Double Integrals over Rectangles, multivariable and vector calculus lesson

A thermal camera shows a rectangular sheet as a grid of colored tiles, one temperature reading per tile. Multiply each reading by the area of its tile, add everything up, and you have a single number that describes the whole sheet. Make the tiles smaller and that number settles down to the double integral, which is what this lesson builds.

Key terms

Double Riemann sums

Definition Let $R = [a, b] \times [c, d]$ and let $f$ be defined on $R$ . A partition of $R$ comes from points $a = x_0 < x_1 < \dots < x_m = b$ and $c = y_0 < y_1 < \dots < y_n = d$ , which cut $R$ into the subrectangles $R_{ij} = [x_{i-1}, x_i] \times [y_{j-1}, y_j]$ with areas $\Delta A_{ij} = \Delta x_i \, \Delta y_j$ . Choose a sample point $(x_{ij}^*, y_{ij}^*)$ in each $R_{ij}$ . The double Riemann sum is

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

In plain words Cut the rectangle into tiles, read the function once on each tile, and weight each reading by the tile’s area.

Example Let $f(x, y) = 16 - x^2 - 2y^2$ on $R = [0, 2] \times [0, 2]$ . Cut it into four unit squares and use their upper right corners as sample points, so every $\Delta A_{ij} = 1$ .

Sample point	$(1, 1)$	$(1, 2)$	$(2, 1)$	$(2, 2)$
Value of $f$	$13$	$7$	$10$	$4$

The Riemann sum is $13 + 7 + 10 + 4 = 34$ . The exact integral is $48$ , as the iterated integral section shows. Since $f$ decreases as $x$ or $y$ grows, upper right corners give low readings and the sum comes out too small.

Integrability

Definition The norm $\lVert P \rVert$ of a partition is the length of the longest diagonal among its subrectangles. A bounded function $f$ on $R$ is integrable on $R$ if there is a number $I$ with the following property. For every $\varepsilon > 0$ there is a $\delta > 0$ such that $\lvert S - I \rvert < \varepsilon$ for every partition with $\lVert P \rVert < \delta$ and every choice of sample points. Then $I$ is the double integral,

\begin{aligned} &\iint_R f(x, y) \, dA \\[0.4em] &\quad = \lim_{\lVert P \rVert \to 0} \sum_{i=1}^{m} \sum_{j=1}^{n} f(x_{ij}^*, y_{ij}^*) \, \Delta A_{ij}. \end{aligned}

Every function that is continuous on $R$ is integrable. More generally, $f$ is integrable on $R$ if it is bounded on $R$ and every point where it fails to be continuous lies on one of finitely many smooth curves. Here a smooth curve means the graph of a continuously differentiable function $y = g(x)$ or $x = h(y)$ over a closed interval, or a curve with a continuously differentiable parametrization over a closed interval. Such curves can be covered by tiles of arbitrarily small total area, which is why they do not matter. A function that is unbounded on $R$ is not integrable in this sense, and needs an improper integral instead.

In plain words The double integral exists when all fine enough Riemann sums agree, no matter how the tiles are shaped or where the readings are taken. Continuity is enough for that, and a few clean jumps along curves do no harm.

Example On $R = [0, 2] \times [0, 1]$ let $f(x, y) = 1$ for $x \le 1$ and $f(x, y) = 2$ for $x > 1$ . It is bounded and jumps only along the segment $x = 1$ , so it is integrable, and the integral is $1 \cdot 1 + 2 \cdot 1 = 3$ . By contrast, the function equal to $1$ at points with both coordinates rational and $0$ elsewhere is not integrable on $R$ . Sums with rational sample points give $2$ and sums with irrational ones give $0$ , however fine the partition.

Signed volume

Definition If $f$ is integrable and $f(x, y) \ge 0$ on $R$ , then $\iint_R f \, dA$ is the volume of the solid above $R$ and below the graph $z = f(x, y)$ . For a general integrable $f$ the integral is a signed volume, the volume above the $xy$ plane minus the volume below it. The geometric volume between the graph and $R$ is

\iint_R \lvert f(x, y) \rvert \, dA.

In plain words Parts of the graph below the plane count as negative. If you want the actual amount of space between the surface and the plane, integrate the absolute value.

