Double Integrals over General Regions, multivariable and vector calculus lesson

Real regions are rarely rectangles. A metal plate may be cut along a parabola, and a field may be bounded by a river and a road. The trick is to place the region inside a rectangle you already know how to handle and declare the integrand to be zero everywhere outside the region. What remains is careful bookkeeping, reading the bounds of each slice off a sketch.

Key terms

Integral over a bounded region

Definition Let $D$ be a bounded region in the plane whose boundary has zero area (for example, a boundary made of finitely many smooth curves), and let $f$ be defined on $D$ . Choose a closed rectangle $R$ containing $D$ and define

F(x, y) = \begin{cases} f(x, y) & \text{on } D, \\ 0 & \text{on } R \setminus D. \end{cases}

If $F$ is integrable on $R$ , then $f$ is integrable on $D$ and

\iint_D f \, dA = \iint_R F \, dA.

The value does not depend on which rectangle $R$ is chosen.

In plain words Draw a faint box around the region and set the height to zero outside the region. The integral over the box then only collects what sits above $D$ .

Example Let $D$ be the triangle $0 \le y \le x \le 1$ inside $R = [0, 1] \times [0, 1]$ and $f(x, y) = xy$ . For a fixed $x$ , the extended function is $xy$ for $0 \le y \le x$ and $0$ for $x < y \le 1$ , so the zero part adds nothing and

\begin{aligned} \iint_D xy \, dA &= \int_0^1 \int_0^x xy \, dy \, dx \\[0.4em] &= \int_0^1 \frac{x^3}{2} \, dx = \frac{1}{8}. \end{aligned}

Type I and type II descriptions

Definition A region is of type I if it can be written as

\begin{aligned} D = \{ (x, y) \mid{} & a \le x \le b, \\ & g_1(x) \le y \le g_2(x) \} \end{aligned}

with $g_1$ and $g_2$ continuous on $[a, b]$ and $g_1 \le g_2$ there. It is of type II if it can be written as

\begin{aligned} D = \{ (x, y) \mid{} & c \le y \le d, \\ & h_1(y) \le x \le h_2(y) \} \end{aligned}

with $h_1$ and $h_2$ continuous on $[c, d]$ and $h_1 \le h_2$ . If $f$ is continuous on a type I region $D$ , then

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

and if $f$ is continuous on a type II region $D$ , then

\iint_D f \, dA = \int_c^d \int_{h_1(y)}^{h_2(y)} f \, dx \, dy.

In plain words Type I means every vertical line through the region enters along one curve and leaves along one curve. Type II is the same with horizontal lines. The inner bounds may depend on the outer variable, but the outer bounds must be constants.

Example The region between $y = x^2$ and $y = 2x$ is of both types. The curves meet where $x^2 = 2x$ , at $x = 0$ and $x = 2$ , and on $[0, 2]$ the line lies above the parabola. As type I,

\begin{aligned} &\int_0^2 \int_{x^2}^{2x} x \, dy \, dx \\[0.4em] &\quad = \int_0^2 (2x^2 - x^3) \, dx = \frac{4}{3}. \end{aligned}

As type II, solve each curve for $x$ to get $x = y/2$ on the left and $x = \sqrt{y}$ on the right, for $0 \le y \le 4$ . Then

\begin{aligned} &\int_0^4 \int_{y/2}^{\sqrt{y}} x \, dx \, dy \\[0.4em] &\quad = \int_0^4 \left( \frac{y}{2} - \frac{y^2}{8} \right) dy \\[0.4em] &\quad = 4 - \frac{8}{3} = \frac{4}{3}. \end{aligned}

Both orders give the same answer, as they must.

Bounds from geometry

Definition To find iterated bounds for a region given by its boundary curves, sketch the curves, solve the boundary equations simultaneously to find where they meet, and choose the order. For a $dy \, dx$ integral, the outer bounds are the least and greatest $x$ in $D$ , and for each fixed $x$ the inner bounds are the $y$ values where a vertical line at that $x$ enters and leaves $D$ . For $dx \, dy$ , swap the roles of $x$ and $y$ .

In plain words Sweep a strip across the region and only through points that belong to it. The ends of the strip are the inner bounds and the range of the sweep gives the outer bounds.

