Math Core

In single-variable calculus, $\int_a^b f(x)\,dx$ adds up thin strips to find the area under a curve. Now the function is $z = f(x, y)$ , its graph is a surface, and the natural question is the volume under that surface. The double integral answers it, and the same machinery will later give mass, probability, averages and much more.

Riemann sums in two dimensions

Start with a rectangle $R = [a, b] \times [c, d]$ , meaning $a \le x \le b$ and $c \le y \le d$ . Cut $[a, b]$ into $m$ pieces of width $\Delta x$ and $[c, d]$ into $n$ pieces of width $\Delta y$ . This slices $R$ into $mn$ small rectangles $R_{ij}$ , each of area $\Delta A = \Delta x\,\Delta y$ .

Pick a sample point $(x_{ij}^*, y_{ij}^*)$ in each small rectangle. Over $R_{ij}$ , the solid under the surface is roughly a thin box with base $\Delta A$ and height $f(x_{ij}^*, y_{ij}^*)$ . Adding the boxes gives the double Riemann sum

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

Definition

Double integral over a rectangle

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

\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,

when this limit exists. It always exists when $f$ is continuous on $R$ . If $f \ge 0$ , it equals the volume of the solid above $R$ and below the surface $z = f(x, y)$ .

When $f$ takes negative values, the integral counts volume below the $xy$ -plane as negative, exactly like signed area in one variable.

A common choice of sample point is the center of each small rectangle. That gives the midpoint rule, which is usually a good estimate even with a coarse grid.

The square R = [0, 2] × [0, 2] cut into four unit squares. The dots are the midpoints used by the midpoint rule.Open in grapher →

Worked example: The midpoint rule

Estimate $\iint_R (x^2 + y)\,dA$ over $R = [0, 2] \times [0, 2]$ using the midpoint rule with four equal squares.

Solution. Each square has area $\Delta A = 1$ , and the midpoints are $(0.5, 0.5)$ , $(1.5, 0.5)$ , $(0.5, 1.5)$ and $(1.5, 1.5)$ . The function values there are $0.75$ , $2.75$ , $1.75$ and $3.75$ . So

\iint_R (x^2 + y)\,dA \approx (0.75 + 2.75 + 1.75 + 3.75)(1) = 9.

Below you will find the exact value is $\dfrac{28}{3} \approx 9.33$ , so the estimate is already close.

Iterated integrals

Computing a limit of double sums directly is painful. Instead, you integrate one variable at a time. In the iterated integral

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

you first compute the inner integral $\int_c^d f(x, y)\,dy$ , treating $x$ as a constant (partial integration, the reverse of partial differentiation). The result is a function of $x$ alone, which you then integrate from $a$ to $b$ .

Geometrically, the inner integral is the area $A(x)$ of the cross-section of the solid at a fixed $x$ . Integrating cross-sectional area gives volume, which is exactly the slicing method from Calculus I.

Fubini's theorem

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

\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.

On a rectangle, the order of integration does not change the answer, and every limit is a constant.

Read the differentials from the inside out: in $\int_a^b \int_c^d \cdots\,dy\,dx$ , the inner $dy$ goes with the inner limits $c$ to $d$ .

Worked example: Computing an iterated integral

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

Solution. Integrate with respect to $x$ first, holding $y$ fixed:

\int_0^3 (x^2 + 3y)\,dx = \left[\frac{x^3}{3} + 3xy\right]_{x=0}^{x=3} = 9 + 9y.

Now integrate that in $y$ :

\int_1^2 (9 + 9y)\,dy = \left[9y + \frac{9y^2}{2}\right]_1^2 = (18 + 18) - \left(9 + \frac{9}{2}\right) = \frac{45}{2}.

As a check, integrating in $y$ first gives $\int_0^3 \left(x^2 + \tfrac{9}{2}\right)dx = 9 + \tfrac{27}{2} = \tfrac{45}{2}$ , as Fubini promises.

Separable integrands

If the integrand factors as $g(x)\,h(y)$ and the region is a rectangle, the double integral splits into a product of two single integrals:

\iint_{[a,b] \times [c,d]} g(x)\,h(y)\,dA = \left(\int_a^b g(x)\,dx\right)\left(\int_c^d h(y)\,dy\right).

This works because $h(y)$ is a constant during the $x$ -integration, so it pulls out.

Worked example: A product of one-variable functions

Evaluate $\iint_R y \sin x\,dA$ over $R = \left[0, \tfrac{\pi}{2}\right] \times [0, 1]$ .

Solution. The integrand is $\sin x$ times $y$ , so

\iint_R y \sin x\,dA = \left(\int_0^{\pi/2} \sin x\,dx\right)\left(\int_0^1 y\,dy\right) = (1)\left(\frac{1}{2}\right) = \frac{1}{2}.

Choosing the order

Fubini says both orders give the same number, but one order is often much easier to carry out.

Worked example: Let the easy order do the work

Evaluate $\iint_R x e^{xy}\,dA$ over $R = [0, 1] \times [0, 1]$ .

Solution. Integrating in $x$ first would need integration by parts with $y$ floating around. Integrating in $y$ first is clean, because $x$ is exactly the factor that the chain rule produces:

\int_0^1 x e^{xy}\,dy = \Big[e^{xy}\Big]_{y=0}^{y=1} = e^{x} - 1.

Then $\displaystyle\int_0^1 (e^x - 1)\,dx = (e - 1) - 1 = e - 2$ .

Common mistake

During the inner integration, only the inner variable changes. Treat the other variable as a constant, and do not substitute numbers for it until the outer integration. Also keep the limits paired with the right differential: the inner limits belong to the inner variable.

Average value

Dividing a double integral by the area of the region gives the average height of the surface:

f_{\text{avg}} = \frac{1}{A(R)} \iint_R f(x, y)\,dA.

The volume under the surface equals the volume of a box with base $R$ and height $f_{\text{avg}}$ .

Tip

A quick sanity check: if $m \le f(x, y) \le M$ on $R$ , then $m\,A(R) \le \iint_R f\,dA \le M\,A(R)$ . For $x^2 + y$ on $[0, 2] \times [0, 2]$ the values run from $0$ to $6$ , so the integral must lie between $0$ and $24$ .