Double Integrals over Rectangles

The single-variable definite integral measures the area under a curve. The double integral extends that idea to two dimensions: it measures the volume under a surface.

1. Warm-up: Riemann sums in one dimension

Recall that the definite integral $\int_a^b f(x)\,dx$ is built by slicing the interval $[a,b]$ into $n$ sub-intervals of width $\Delta x = (b-a)/n$, picking a sample point $x_i^*$ in each sub-interval, and summing up rectangle areas:

$\sum_{i=1}^{n} f(x_i^*)\,\Delta x.$

As $n \to \infty$ (i.e.\ $\Delta x \to 0$), this sum converges to the exact area:

$\int_a^b f(x)\,dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*)\,\Delta x.$

The graph below shows $y = x^2$ on $[1, 2]$. Use the slider to increase $n$ and watch the Riemann rectangles fill the area more and more accurately. Try all three sampling rules (left, right, midpoint) and observe how the error shrinks as $\Delta x \to 0$.

Key formula for reference:

$\boxed{\int_a^b f(x)\,dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*)\,\Delta x.}$

2. Definition of the definite integral

Let $f$ be a bounded function on $[a,b]$. Partition $[a,b]$ into $n$ equal sub-intervals $[x_{i-1}, x_i]$ where $x_i = a + i\,\Delta x$ and $\Delta x = (b-a)/n$. Choose any sample point $x_i^* \in [x_{i-1},x_i]$.

The definite integral is defined as

$\int_a^b f(x)\,dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*)\,\Delta x,$

provided this limit exists and is independent of the choice of sample points. When $f \geq 0$, this limit equals the area of the region between the graph of $f$ and the $x$-axis.

3. From one dimension to two: volume under a surface

Now suppose $f(x,y) \geq 0$ is a function of two variables and $R$ is a rectangle in the $xy$-plane. We want to find the volume of the solid that lies above $R$ and below the surface $z = f(x,y)$.

The 2D construction. Partition $R$ into an $m \times n$ grid of small sub-rectangles $R_{ij}$, each with area $\Delta A = \Delta x\,\Delta y$. Pick a sample point $(x_{ij}^*, y_{ij}^*)$ in each $R_{ij}$. The column above $R_{ij}$ with height $f(x_{ij}^*, y_{ij}^*)$ has volume

$f(x_{ij}^*, y_{ij}^*)\,\Delta A.$

Adding up all columns gives the double Riemann sum

$\sum_{i=1}^{m}\sum_{j=1}^{n} f(x_{ij}^*, y_{ij}^*)\,\Delta A.$

The interactive graph below shows the surface $z = 6 - x^2 - y^2$ on the rectangle $R = [-1.5, 1.5] \times [-1.5, 1.5]$. The blue columns approximate the volume under the surface. Increase $n$ to see the approximation improve.

4. Definition of the double integral

Let $f$ be a bounded function on a rectangle $R = [a,b] \times [c,d]$. Partition $R$ into an $m \times n$ grid of sub-rectangles $R_{ij} = [x_{i-1},x_i] \times [y_{j-1},y_j]$, each of area $\Delta A = \Delta x\,\Delta y$.

The double integral of $f$ over $R$ is

$\boxed{\iint_R f(x,y)\,dA = \lim_{m,n \to \infty} \sum_{i=1}^{m}\sum_{j=1}^{n} f(x_{ij}^*, y_{ij}^*)\,\Delta A,}$

provided the limit exists independently of the choice of sample points. When $f(x,y) \geq 0$, the double integral equals the volume of the solid below $z = f(x,y)$ and above $R$.

Existence. If $f$ is continuous on $R$ (or has only finitely many jump discontinuities), the double integral exists.

5. Iterated integrals on a rectangle

For a rectangle $R = [a,b] \times [c,d]$ we almost never compute double integrals directly from the limit definition. Instead, we reduce them to two ordinary one-variable integrals done one at a time — called iterated integrals.

The inner integral (holding $x$ fixed) integrates $f(x,y)$ with respect to $y$ over $[c,d]$:

$A(x) = \int_c^d f(x,y)\,dy.$

$A(x)$ is the area of a cross-sectional slice of the solid at a fixed value of $x$.

The outer integral then integrates $A(x)$ over $[a,b]$:

$\int_a^b A(x)\,dx = \int_a^b \!\left[\int_c^d f(x,y)\,dy\right]\!dx.$

Fubini's Theorem. If $f$ is continuous on $R = [a,b]\times[c,d]$, then

$\boxed{\iint_R f(x,y)\,dA = \int_a^b \!\int_c^d f(x,y)\,dy\,dx = \int_c^d \!\int_a^b f(x,y)\,dx\,dy.}$

Both iterated integrals give the same value as the double integral.

Example. Compute $\displaystyle\iint_R (x + 2y)\,dA$ where $R = [1,3] \times [0,2]$.

$\int_1^3 \!\int_0^2 (x + 2y)\,dy\,dx = \int_1^3 \bigl[xy + y^2\bigr]_0^2\,dx = \int_1^3 (2x + 4)\,dx = \bigl[x^2 + 4x\bigr]_1^3 = 20 - 5 = 15.$

6. Switching the order of integration

Fubini's theorem says the two orders are equal for continuous $f$:

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

For rectangles this is automatic. The order can always be reversed without changing the answer.

7. When one order is easy and the other is not

Switching order of integration can make a hard integral easy. Consider

$\iint_R x\sin(xy)\,dA, \qquad R = [0,1]\times[0,\pi].$

Order $dy\,dx$ (integrate $y$ first):

$\int_0^1 \!\int_0^{\pi} x\sin(xy)\,dy\,dx = \int_0^1 \bigl[-\cos(xy)\bigr]_0^{\pi}\,dx = \int_0^1 \bigl(1 - \cos(\pi x)\bigr)\,dx = \Bigl[x - \tfrac{\sin(\pi x)}{\pi}\Bigr]_0^1 = 1.$

Order $dx\,dy$ (integrate $x$ first):

$\int_0^{\pi} \!\int_0^1 x\sin(xy)\,dx\,dy.$

Integrating $x\sin(xy)$ with respect to $x$ by parts gives $\dfrac{\sin(xy) - xy\cos(xy)}{y^2}$, and the outer integral $\int_0^{\pi}$ of that expression involves the sine integral $\operatorname{Si}(y)$, which has no elementary closed form.

Lesson. Always check both orders. When the "natural" order is difficult, try the other — Fubini guarantees the same answer.

8. Exercises