell count to approach 2/3. Compare lower-left, midpoint and upper-right sampling; the corner sums bracket the integral for this increasing function.

Cells per side4Sample in each cell

f(x,y) = x² + y² on [0,1] × [0,1]. 4 × 4 cells; area per cell = 0.063. midpoint sum = 0.656; exact integral = 2/3 ≈ 0.667. Signed error = -0.01. Darker cells have larger sampled height (scale 0 to 2). Left/right sample the lower-left/upper-right corner.

This explorer uses the different model $f=x^2+y^2$ on the unit square, whose exact integral is $2/3$ . Cell area changes with resolution; sample height alone is not a cell's contribution. For this coordinatewise increasing function, lower-left and upper-right samples bracket the integral.

Worked example: symmetry and sign

For $f(x,y)=x$ on $[-1,1]\times[0,2]$ , symmetric positive and negative contributions cancel, so the integral is zero. The integral of $|x|$ is instead $2$ . A zero signed integral does not imply the surface has zero height or the geometric volume is zero.

The average of $f$ over a region of positive area is $\bar f=(1/\operatorname{area}(R))\iint_Rf\,dA$ . If density varies, a mass-weighted average uses density in both numerator and denominator; it is not the same as an unweighted area average.

Practice

Integrate $xy$ on $[0,1]\times[0,2]$ .
Integrate the constant $3$ over a region of area $5$ .
What are the units of a double integral of density in kg/m² over a region measured in m²?

Show worked solutions

The separated product is $(\int_0^1x dx)(\int_0^2y dy)=1$ .
$3\cdot5=15$ .
Kilograms, because density is multiplied by area.

Further study

MIT OpenCourseWare: Multivariable Calculus provides a full university course with additional lectures and exercises.