Double integrals and iterated integration

All rectangle sides here are ordered with $a<b$ and $c<d$; for a simple region its continuous bounds satisfy $g_1\le g_2$ everywhere. The complete Darboux and slicing proofs below preserve these assumptions.

From area sums to accumulated quantity

If $f(x,y)$ is a density per unit area, multiplying a sampled density by a small area estimates the quantity in that cell. Add over cells and refine. The result is signed if $f$ can be negative: it is not automatically a physical mass or an unsigned volume.

Definition

Double integral on a rectangle

Let $R=[a,b]\times[c,d]$. Partition $R$ into subrectangles $R_i$ of areas $\Delta A_i$ and choose $p_i\in R_i$. If the sums $\sum_i f(p_i)\Delta A_i$ approach the same limit as the maximum subrectangle diameter tends to zero, regardless of the choices, that limit is $\iint_R f\,dA$. Every continuous function on a closed rectangle is integrable.

For a bounded region $D\subset R$ with suitable boundary, define the integral over $D$ as the integral of $f\mathbf1_D$ over $R$. This accommodates regions described by inequalities; sufficiently regular boundaries have area zero, so changing values only on the boundary does not alter the integral. The integral of $1$ over $D$ gives its area. For $f\ge0$, $\iint_D f\,dA$ is the volume under the graph above $D$.

Theorem

Fubini and slicing

If $f$ is continuous on a rectangle $R=[a,b]\times[c,d]$, then

$$\iint_R f\,dA=\int_a^b\left(\int_c^d f(x,y)\,dy\right)dx =\int_c^d\left(\int_a^b f(x,y)\,dx\right)dy.$$

For continuous $f$ on a regular region $D=\{a\le x\le b,\ g_1(x)\le y\le g_2(x)\}$ bounded by continuous functions, the first formula becomes

$$\iint_D f\,dA=\int_a^b\int_{g_1(x)}^{g_2(x)}f(x,y)\,dy\,dx.$$

The analogous horizontal-slice formula applies when $D$ is described by $c\le y\le d$, $h_1(y)\le x\le h_2(y)$.

The inner differential is integrated first. Its bounds may depend on variables held fixed, while the outer bounds are constants in this two-variable setup. A region may need splitting into pieces before one description is valid. These convenient sufficient assumptions do not cover every possible integrable function or irregular region.

Proof

Why slicing is natural

For a rectangle, group the two-dimensional Riemann sums by vertical columns. Within each column, the inner sum approximates the integral in $y$ for a fixed $x$, and the remaining sum approximates the outer integral. Continuity on the compact rectangle supplies the uniform control needed for the limiting argument. For a simple region, integrate the indicator-extended function over a rectangle; slices outside the lower and upper boundaries contribute zero. This explains the formulas; the theorem supplies the full convergence justification.

Worked example

Draw the region before computing

Let $D$ be between $y=x^2$ and $y=x$ for $0\le x\le1$. For $f(x,y)=x+y$,

$$\iint_D(x+y)\,dA=\int_0^1\int_{x^2}^x(x+y)\,dy\,dx =\int_0^1\left(\frac32x^2-x^3-\frac12x^4\right)dx=\frac3{20}.$$

Horizontal slices have $0\le y\le1$ and $y\le x\le\sqrt y$, so the same integral is $\int_0^1\int_y^{\sqrt y}(x+y)\,dx\,dy$. The inequalities reverse roles; simply exchanging the differentials without changing the bounds changes the region.

Worked example

Switch order to make an integral elementary

Consider $I=\int_0^1\int_x^1 e^{y^2}\,dy\,dx$. The inner antiderivative is not elementary. The region is $0\le x\le y\le1$, so

$$I=\int_0^1\int_0^y e^{y^2}\,dx\,dy =\int_0^1 ye^{y^2}\,dy=\frac{e-1}{2}.$$

The order change follows from continuity and the correct region description, not a formal interchange of symbols.

