From a double Riemann sum to a volume
Just as the definite integral was born from
Riemann sums,
the double integral is the limit of a double sum. Chop a rectangle
R = [a,b] \times [c,d] into a grid of
m \times n small tiles, each of area
\Delta A = \Delta x\, \Delta y.
Step 1 — build one matchstick. Over the tile with corner
(x_i^*, y_j^*), erect a thin box of height
f(x_i^*, y_j^*). Its volume is height times base:
\text{box}_{ij} = f(x_i^*, y_j^*)\, \Delta A.
Step 2 — stack every matchstick. Add the boxes over the whole grid to
approximate the volume under the surface:
V \approx \sum_{i=1}^{m} \sum_{j=1}^{n} f(x_i^*, y_j^*)\, \Delta A.
Step 3 — refine the grid. Let the tiles shrink
(m, n \to \infty). The staircase of boxes settles onto the smooth
volume, and the double sum becomes the double integral:
\iint_R f(x, y)\, dA = \lim_{m, n \to \infty} \sum_{i=1}^{m} \sum_{j=1}^{n} f(x_i^*, y_j^*)\, \Delta A.
Elegant, but a double limit is no way to compute. The rescue is Fubini's
theorem: do one variable at a time.
Fubini's theorem: integrate one variable, then the other
Slice the solid with planes parallel to the y-axis. Each slice at
a fixed x has cross-sectional area
A(x) = \int_c^d f(x, y)\, dy — an ordinary single integral in
which x is held constant. Sweeping the slices across
[a, b] recovers the volume, exactly the
volumes-by-slicing
principle:
\iint_R f\, dA = \int_a^b A(x)\, dx = \int_a^b \!\left( \int_c^d f(x, y)\, dy \right) dx.
Fubini says the order does not matter: slice the other way and you get the same number.
Worked example — \iint_R xy\, dA over [0,1]\times[0,2]
Step 1 — write the iterated integral. Inner variable
y runs 0 \to 2, outer variable
x runs 0 \to 1:
\iint_R xy\, dA = \int_0^1 \!\left( \int_0^2 xy\, dy \right) dx.
Step 2 — do the inner integral, treating x as a
constant. Pull x out and integrate
y:
\int_0^2 xy\, dy = x \int_0^2 y\, dy = x \left[ \frac{y^2}{2} \right]_0^2 = x \cdot \frac{4}{2} = 2x.
Step 3 — feed that into the outer integral. The inner slice-area is
A(x) = 2x:
\int_0^1 2x\, dx = \left[ x^2 \right]_0^1 = 1.
So \iint_R xy\, dA = 1. (Reverse the order — inner
x, outer y — and you land on
1 again; Fubini guarantees it.)
Non-rectangular regions: Type I
Real regions are rarely tidy rectangles. A Type I region sits between two
curves of x: for a \le x \le b,
y runs from a floor g_1(x) to a ceiling
g_2(x). The inner limits now depend on
x:
\iint_R f\, dA = \int_a^b \!\left( \int_{g_1(x)}^{g_2(x)} f(x, y)\, dy \right) dx.
Worked example — the region under y = x^2 on
[0,1], with f = 1 (so we get its
area).
Step 1 — read the limits off the region. For each
x \in [0,1] the strip runs y: 0 \to x^2:
\text{Area} = \int_0^1 \!\left( \int_0^{x^2} 1\, dy \right) dx.
Step 2 — inner integral. The length of the strip is the ceiling minus the
floor:
\int_0^{x^2} 1\, dy = x^2 - 0 = x^2.
Step 3 — outer integral. Sum the strip lengths across
x:
\int_0^1 x^2\, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3}.
The region has area \tfrac13. (A Type II region
flips the roles — y outer, with
x running between curves h_1(y) and
h_2(y) — and you slice horizontally instead.)
Let f be continuous on the region R.
Then the double integral equals either iterated integral, and the order may be swapped:
-
Rectangle R = [a,b]\times[c,d]:
\iint_R f\, dA = \int_a^b \!\int_c^d f\, dy\, dx = \int_c^d \!\int_a^b f\, dx\, dy.
-
Type I (g_1(x) \le y \le g_2(x)):
\displaystyle \iint_R f\, dA = \int_a^b \!\int_{g_1(x)}^{g_2(x)} f\, dy\, dx.
-
Type II (h_1(y) \le x \le h_2(y)):
\displaystyle \iint_R f\, dA = \int_c^d \!\int_{h_1(y)}^{h_2(y)} f\, dx\, dy.
-
Special case f \equiv 1:
\iint_R 1\, dA = \operatorname{area}(R).
Fubini's promise — both orders give the same answer — is a practical superpower, because
one order can be elementary while the other is impossible. The classic is
\int_0^1 \!\int_x^1 e^{y^2}\, dy\, dx.
The inner integral \int e^{y^2}\, dy has no elementary
antiderivative — dead on arrival. But the region
\{ 0 \le x \le 1,\ x \le y \le 1 \} is the triangle below the
line y = x's mirror; redescribe it as Type II
(0 \le y \le 1, 0 \le x \le y) and
swap:
\int_0^1 \!\int_0^y e^{y^2}\, dx\, dy = \int_0^1 y\, e^{y^2}\, dy = \left[ \tfrac12 e^{y^2} \right]_0^1 = \tfrac12 (e - 1).
Now the inner integral is trivial (x is absent, so it just
contributes a factor y) and the
substitution
u = y^2 finishes it. Same region, same volume — only the order
changed. Reading the region correctly is the whole game.