Type I and Type II Regions
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 5.2: “Double Integrals over General Regions” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/5-2-double-integrals-over-general-regions |
| Textbook used in class | Stewart, Calculus, Section 15.3: “Double Integrals over General Regions” (Examples 1, 2) |
Quick Reference
Type I region ($x$-simple): $D = \{(x,y): a \leq x \leq b,\; g_1(x) \leq y \leq g_2(x)\}$. $$\iint_D f\,dA = \int_a^b\int_{g_1(x)}^{g_2(x)} f(x,y)\,dy\,dx.$$
Type II region ($y$-simple): $D = \{(x,y): c \leq y \leq d,\; h_1(y) \leq x \leq h_2(y)\}$. $$\iint_D f\,dA = \int_c^d\int_{h_1(y)}^{h_2(y)} f(x,y)\,dx\,dy.$$
Motivation
Rectangles are special. Most regions of integration -- triangles, disks, regions between curves -- are not rectangles. To integrate over such regions, the limits of the inner integral become functions of the outer variable rather than constants. The key skill is reading the geometry: which variable has the “simple” range (constant limits) and which has a range that depends on the other?
Key Concepts
1. Type I (Vertical Slices)
In a Type I region, for each fixed $x \in [a,b]$, $y$ runs from the lower boundary curve $g_1(x)$ to the upper boundary curve $g_2(x)$. The outer integral is then over the full $x$-range $[a,b]$.
Sketch strategy: draw the region, draw a vertical slice at a generic $x$, read off the bottom and top of the slice.
2. Type II (Horizontal Slices)
In a Type II region, for each fixed $y \in [c,d]$, $x$ runs from the left boundary $h_1(y)$ to the right boundary $h_2(y)$.
Sketch strategy: draw the region, draw a horizontal slice at a generic $y$, read off the left and right edges.
3. Choosing the Order
Many regions can be described as either Type I or Type II. Choose the type that leads to an integral you can evaluate. Sometimes one order produces an antiderivative that can be found; the other does not.
Worked Example
Evaluate $\iint_D x\,dA$ where $D$ is the region bounded by $y = x^2$ and $y = x + 2$. (Stewart 15.3, Example 2.)
Find the intersection: $x^2 = x+2 \implies x^2-x-2 = 0 \implies (x-2)(x+1) = 0$, so $x = -1$ and $x = 2$.
For $-1 \leq x \leq 2$, the parabola $y = x^2$ is below the line $y = x+2$.
Set up as Type I: $$\iint_D x\,dA = \int_{-1}^2\int_{x^2}^{x+2} x\,dy\,dx.$$
Inner integral: $\int_{x^2}^{x+2} x\,dy = x(x+2-x^2) = x^2 + 2x - x^3$.
Outer integral: $$\int_{-1}^2 (x^2+2x-x^3)\,dx = \left[\frac{x^3}{3}+x^2-\frac{x^4}{4}\right]_{-1}^2.$$ $$= \left(\frac{8}{3}+4-4\right) - \left(-\frac{1}{3}+1-\frac{1}{4}\right) = \frac{8}{3} - \frac{1}{3} + \frac{1}{4} - 1 = \frac{7}{3} - \frac{3}{4} = \frac{28}{12} - \frac{9}{12} = \frac{19}{12}.$$
the inner limits are constants in any iterated integral. Over a rectangle, inner limits are constants. Over a general region, the inner limits are functions of the outer variable. For a Type I region $g_1(x) \leq y \leq g_2(x)$, the limits of the inner $dy$ integral are $g_1(x)$ and $g_2(x)$, which depend on $x$. Treating them as constants is the most common setup error.
Leveled Practice
Problem 1. Set up (do not evaluate) $\iint_D y\,dA$ where $D$ is the triangle with vertices $(0,0)$, $(1,0)$, $(0,2)$.
Show answer
The boundary lines are $x = 0$, $y = 0$, and $y = 2-2x$ (line through $(1,0)$ and $(0,2)$).
Type I: $0 \leq x \leq 1$, $0 \leq y \leq 2-2x$.
$$\int_0^1\int_0^{2-2x} y\,dy\,dx.$$