Let $R=[a,b]\times[c,d]$ with $a<b,c<d$. A grid consists of partitions of each coordinate interval. Its cells are closed subrectangles with disjoint interiors. For bounded $f:R\to\mathbb R$, let $m_{ij},M_{ij}$ be its infimum and supremum on cell $R_{ij}$, of area $\Delta A_{ij}$. Define $L(f,P)=\sum m_{ij}\Delta A_{ij}$ and $U(f,P)=\sum M_{ij}\Delta A_{ij}$.

1. The definition and its useful criterion

Refining a grid raises lower sums and lowers upper sums. Given any two grids, their common refinement proves every lower sum is at most every upper sum. Thus the lower integral $\underline I=\sup_P L(f,P)$ is at most the upper integral $\overline I=\inf_P U(f,P)$, and both are finite by boundedness. The function is Riemann integrable when these numbers agree, with common value $\iint_R f$.

Theorem: rectangular Darboux criterion

The bounded function is integrable if and only if for every $\varepsilon>0$ some grid has $U(f,P)-L(f,P)<\varepsilon$.

Proof. If both integrals equal $I$, choose a lower sum above $I-\varepsilon/2$ and an upper sum below $I+\varepsilon/2$, then refine both grids. The new gap is below epsilon. Conversely any such grid bounds $0\le\overline I-\underline I\le U-L$. Arbitrarily small gaps force equality. The proof is the supremum argument in two dimensions, not an assumption of Fubini's theorem.

Theorem: continuity implies proper integrability

If $f$ is continuous on $R$, it is bounded and uniformly continuous there. Write $A=(b-a)(d-c)>0$. Choose a distance scale making output differences less than $\varepsilon/(2A)$. A sufficiently fine grid has every cell diameter smaller than that scale, so each oscillation $M_{ij}-m_{ij}$ is at most $\varepsilon/(2A)$. Therefore $U-L\le\varepsilon/2<\varepsilon$, proving integrability.

For any tags $\xi_{ij}\in R_{ij}$, the corresponding finite sum lies between the grid's lower and upper sums. Thus whenever the grid gap is small, its tagged approximation is close to the integral. This statement is a consequence of the definition, not a new limiting convention.

2. Prove the continuous rectangle Fubini theorem

Theorem

For continuous $f$ on $R$,

$$\iint_R f(x,y)\,dA =\int_a^b\left(\int_c^df(x,y)\,dy\right)dx =\int_c^d\left(\int_a^bf(x,y)\,dx\right)dy.$$

First, the iterated integrals exist. Each one-variable slice is continuous and integrable. Let $g(x)=\int_c^df(x,y)dy$. Uniform continuity of $f$ gives

$$|g(x)-g(x_0)|\le(d-c)\sup_{y\in[c,d]}|f(x,y)-f(x_0,y)|\longrightarrow0.$$

Hence $g$ is continuous and integrable. The reversed inner-integral function is continuous by the same argument.

Next, squeeze them between the two-dimensional sums. On every cell, $m_{ij}\le f(x,y)\le M_{ij}$. For a fixed $x$ in its $i$th strip, integrate these bounds over each $y$ interval and add:

$$\sum_jm_{ij}\Delta y_j\le g(x)\le\sum_jM_{ij}\Delta y_j.$$

Integrate over that $x$ strip and sum over $i$. The result is $L(f,P)\le\int_a^bg(x)dx\le U(f,P)$. The double integral lies between the same bounds. Grids with arbitrarily small gap therefore force equality. Repeating with coordinates reversed proves the second equality. No order of an unproved infinite double sum was exchanged; only finite sums, ordinary integral inequalities and the Darboux criterion were used.

A separable continuous integrand $f(x,y)=p(x)q(y)$ consequently has integral $(\int p)(\int q)$. This is an integration identity under the stated hypotheses, not a claim about probabilistic independence in an unspecified experiment.

3. Why ordinary curved boundaries do not destroy proper integrability

Let $D=\{a\le x\le b,\ell(x)\le y\le u(x)\}$ with continuous $\ell\le u$, and let $f$ be continuous on $D$. This is a compact set. Choose a bounding rectangle $R=[a,b]\times[c,d]$ with $c\le\ell(x)\le u(x)\le d$. The clamping map

$$p(x,y)=\bigl(x,\min(u(x),\max(\ell(x),y))\bigr)$$

is continuous, maps $R$ into $D$ and fixes every point of $D$. Consequently $\widetilde f=f\circ p$ is a continuous extension to $R$. In the estimates below write $f$ for this extension. Now set it to zero outside $D$ within $R$. This zero extension may jump across the boundary, so one cannot cite continuity everywhere.