Example Take the region bounded by the parabola $x = y^2$ and the line $x = y + 2$ . They meet where $y^2 = y + 2$ , that is $(y - 2)(y + 1) = 0$ , so at $y = -1$ and $y = 2$ . A horizontal strip at height $y$ starts on the parabola and ends on the line, since $y^2 \le y + 2$ for $-1 \le y \le 2$ . So

\begin{aligned} \iint_D x \, dA &= \int_{-1}^{2} \int_{y^2}^{y + 2} x \, dx \, dy \\[0.4em] &= \int_{-1}^{2} \frac{(y + 2)^2 - y^4}{2} \, dy \\[0.4em] &= \frac{36}{5}. \end{aligned}

Splitting regions

Definition If $D = D_1 \cup D_2$ , where $D_1$ and $D_2$ overlap at most along their boundaries (a set of zero area), and $f$ is integrable on $D_1$ and on $D_2$ , then

\begin{aligned} \iint_D f \, dA &= \iint_{D_1} f \, dA \\[0.4em] &\quad + \iint_{D_2} f \, dA. \end{aligned}

In plain words When the strip in your chosen order would hit different curves in different parts of the region, cut the region where the switch happens and add the pieces.

Example Use the parabola and line region again, but with vertical strips. For $0 \le x \le 1$ a vertical line meets the region from the lower half of the parabola $y = -\sqrt{x}$ to the upper half $y = \sqrt{x}$ . For $1 \le x \le 4$ it starts on the line $y = x - 2$ instead. So the region splits at $x = 1$ and

\begin{aligned} \iint_D x \, dA &= \int_0^1 \int_{-\sqrt{x}}^{\sqrt{x}} x \, dy \, dx \\[0.4em] &\quad + \int_1^4 \int_{x - 2}^{\sqrt{x}} x \, dy \, dx \\[0.4em] &= \frac{36}{5}. \end{aligned}

This matches the single type II integral, which shows why the horizontal order was the better choice here.

Changing the order of integration

Definition Reversing the order of an iterated integral means rewriting the same region in the other description. From the given bounds, reconstruct the region $D$ as a set, sketch it, and describe it again with the other variable outside. When $f$ is continuous on a region that is both type I and type II,

\begin{aligned} &\int_a^b \int_{g_1(x)}^{g_2(x)} f \, dy \, dx \\[0.4em] &\quad = \int_c^d \int_{h_1(y)}^{h_2(y)} f \, dx \, dy. \end{aligned}

Simply swapping the bounds is wrong, since the new outer bounds must be constants.

In plain words The bounds are a coded picture of the region. Decode them into a sketch, then read the sketch the other way.

Example The inner integral of

\int_0^1 \int_{\sqrt{y}}^{1} \sqrt{x^3 + 1} \, dx \, dy

has no elementary antiderivative. The bounds say $0 \le y \le 1$ and $\sqrt{y} \le x \le 1$ , which is the region under $y = x^2$ for $0 \le x \le 1$ . In the other order,

\begin{aligned} &\int_0^1 \int_0^{x^2} \sqrt{x^3 + 1} \, dy \, dx \\[0.4em] &\quad = \int_0^1 x^2 \sqrt{x^3 + 1} \, dx \\[0.4em] &\quad = \frac{2}{9} \left( 2\sqrt{2} - 1 \right), \end{aligned}

using the substitution $u = x^3 + 1$ .

Bounded piecewise continuous integrands

Definition Let $D$ be a bounded region whose boundary has zero area. If $f$ is bounded on $D$ and continuous on $D$ except on a set of zero area, such as finitely many smooth curves, then $f$ is integrable on $D$ . Continuity is not needed for the iterated formula either. If $f$ is integrable on a type I region $D$ and, for every $x$ in $[a, b]$ , the inner integral

A(x) = \int_{g_1(x)}^{g_2(x)} f(x, y) \, dy

exists, then $A$ is integrable on $[a, b]$ and the double integral equals $\int_a^b A(x) \, dx$ . The inner integrals do exist when $f$ is bounded and its discontinuities lie on finitely many curves that meet each vertical line in finitely many points, because then each slice is a bounded function with finitely many discontinuities. The type II formula holds in the same way with horizontal lines.

In plain words A jump along a curve does not spoil the integral, because a curve has no area. Unbounded functions are a different matter and need the next section.

