Double integral

Definition

Let $R = [a, b] \times [c, d]$ with $a < b$ and $c < d$, and let $f \colon R \to \mathbb{R}$ be bounded. Choose partitions $a = x_0 < \dots < x_k = b$ and $c = y_0 < \dots < y_l = d$, and let $R_{ij} = [x_{i-1}, x_i] \times [y_{j-1}, y_j]$ with area $\Delta A_{ij} = (x_i - x_{i-1})(y_j - y_{j-1})$. For points $p_{ij} \in R_{ij}$, the sum $S = \sum_{i, j} f(p_{ij}) \, \Delta A_{ij}$ is a Riemann sum of $f$. The function $f$ is Riemann integrable on $R$ if there is $I \in \mathbb{R}$ with the following property. For each $\varepsilon > 0$ there is $\delta > 0$ such that $|S - I| < \varepsilon$ for all such sums whose rectangles have diameter less than $\delta$. Then $I$ is the double integral of $f$ over $R$, written $\iint_R f(x, y) \, dA$ or $\iint_R f(x, y) \, dx \, dy$.

Remarks

Each continuous function on $R$ is Riemann integrable on $R$. For a bounded region $D \subseteq R$, one integrates $f \mathbf{1}_D$ over $R$, where $\mathbf{1}_D$ is $1$ on $D$ and $0$ elsewhere. This gives integrals over non-rectangular regions. In practice one computes double integrals as iterated integrals, by Fubini's theorem. For example, $\iint_{[0, 1]^2} x y \, dA = \int_0^1 \int_0^1 x y \, dy \, dx = 1/4$. The integral is linear in $f$, and $\iint_R 1 \, dA = (b - a)(d - c)$ is the area of $R$.

Explanations

L1 human

Cut the rectangle into small pieces, multiply the area of each piece by a value of $f$ in it, and add. As the pieces get smaller, these sums approach the double integral. For $f \ge 0$ this is the volume under the graph of $f$.