Double integral

If \(f\ge0\), the double integral is the volume of the solid below the graph of \(z=f(x,y)\) and above \(D\). It is approximated by splitting \(D\) into cells of area \(\Delta A_i\), choosing a sample point \((x_i^*,y_i^*)\) in each, and adding prisms of volume \(f(x_i^*,y_i^*)\,\Delta A_i\). If \(f\) is continuous, the limit exists and does not depend on how the points are chosen.

Where \(f\) is negative, the volume counts with a negative sign. With \(f=1\) the integral is the area of \(D\), and dividing by that area gives the average value of \(f\) on \(D\). The integral is linear, \(\iint_D(af+bg)\,dA=a\iint_D f\,dA+b\iint_D g\,dA\); additive, \(\iint_D=\iint_{D_1}+\iint_{D_2}\) when \(D_1\) and \(D_2\) do not overlap; and order-preserving: if \(f\le g\), then \(\iint_D f\,dA\le\iint_D g\,dA\).

Guiding example: for \(f(x,y)=x+y\) on \([0,2]\times[0,2]\), the sum with sample points at the centers of four squares gives \(8\), the exact value. Corner sums give \(4\) or \(12\), a lower and an upper bound that approach \(8\) as the partition is refined. Computing the limit from the definition is impractical; Fubini's theorem reduces it to two single-variable integrals.

Typical mistakes: forgetting the area element \(dA\), confusing the double integral with a product of two single integrals (valid only for functions of the form \(g(x)h(y)\) on a rectangle), or believing the integral is always positive.