Double Integrals over General Regions

Quick answer

Over a region bounded by curves, the inner limits of an iterated integral become functions of the outer variable. A type I region runs between two graphs y = g₁(x) and y = g₂(x); a type II region runs between two graphs x = h₁(y) and x = h₂(y). Sketching the region and drawing one strip is the surest way to find the limits. Reversing the order of integration describes the same region the other way and can turn an impossible inner integral into a routine one.

What you'll learn

Describe a region as type I or type II
Set up iterated integrals with variable limits
Compute areas and volumes with double integrals
Reverse the order of integration

Regions bounded by curves

Most regions are not rectangles. The double integral over any bounded region $D$ still means the same thing, a signed volume, and the Riemann-sum definition carries over: enclose $D$ in a rectangle and treat $f$ as zero outside $D$ . What changes is the iterated integral. Its inner limits become functions of the outer variable.

Type I region. If $D$ is $a \le x \le b$ , $g_1(x) \le y \le g_2(x)$ , then

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

Type II region. If $D$ is $c \le y \le d$ , $h_1(y) \le x \le h_2(y)$ , then

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

The outer limits are always constants. Many regions, such as triangles and disks, are both types, and then either order works.

Why the inner limits move

Fubini’s theorem sliced a solid over a rectangle with planes $x = \text{constant}$ . Slice a solid over $D$ the same way. At a particular $x$ , the slice sits over the segment of $D$ that runs from the lower boundary $y = g_1(x)$ to the upper one $y = g_2(x)$ . Its area is $A(x) = \int_{g_1(x)}^{g_2(x)} f(x, y)\,dy$ , and the volume is $\int_a^b A(x)\,dx$ as before. The inner limits are the ends of one strip, and they move because the strip’s ends move. That is why the surest way to set up the limits is to sketch $D$ and draw one strip.

y = 2x²
y = 1 + x²

A type I region with one vertical strip

Example 1 — Between two parabolas

Evaluate $\iint_D (x + 2y)\,dA$ , where $D$ is the region between $y = 2x^2$ and $y = 1 + x^2$ .

The curves meet where $2x^2 = 1 + x^2$ , at $x = \pm 1$ . Between them $1 + x^2$ is on top, so a vertical strip at $x$ runs from $y = 2x^2$ to $y = 1 + x^2$ :

\int_{-1}^{1} \int_{2x^2}^{1 + x^2} (x + 2y)\,dy\,dx = \int_{-1}^{1} \Big[xy + y^2\Big]_{2x^2}^{1 + x^2} dx

Expanding gives $\int_{-1}^{1} (-3x^4 - x^3 + 2x^2 + x + 1)\,dx$ . The odd powers integrate to zero over $[-1, 1]$ , leaving $-\tfrac{6}{5} + \tfrac{4}{3} + 2 = \tfrac{32}{15}$ .

Worked examples

Example 5 — Reversing the order

Evaluate $\displaystyle\int_0^1 \int_x^1 e^{y^2}\,dy\,dx$ .

The inner integral has no elementary antiderivative, so reverse the order. The limits say $0 \le x \le 1$ and $x \le y \le 1$ : the triangle with corners $(0, 0)$ , $(0, 1)$ and $(1, 1)$ . Read horizontally, the same triangle is $0 \le y \le 1$ and $0 \le x \le y$ :

\int_0^1 \int_0^y e^{y^2}\,dx\,dy = \int_0^1 y\,e^{y^2}\,dy = \Big[\tfrac{1}{2}e^{y^2}\Big]_0^1 = \frac{e - 1}{2} \approx 0.859

The inner integral in $x$ produced the factor $y$ that the substitution $u = y^2$ needs.

y = x

The triangle of Example 5, read with horizontal strips

The tetrahedron under x + y + z = 1

Common mistakes

Practice problems

Use a double integral to find the area between $y = x^2$ and $y = 2x$ .

Answer

$\tfrac{4}{3}$

Full solution

The curves meet at $x = 0$ and $x = 2$ , with $y = 2x$ on top. $\int_0^2 \int_{x^2}^{2x} 1\,dy\,dx = \int_0^2 (2x - x^2)\,dx = 4 - \tfrac{8}{3} = \tfrac{4}{3}$ .
Evaluate $\displaystyle\int_0^1 \int_0^x (x + y)\,dy\,dx$ .

