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Mom's Basement University

Double Integrals over General Regions

What You’ll Learn

Section titled “What You’ll Learn”

In this lesson you’ll learn how to set up double integrals when the region isn’t a nice rectangle. The key skill is reading the region’s shape and translating it into the correct limits of integration.

The Concept

Section titled “The Concept”

Beyond Rectangles

Section titled “Beyond Rectangles”

In the previous lesson, the region was always a rectangle, so the limits were just constants. But most real regions aren’t rectangles. A region might be bounded by curves, lines, or a mix of both. The limits of integration now depend on the other variable.

Type I Regions (dy dx)

Section titled “Type I Regions (dy dx)”

A Type I region is described by fixing x and letting y range between two curves

∫ab∫g1(x)g2(x)f(x,y) dy dx\int_a^b \int_{g_1(x)}^{g_2(x)} f(x, y) \, dy \, dx

For each value of x between a and b, y goes from the lower curve g₁(x) to the upper curve g₂(x). Think of it as sweeping vertical slices across the region from left to right.

Type II Regions (dx dy)

Section titled “Type II Regions (dx dy)”

A Type II region is described by fixing y and letting x range between two curves

∫cd∫h1(y)h2(y)f(x,y) dx dy\int_c^d \int_{h_1(y)}^{h_2(y)} f(x, y) \, dx \, dy

For each value of y between c and d, x goes from the left curve h₁(y) to the right curve h₂(y). Think of it as sweeping horizontal slices from bottom to top.

How to Choose

Section titled “How to Choose”

Always sketch the region first. Then ask: “Is it easier to describe the y-limits as functions of x, or the x-limits as functions of y?” Pick whichever gives simpler expressions. Sometimes one order requires splitting the region into pieces while the other doesn’t.

Worked Examples

Section titled “Worked Examples”

Example 1: Area between two curves

Find the area of the region bounded by y = x² and y = x + 2.

First, find where the curves intersect: x² = x + 2 gives x² - x - 2 = 0, so (x - 2)(x + 1) = 0. The curves meet at x = -1 and x = 2.

For each x from -1 to 2, y ranges from x² (bottom, blue curve) to x + 2 (top, green line). Setting up as a Type I integral

Area=∫−12[(x+2)−x2]dx\text{Area} = \int_{-1}^{2} \left[ (x + 2) - x^2 \right] dx =∫−12(x+2−x2) dx=[x22+2x−x33]−12= \int_{-1}^{2} (x + 2 - x^2) \, dx = \left[ \frac{x^2}{2} + 2x - \frac{x^3}{3} \right]_{-1}^{2} =(2+4−83)−(12−2+13)=103−(−76)=103+76=276=92= \left( 2 + 4 - \frac{8}{3} \right) - \left( \frac{1}{2} - 2 + \frac{1}{3} \right) = \frac{10}{3} - \left( -\frac{7}{6} \right) = \frac{10}{3} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2}

The area is 9/2 = 4.5 square units.

Example 2: Switching the order of integration

Sometimes the integral is easier (or only possible) in one order. Consider

∫01∫x1ey2 dy dx\int_0^1 \int_x^1 e^{y^2} \, dy \, dx

The inner integral has no closed-form antiderivative with respect to y. But if we switch the order, the region is the triangle where 0 ≤ x ≤ y and 0 ≤ y ≤ 1. In dx dy order

∫01∫0yey2 dx dy=∫01yey2 dy=[12ey2]01=e−12\int_0^1 \int_0^y e^{y^2} \, dx \, dy = \int_0^1 ye^{y^2} \, dy = \left[ \frac{1}{2} e^{y^2} \right]_0^1 = \frac{e - 1}{2}

Switching the order turned an impossible integral into a straightforward one.

Example 3: Integrating over a triangle

Evaluate the double integral of f(x, y) = x + 2y over the triangle with vertices (0, 0), (1, 0), and (0, 1).

The hypotenuse is y = 1 - x. For each x from 0 to 1, y ranges from 0 to 1 - x

∫01∫01−x(x+2y) dy dx\int_0^1 \int_0^{1-x} (x + 2y) \, dy \, dx

Inner integral (with respect to y)

∫01−x(x+2y) dy=[xy+y2]01−x=x(1−x)+(1−x)2\int_0^{1-x} (x + 2y) \, dy = \left[ xy + y^2 \right]_0^{1-x} = x(1-x) + (1-x)^2 =x−x2+1−2x+x2=1−x= x - x^2 + 1 - 2x + x^2 = 1 - x

Outer integral

∫01(1−x) dx=[x−x22]01=1−12=12\int_0^1 (1 - x) \, dx = \left[ x - \frac{x^2}{2} \right]_0^1 = 1 - \frac{1}{2} = \frac{1}{2}

The blue surface is z = x + 2y rising over the green triangular base. The volume under this surface over the triangle is 1/2 cubic unit, which is exactly what the double integral computed.

Real-World Application

Section titled “Real-World Application”

Double integrals over general regions show up whenever the domain isn’t a rectangle:

  • Physics uses them to compute mass and center of mass for irregularly shaped plates with varying density
  • Game engines compute average terrain properties over non-rectangular regions like river basins, forest boundaries, or irregular map zones
  • Engineering calculates stress and heat flow over cross-sections that are circular, triangular, or bounded by curves
  • Environmental science averages rainfall or pollution over watershed boundaries that follow natural terrain

Quiz

Section titled “Quiz”
For a Type I region, the inner integral is with respect to
The first step when setting up a double integral over a general region is
Switching the order of integration is useful when
The area between $y = x^2$ and $y = x + 2$ from $x = -1$ to $x = 2$ is
For the triangle with vertices $(0,0)$, $(1,0)$, $(0,1)$, the upper y-limit in dy dx order is