Example On the square $[0, 1] \times [0, 1]$ let $f(x, y) = 1$ when $y \le x$ and $f(x, y) = 2$ when $y > x$ . The function jumps along the diagonal, which has zero area and meets each vertical line once. For fixed $x$ the inner integral is

\int_0^x 1 \, dy + \int_x^1 2 \, dy = 2 - x,

and integrating from $0$ to $1$ gives $2 - 1/2 = 3/2$ . This agrees with adding the two triangles, $1 \cdot \tfrac{1}{2} + 2 \cdot \tfrac{1}{2}$ .

Improper cases

Definition A double integral is improper when the region is unbounded or the integrand is unbounded near some point or curve of $D$ . Suppose $f \ge 0$ is continuous on $D$ except on a bad set of zero area (a point, a curve, or nothing when only the region is unbounded). Take closed bounded regions $D_1 \subseteq D_2 \subseteq \dots$ inside $D$ , each with a boundary of zero area and each avoiding the bad set, so $f$ is continuous on every $D_n$ , and suppose together they fill up $D$ apart from a set of zero area. Then

\iint_D f \, dA = \lim_{n \to \infty} \iint_{D_n} f \, dA,

which is either finite (the integral converges) or $\infty$ (it diverges). The limit exists because the integrals increase with $n$ , and it is the same for every such sequence. For nonnegative $f$ the iterated integrals in either order, computed as improper one variable integrals, also give this same value, finite or infinite. This is Tonelli’s theorem. If $f$ changes sign, the double integral is called convergent only when the integral of $\lvert f \rvert$ is finite. Then every such sequence gives the same limit, and both iterated integrals equal it. Without that condition the two orders can disagree.

In plain words Cut away a thin piece around the trouble spot, or keep only a large bounded part of an unbounded region, integrate, and let the part you left out shrink away. Nonnegative integrands behave well. Sign changing ones are safe only when the integral of the absolute value is finite.

Example On the triangle $0 < y \le x \le 1$ , the function $1/\sqrt{x}$ blows up near the origin. The inner integral gives $\sqrt{x}$ , and

\begin{aligned} \int_0^1 \int_0^x \frac{1}{\sqrt{x}} \, dy \, dx &= \int_0^1 \sqrt{x} \, dx \\[0.4em] &= \frac{2}{3}, \end{aligned}

so the integral converges. With $1/x^2$ instead, the inner integral is $1/x$ and $\int_0^1 dx/x$ diverges. For a warning about sign changes, the iterated integrals of

\frac{x^2 - y^2}{(x^2 + y^2)^2}

over $(0, 1] \times (0, 1]$ equal $\pi/4$ in the order $dy \, dx$ and $-\pi/4$ in the order $dx \, dy$ . The inner integral in $y$ is $1/(1 + x^2)$ , and the other order gives $-1/(1 + y^2)$ . This does not contradict the theorem, because near the origin the absolute value behaves like $1/r^2$ in polar coordinates and its double integral is infinite. The one-variable background is in improper integrals with singularities.

Check yourself

Reverse the order and evaluate

\int_0^2 \int_{y/2}^{1} e^{x^2} \, dx \, dy.

Show the answer

The bounds describe $0 \le y \le 2$ and $y/2 \le x \le 1$ , which is the triangle $0 \le x \le 1$ , $0 \le y \le 2x$ . Then

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

Let $D$ be the region between $y = x^2$ and $y = x$ . Evaluate the integral of $y$ over $D$ .

Show the answer

The curves meet at $x = 0$ and $x = 1$ , and $x^2 \le x$ there. As a type I region,

\begin{aligned} \int_0^1 \int_{x^2}^{x} y \, dy \, dx &= \int_0^1 \frac{x^2 - x^4}{2} \, dx \\[0.4em] &= \frac{1}{15}. \end{aligned}

For which $p > 0$ does the integral of $x^{-p}$ over the triangle $0 < y \le x \le 1$ converge, and what is its value then?

Show the answer

The integrand is positive, so iterated integration is valid. The inner integral is $x \cdot x^{-p} = x^{1 - p}$ , and $\int_0^1 x^{1 - p} \, dx$ converges exactly when $1 - p > -1$ , that is $p < 2$ . Its value is then $1/(2 - p)$ .

Double Integrals over General Regions

Key terms

Integral over a bounded region

Type I and type II descriptions

Bounds from geometry

Splitting regions

Changing the order of integration

Bounded piecewise continuous integrands

Improper cases

Check yourself

How this lesson connects