Answer

$\tfrac{1}{2}$

Full solution

The inner integral is $x^2 + \tfrac{x^2}{2} = \tfrac{3}{2}x^2$ , and $\int_0^1 \tfrac{3}{2}x^2\,dx = \tfrac{1}{2}$ .
Evaluate $\displaystyle\int_0^2 \int_0^y xy\,dx\,dy$ .

Answer

$2$

Full solution

The inner integral, in $x$ , is $y \cdot \tfrac{y^2}{2} = \tfrac{y^3}{2}$ , and $\int_0^2 \tfrac{y^3}{2}\,dy = \tfrac{16}{8} = 2$ .
Describe the triangle bounded by $y = x$ , $y = 0$ and $x = 2$ as a type I region and as a type II region.

Answer

Type I: $0 \le x \le 2$ , $0 \le y \le x$ . Type II: $0 \le y \le 2$ , $y \le x \le 2$ .

Full solution

A vertical strip at $x$ runs from the $x$ -axis up to the line $y = x$ . A horizontal strip at height $y$ runs from the line, where $x = y$ , right to $x = 2$ .
Find the area of the region bounded by $x = y^2$ , $x = 4$ and $y = 0$ , with $y \ge 0$ .

Answer

$\tfrac{16}{3}$

Full solution

With horizontal strips as in Example 4: $\int_0^2 (4 - y^2)\,dy = 8 - \tfrac{8}{3} = \tfrac{16}{3}$ .
Reverse the order and evaluate $\displaystyle\int_0^1 \int_y^1 \sin(x^2)\,dx\,dy$ .

Hint

The region is the triangle $0 \le y \le x \le 1$ .

Answer

$\tfrac{1 - \cos 1}{2} \approx 0.2298$

Full solution

Read with vertical strips, the triangle is $0 \le x \le 1$ , $0 \le y \le x$ . Then $\int_0^1 \int_0^x \sin(x^2)\,dy\,dx = \int_0^1 x\sin(x^2)\,dx = \Big[-\tfrac{1}{2}\cos(x^2)\Big]_0^1 = \tfrac{1 - \cos 1}{2}$ .
Find the volume of the solid under $z = xy$ and over the region between $y = x^2$ and $y = x$ .

Answer

$\tfrac{1}{24}$

Full solution

The height $xy$ is positive on the region, so the volume is the integral of Example 3, $\tfrac{1}{24}$ .
Rewrite $\displaystyle\int_0^4 \int_{\sqrt{x}}^{2} f(x, y)\,dy\,dx$ with the order reversed.

Answer

$\displaystyle\int_0^2 \int_0^{y^2} f(x, y)\,dx\,dy$

Full solution

The region is $0 \le x \le 4$ , $\sqrt{x} \le y \le 2$ : above the curve $y = \sqrt{x}$ and below $y = 2$ . Horizontally, $y$ runs from $0$ to $2$ , and at height $y$ a strip runs from the $y$ -axis to the curve, where $x = y^2$ .
A student reverses $\displaystyle\int_0^1 \int_x^1 e^{y^2}\,dy\,dx$ to $\displaystyle\int_x^1 \int_0^1 e^{y^2}\,dx\,dy$ . What went wrong?

Answer

The limits were swapped instead of re-read from the region; an outer limit now contains $x$ .

Full solution

The outer limits must be constants. Sketching the triangle $0 \le x \le y \le 1$ and reading it with horizontal strips gives $\int_0^1 \int_0^y e^{y^2}\,dx\,dy$ , as in Example 5.

Frequently asked questions

What is a type I region?

A region of the form a ≤ x ≤ b, g₁(x) ≤ y ≤ g₂(x): between two graphs of functions of x. Integrate in y first, from the lower curve to the upper one, then in x.

What is a type II region?

A region of the form c ≤ y ≤ d, h₁(y) ≤ x ≤ h₂(y): between two graphs of functions of y. Integrate in x first, from the left curve to the right one, then in y.

How do you find the limits of integration?

Sketch the region and draw one strip in the direction of the inner variable. The strip's ends give the inner limits, as functions of the outer variable. The outer limits are constants that cover every strip.

How do you reverse the order of integration?

Sketch the region from the given limits, then describe the same region the other way: read the new outer limits as constants and the new inner limits from a strip in the other direction.

How do you find an area with a double integral?

Integrate the constant 1: the double integral of 1 over D is the volume of a slab of height 1, numerically equal to the area of D.

What to learn next

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