In Calculus 1 the definite integral was built from thin rectangles under a curve. A surface $z = f(x, y)$ over a rectangle $R$ in the plane encloses a solid, and its volume is built the same way from thin boxes.
Key idea
Divide $R$ into small rectangles of area $\Delta A$, pick a sample point $(x_{ij}^*, y_{ij}^*)$ in each, and add up box volumes:
$$\iint_R f(x, y)\,dA = \lim_{m, n \to \infty} \sum_{i=1}^{m}\sum_{j=1}^{n} f(x_{ij}^*, y_{ij}^*)\,\Delta A.$$
If $f \ge 0$ this is the volume under the surface; parts below the $xy$-plane count as negative. The average value of $f$ on $R$ is $\frac{1}{\text{area}(R)}\iint_R f\,dA$.
Why it works
Over one small rectangle the surface is nearly flat, so the solid above it is nearly a box with base $\Delta A$ and height $f$ at the sample point. Each box is slightly too tall or too short, but for a continuous $f$ the surface varies less and less across a smaller rectangle, so the total error goes to ${}0$ as the grid gets finer. The sums settle on one number, the double integral. Sums split and constants factor out, so the integral is linear: $$\iint (af + bg)\,dA = a\iint f\,dA + b\iint g\,dA.$$
Worked examples
Worked example: The midpoint rule
Estimate $\iint_R (x + 2y)\,dA$ on $R = [0, 2] \times [0, 2]$ with a ${}2 \times 2$ grid and midpoints.
- Four unit squares, so $\Delta A = 1$. Midpoints: $(0.5, 0.5)$, $(1.5, 0.5)$, $(0.5, 1.5)$, $(1.5, 1.5)$.
- Values: ${}1.5$, ${}2.5$, ${}3.5$, ${}4.5$, with sum ${}12$.
- Estimate $= 12 \cdot 1 = 12$. (For a function like this, whose graph is a plane, the midpoint rule is exact.)
Answer: ${}12$
Worked example: Average value
Find the average value of $f(x, y) = xy$ on $R = [0, 2] \times [0, 4]$.
- $$\iint_R xy\,dA = \left(\int_0^2 x\,dx\right)\left(\int_0^4 y\,dy\right) = 2 \cdot 8 = 16$$ (the integral of a product $g(x)h(y)$ over a rectangle splits).
- Area $= 8$, so the average is $\frac{16}{8} = 2$.
- A flat box of height ${}2$ over $R$ holds the same volume as the region under the surface.
Answer: ${}2$
Worked example: Linearity
If $\iint_R f\,dA = 5$ and $\iint_R g\,dA = -2$ on a rectangle of area ${}3$, find $\iint_R (3f - 2g + 1)\,dA$.
- Split: ${}3(5) - 2(-2) + \iint_R 1\,dA$.
- $\iint_R 1\,dA$ is the area, ${}3$. Total: ${}15 + 4 + 3 = 22$.
Answer: ${}22$
Common mistake
Forgetting $\Delta A$ in a Riemann sum. Adding the heights alone gives a number that changes when you refine the grid. Each term is a volume: height times base area. Check: doubling $m$ and $n$ should give a similar estimate, not one four times as big. Also remember that a double integral is a signed volume: if $f$ dips below zero, that part subtracts.