Example Take $f(x, y) = x - y$ on $[0, 2] \times [0, 2]$ . The graph is a tilted plane, above the $xy$ plane where $x > y$ and below it where $x < y$ , and the two parts are mirror images. So the signed volume is $0$ . For the geometric volume, look at the part where $x > y$ . Over the triangle with corners $(0, 0)$ , $(2, 0)$ and $(2, 2)$ , of area $2$ , the graph rises from height $0$ along the diagonal to height $2$ above $(2, 0)$ . That solid is a tetrahedron with apex $(2, 0, 2)$ straight above a corner of its base, so its volume is

\frac{1}{3} \cdot 2 \cdot 2 = \frac{4}{3}.

The part below the plane is its mirror image, so the geometric volume is $\tfrac{8}{3}$ while the signed volume is $0$ .

Iterated integrals

Definition For a function $f$ on $R = [a, b] \times [c, d]$ , the iterated integral

\begin{aligned} &\int_a^b \int_c^d f(x, y) \, dy \, dx \\[0.4em] &\quad = \int_a^b \left[ \int_c^d f(x, y) \, dy \right] dx \end{aligned}

means the following. For each fixed $x$ , integrate $f(x, y)$ in $y$ from $c$ to $d$ , which gives a function $A(x)$ , and then integrate $A(x)$ from $a$ to $b$ . It is defined when $A(x)$ exists for every $x$ in $[a, b]$ and $A$ is integrable on $[a, b]$ . The other order, $\int_c^d \int_a^b f(x, y) \, dx \, dy$ , integrates in $x$ first. The inner integral is a partial integration, with the other variable held constant.

In plain words Work from the inside out. The inner integral collapses one direction and leaves an ordinary function of the other variable. When $f \ge 0$ , the number $A(x)$ is the area of the slice of the solid at that $x$ .

Example Compute $\int_0^3 \int_1^2 x^2 y \, dy \, dx$ . Holding $x$ fixed,

\int_1^2 x^2 y \, dy = x^2 \left[ \frac{y^2}{2} \right]_1^2 = \frac{3}{2} x^2.

Then $\int_0^3 \tfrac{3}{2} x^2 \, dx = \tfrac{1}{2} \cdot 27 = \tfrac{27}{2}$ . Doing $x$ first gives $\int_1^2 9y \, dy = \tfrac{27}{2}$ as well.

The same steps give the exact value for the Riemann sum example. With $f(x, y) = 16 - x^2 - 2y^2$ ,

\begin{aligned} &\int_0^2 \int_0^2 f \, dy \, dx \\[0.4em] &\quad = \int_0^2 \left( \frac{80}{3} - 2x^2 \right) dx \\[0.4em] &\quad = \frac{160}{3} - \frac{16}{3} = 48. \end{aligned}

Fubini’s theorem

Definition Fubini’s theorem for rectangles says the following. If $f$ is continuous on $R = [a, b] \times [c, d]$ , then

\begin{aligned} \iint_R f \, dA &= \int_a^b \int_c^d f(x, y) \, dy \, dx \\[0.4em] &= \int_c^d \int_a^b f(x, y) \, dx \, dy. \end{aligned}

More generally, if $f$ is integrable on $R$ and the inner integral $\int_c^d f(x, y) \, dy$ exists for every $x$ in $[a, b]$ , then the first equality holds, and the same goes for the other order. This covers bounded functions that are continuous except on finitely many smooth curves, as long as the inner integrals exist.

For improper integrals, where $f$ is unbounded near an edge or the rectangle is unbounded, the hypothesis is absolute integrability, a finite integral of $\lvert f \rvert$ . Here is a version that needs only one variable calculus. Let $f$ be continuous on $Q = [a, \infty) \times [c, \infty)$ and suppose

\lvert f(x, y) \rvert \le g(x) \, h(y) \quad \text{on } Q,

where $g$ and $h$ are continuous, nonnegative, and have convergent improper integrals on $[a, \infty)$ and $[c, \infty)$ . Then both iterated integrals of $f$ and of $\lvert f \rvert$ converge, and the two iterated integrals of $f$ are equal. Both also equal the limit of the integrals of $f$ over $[a, M] \times [c, N]$ as $M$ and $N$ go to $\infty$ , which is the improper double integral. The same holds on $(a, b] \times (c, d]$ for $f$ continuous there and possibly unbounded near the edges $x = a$ and $y = c$ . The fully general statement, that the two orders agree whenever the integral of $\lvert f \rvert$ is finite, is true but needs the Lebesgue integral, which this course does not develop. Without some such hypothesis the two orders can give different answers.

