Fubini's theorem

1. Context: From Definition to Calculation

The definition of the double integral over a region is based on the Riemann Sum limit.
For a function $f (x, y)$ over a rectangle $R = [a, b] \times [c, d]$ , we partition the area into small sub-rectangles $Δ A_{i j}$ and take the limit:

\iint_{R} f (x, y) d A = lim_{∥ P ∥ \to 0} \sum_{i, j} f (x_{i}^{*}, y_{j}^{*}) Δ A_{i j}

Handling General Regions

If the region of integration $Ω$ is not a rectangle (e.g., a circle or triangle), we enclose it in a larger rectangle $R$ and extend the function definition:

\tilde{f} (x, y) = {\begin{cases} f (x, y) & if (x, y) \in Ω \\ 0 & if (x, y) \in R ∖ Ω \end{cases}

Thus, $\iint_{Ω} f d A = \iint_{R} \tilde{f} d A$ .

2. Fubini's Theorem

While Riemann sums define what the integral is, Fubini's Theorem tells us how to calculate it. It states that the double integral can be computed as iterated integrals.
Statement:
If $f$ is bounded and integrable (continuous almost everywhere) on the rectangle $R = [a, b] \times [c, d]$ , then:

\iint_{R} f (x, y) d A = \int_{a}^{b} [\int_{c}^{d} f (x, y) d y] d x = \int_{c}^{d} [\int_{a}^{b} f (x, y) d x] d y

Technical Nuance: Continuity vs. Integrability

Introductory texts often state Fubini requires $f$ to be continuous. However, strictly speaking, $f$ only needs to be Riemann integrable.
When we extend $f$ to a rectangle (setting it to 0 outside $Ω$ ), the new function $\tilde{f}$ usually becomes discontinuous at the boundary of $Ω$ .

Since the boundary is a curve (1D object in 2D space), it has zero area (measure zero).
Riemann integration ignores discontinuities that occur on sets of measure zero. Therefore, Fubini still applies.

Key Implications

Reduction of Dimension: We reduce a 2D problem into two 1D problems.
Order Independence: We can swap the order of integration ( $d y d x$ vs $d x d y$ ) without changing the result, provided the limits of integration are adjusted accordingly.

3. Geometric Intuition: The "Slicing" Method

We can interpret the inner integral as the cross-sectional area function $A (x)$ .

Fix $x$ constant. The inner integral $\int_{c}^{d} f (x, y) d y$ calculates the area of the "slice" of the solid at that position $x$ . Let's call this $A (x)$ .
The outer integral $\int_{a}^{b} A (x) d x$ sums up all these areas to form the total volume.