Math Core

Real regions are rarely rectangles: a plate may be triangular, a field may be bounded by a river's curve. To integrate over such a region you let the inner limits vary with the outer variable. Setting up those limits correctly is the main skill of this lesson, and it comes down to drawing the region.

Two standard shapes of region

The trick is to describe the region $D$ with inequalities in which one variable has constant limits and the other is trapped between two curves.

Definition

Type I and Type II regions

A Type I region lies between two graphs of functions of $x$ :

D = \{(x, y) : a \le x \le b,\ g_1(x) \le y \le g_2(x)\}.

A Type II region lies between two graphs of functions of $y$ :

D = \{(x, y) : c \le y \le d,\ h_1(y) \le x \le h_2(y)\}.

Integrating over a general region

For a Type I region, integrate in $y$ first:

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

For a Type II region, integrate in $x$ first:

\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. The inner limits may depend on the outer variable only.

Here is a reliable way to find limits for a Type I setup. Draw the region. Draw a vertical arrow through it at a typical $x$ . The arrow enters $D$ on the lower curve $y = g_1(x)$ and leaves on the upper curve $y = g_2(x)$ : those are the inner limits. Then slide the arrow left and right to find the smallest and largest $x$ that still hit $D$ : those are the outer limits. For Type II, use a horizontal arrow instead, entering on the left curve and leaving on the right.

Examples of each type

The region between y = x² (below) and y = x (above) for 0 ≤ x ≤ 1. A vertical segment at any x runs from the parabola up to the line.Open in grapher →

Worked example: A Type I region

Evaluate $\iint_D xy\,dA$ , where $D$ is the region between $y = x^2$ and $y = x$ .

Solution. The curves meet where $x^2 = x$ , at $x = 0$ and $x = 1$ . For $0 \le x \le 1$ , the line is on top ( $x \ge x^2$ ). So

\begin{aligned} \iint_D xy\,dA &= \int_0^1 \int_{x^2}^{x} xy\,dy\,dx = \int_0^1 x \cdot \frac{x^2 - x^4}{2}\,dx \\ &= \frac{1}{2}\int_0^1 (x^3 - x^5)\,dx = \frac{1}{2}\left(\frac{1}{4} - \frac{1}{6}\right) = \frac{1}{24}. \end{aligned}

Some regions are awkward as Type I because the top or bottom boundary changes formula partway across. Then Type II can save you from splitting the integral.

The region between the parabola x = y² (left) and the line x = y + 2 (right). Horizontal slices always run from the parabola to the line.Open in grapher →

Worked example: A Type II region

Evaluate $\iint_D y\,dA$ , where $D$ is bounded by $x = y^2$ and $x = y + 2$ .

Solution. The curves meet where $y^2 = y + 2$ , so $y = -1$ or $y = 2$ . A horizontal arrow at height $y$ enters on the parabola and leaves on the line, so $y^2 \le x \le y + 2$ :

\begin{aligned} \iint_D y\,dA &= \int_{-1}^{2} \int_{y^2}^{y+2} y\,dx\,dy = \int_{-1}^{2} \left(y^2 + 2y - y^3\right)dy \\ &= \left[\frac{y^3}{3} + y^2 - \frac{y^4}{4}\right]_{-1}^{2} = \frac{8}{3} - \frac{5}{12} = \frac{9}{4}. \end{aligned}

As Type I, the bottom boundary would switch from $y = -\sqrt{x}$ to $y = x - 2$ at $x = 1$ , forcing two integrals.

Reversing the order of integration

Sometimes the given order leads to an antiderivative you cannot write down, such as $\int e^{y^2}\,dy$ . Describing the same region the other way can make the integral easy.

The procedure: read the region off the given limits, sketch it, then describe it again with the other variable on the outside. Never just swap the limits; the new limits come from the picture.

The triangle 0 ≤ x ≤ 1, x ≤ y ≤ 1. Read the other way, it is 0 ≤ y ≤ 1, 0 ≤ x ≤ y.Open in grapher →

Worked example: An impossible integral made easy

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

Solution. $e^{y^2}$ has no elementary antiderivative, so the inner integral is stuck. The limits say $0 \le x \le 1$ and $x \le y \le 1$ : the triangle with vertices $(0, 0)$ , $(0, 1)$ and $(1, 1)$ . Horizontally, for each $y$ in $[0, 1]$ , $x$ runs from $0$ to $y$ . So

\int_0^1 \int_0^y e^{y^2}\,dx\,dy = \int_0^1 y e^{y^2}\,dy = \left[\frac{1}{2}e^{y^2}\right]_0^1 = \frac{e - 1}{2}.

Common mistake

The outer limits must be numbers. If your answer to a double integral still contains $x$ or $y$ , a variable limit ended up on the outside, or you reversed the order by swapping limits instead of redrawing the region.

Area and volume

Integrating the constant $1$ gives the area of the region: $A(D) = \iint_D 1\,dA$ . For a Type I region this reduces to $\int_a^b \big(g_2(x) - g_1(x)\big)\,dx$ , the familiar area between curves.

If $f \ge 0$ on $D$ , then $\iint_D f\,dA$ is the volume of the solid over $D$ and under $z = f(x, y)$ . The linearity and additivity properties you know still hold: you can split an integrand into pieces, pull out constants, and split a region into non-overlapping parts and add.

Worked example: Volume of a tetrahedron

Find the volume of the solid in the first octant under the plane $2x + 3y + z = 6$ .

Solution. The height is $z = 6 - 2x - 3y$ . The solid sits over the triangle in the $xy$ -plane cut off by $2x + 3y = 6$ , the trace where $z = 0$ . That triangle has $0 \le x \le 3$ and $0 \le y \le \dfrac{6 - 2x}{3}$ . Then

\begin{aligned} V &= \int_0^3 \int_0^{(6-2x)/3} (6 - 2x - 3y)\,dy\,dx = \int_0^3 \frac{(6 - 2x)^2}{6}\,dx \\ &= \left[-\frac{(6 - 2x)^3}{36}\right]_0^3 = \frac{216}{36} = 6. \end{aligned}

This matches the tetrahedron formula $\tfrac{1}{6}(3)(2)(6) = 6$ .

Tip

For the inner integral of a Type I setup, $\int_{g_1}^{g_2} (c - ky)\,dy$ with a linear integrand, the result is (length of segment) times (integrand at the midpoint of the segment). This speeds up many volume problems.