In plain words For a continuous function you may slice either way and you will get the same number, so pick the order that is easier. With an improper integral, check absolute integrability first.

Example For $\iint_R x e^{xy} \, dA$ on $R = [0, 1] \times [0, 1]$ , integrating in $y$ first is quick, because $x e^{xy}$ is the $y$ derivative of $e^{xy}$ ,

\begin{aligned} \int_0^1 \int_0^1 x e^{xy} \, dy \, dx &= \int_0^1 (e^x - 1) \, dx \\[0.4em] &= e - 2. \end{aligned}

The other order would need integration by parts, and Fubini guarantees the same value.

For a warning, take $f(x, y) = \dfrac{x^2 - y^2}{(x^2 + y^2)^2}$ on $(0, 1] \times (0, 1]$ . Since the $y$ derivative of $\frac{y}{x^2 + y^2}$ is $f$ , the inner integral in $y$ is $\frac{1}{1 + x^2}$ , and the iterated integral is $\frac{\pi}{4}$ . By the same computation with the roles swapped, the other order gives $-\frac{\pi}{4}$ . The function is unbounded near the origin, and in polar coordinates $\lvert f \rvert = \lvert \cos 2\theta \rvert / r^2$ , so the integral of $\lvert f \rvert$ is infinite. Fubini does not apply.

Average value

Definition If $f$ is integrable on $R$ and $A(R)$ is the area of $R$ , the average value of $f$ on $R$ is

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

Equivalently, $f_{\text{avg}} \cdot A(R) = \iint_R f \, dA$ , so a box of height $f_{\text{avg}}$ over $R$ has the same signed volume as the solid under the graph. The average has the units of $f$ , while the integral has the units of $f$ times area.

In plain words Spread the total evenly over the rectangle. The height you get is the average.

Example A plate occupies $[0, 4] \times [0, 2]$ , measured in meters, and its temperature is $T(x, y) = 20 + 3x + y^2$ degrees Celsius. Integrating in $y$ first,

\begin{aligned} \iint_R T \, dA &= \int_0^4 \left( 40 + 6x + \frac{8}{3} \right) dx \\[0.4em] &= \frac{656}{3}, \end{aligned}

in degrees Celsius times square meters. The area is $8$ square meters, so the average temperature is

\begin{aligned} T_{\text{avg}} &= \frac{656}{3 \cdot 8} = \frac{82}{3} \\[0.4em] &\approx 27.3 \text{ degrees Celsius.} \end{aligned}

Check yourself

Evaluate $\iint_R y \cos(xy) \, dA$ for $R = [0, 1] \times \left[0, \frac{\pi}{2}\right]$ by choosing a good order.

Show the answer

Integrate in $x$ first, because $y \cos(xy)$ is the $x$ derivative of $\sin(xy)$ . The inner integral is $\int_0^1 y \cos(xy) \, dx = \sin y$ , and then $\int_0^{\pi/2} \sin y \, dy = 1$ . Fubini applies since the integrand is continuous on $R$ .

For $f(x, y) = x$ on $R = [-1, 2] \times [0, 1]$ , find the signed volume and the geometric volume between the graph and $R$ .

Show the answer

The signed volume is

\int_0^1 \int_{-1}^2 x \, dx \, dy = \frac{4 - 1}{2} = \frac{3}{2}.

For the geometric volume integrate $\lvert x \rvert$ , which gives $\tfrac{1}{2} + 2 = \tfrac{5}{2}$ . The part below the plane, over $-1 \le x \le 0$ , contributes $\tfrac{1}{2}$ to the geometric volume but $-\tfrac{1}{2}$ to the signed one.

Find the average value of $f(x, y) = xy^2$ on $R = [0, 2] \times [0, 3]$ .

Show the answer

The integrand is a product of a function of $x$ and a function of $y$ , so the integral splits as

\int_0^2 x \, dx \cdot \int_0^3 y^2 \, dy = 2 \cdot 9 = 18.

The area of $R$ is $6$ , so the average value is $\tfrac{18}{6} = 3$ .

Double Integrals over Rectangles

Key terms

Double Riemann sums

Integrability

Signed volume

Iterated integrals

Fubini’s theorem

Average value

Check yourself

How this lesson connects