Instead control boundary cells. In a vertical grid strip, uniform continuity makes the oscillation of either boundary graph as small as desired. The cells meeting that graph can be covered by a rectangle whose height is at most that oscillation plus two vertical mesh widths. Summing strip widths shows the total area of these graph-boundary cells tends to zero as both mesh widths tend to zero. The two vertical end boundaries can likewise be covered by strips of arbitrarily small total area.

If $|f|\le M$, boundary cells contribute at most $2M$ times their total area to the upper-lower gap. Cells away from the boundary lie wholly inside or wholly outside the region; inside, uniform continuity of $f$ makes their oscillations small, and outside the function is zero. Choose the boundary area small first and then the ordinary cell oscillation small. The total gap becomes arbitrarily small, proving integrability of the zero extension and therefore of $f$ over $D$.

The same bounds show that its integral is $\int_a^b\int_{\ell(x)}^{u(x)}f(x,y)dydx$: on each strip, changing the vertical endpoints and the $x$ input changes the inner integral by uniformly small amounts, while the vanishing boundary-cell area controls the endpoints. Explicitly, using the bounding rectangle's vertical span $H$,

$$|g(x)-g(x_0)|\le H\sup_y|f(x,y)-f(x_0,y)|+M(|\ell(x)-\ell(x_0)|+|u(x)-u(x_0)|),$$

so the inner-integral function is continuous, and the finite-strip Darboux squeeze applies after excluding the already controlled boundary cells. This establishes the common continuous-simple-region case. More general region or improper versions require their own hypotheses and are not inferred from this proof.

Properties and method selection

Linearity and additivity over nonoverlapping pieces reduce complicated expressions or regions. If $f\le g$ throughout $D$, then $\iint_Df\le\iint_Dg$; in particular $|\iint_Df|\le\iint_D|f|$. For a continuous density $\rho\ge0$, mass is $M=\iint_D\rho\,dA$. When $M>0$, the centroid is $(\bar x,\bar y)=(M^{-1}\iint_Dx\rho\,dA,M^{-1}\iint_Dy\rho\,dA)$.

Use vertical slices when their limits are simpler; use horizontal slices when they simplify the region or inner antiderivative. Split the region only when a single pair of bounds cannot describe all slices. Polar or other changes of variables can simplify circular or transformed regions, but require a Jacobian and a separate change-of-variables theorem; they are not automatic replacements for $dx\,dy$.

Practice

1. Evaluate $\iint_{[0,2]\times[0,1]}(x+2y)\,dA$.
2. Reverse the order in $\int_0^1\int_x^1 f(x,y)\,dy\,dx$ and state the region.
3. Compute the area between $y=x^2$ and $y=x$ for $0\le x\le1$.
4. A lamina occupies $[0,1]^2$ with density $\rho(x,y)=1+x$. Find its mass and centroid.

Full solutions

1. $\int_0^2\int_0^1(x+2y)\,dy\,dx=\int_0^2(x+1)\,dx=4$.
2. The region is $0\le x\le y\le1$. Horizontal slices give $\int_0^1\int_0^y f(x,y)\,dx\,dy$.
3. The area is $\int_0^1\int_{x^2}^x1\,dy\,dx=\int_0^1(x-x^2)\,dx=1/6$.
4. $M=\int_0^1\int_0^1(1+x)\,dy\,dx=3/2$. The first moment in $x$ is $\int_0^1x(1+x)dx=5/6$, and the first moment in $y$ is $(\int_0^1(1+x)dx)(\int_0^1y\,dy)=3/4$. Therefore $(\bar x,\bar y)=(5/9,1/2)$.

Synthesis

The integral accumulates a quantity per unit area, while an iterated integral computes the same accumulation slice by slice under suitable hypotheses. Region geometry determines bounds; a reversed order must describe exactly the same set. Later coordinate changes and vector integral theorems build on this accounting of geometry and orientation.