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Mom's Basement University

Double Integrals over Rectangular Regions

What You’ll Learn

Section titled “What You’ll Learn”

In this lesson you’ll learn how to set up and evaluate double integrals over rectangular regions. This is the multivariable version of “area under a curve,” except now you’re computing volume under a surface.

The Concept

Section titled “The Concept”

From Single to Double

Section titled “From Single to Double”

In Calculus 1, you computed the area under a curve y = f(x) from a to b using a single integral. In Calculus 3, you have a surface z = f(x, y) floating above the xy-plane, and you want the volume underneath it over some region.

For a rectangular region R = [a, b] x [c, d], the double integral is

∬Rf(x,y) dA=∫ab∫cdf(x,y) dy dx\iint_R f(x, y) \, dA = \int_a^b \int_c^d f(x, y) \, dy \, dx

How Iterated Integrals Work

Section titled “How Iterated Integrals Work”

You evaluate a double integral by doing two single integrals, one at a time. This is called an iterated integral.

  1. Start with the inner integral: integrate f(x, y) with respect to y (treating x as a constant), using the y-limits
  2. The result is a function of x only
  3. Then integrate that result with respect to x, using the x-limits

You can also switch the order and do dx first, then dy. For rectangular regions, both orders give the same answer (Fubini’s theorem)

∫ab∫cdf(x,y) dy dx=∫cd∫abf(x,y) dx dy\int_a^b \int_c^d f(x,y) \, dy \, dx = \int_c^d \int_a^b f(x,y) \, dx \, dy

What Does a Double Integral Actually Represent?

Section titled “What Does a Double Integral Actually Represent?”

The short answer: it depends on what f(x, y) means.

When f(x, y) is a height (like a surface z = f(x, y) above the xy-plane), the double integral gives the volume of the solid between the surface and the floor over the region R. This is the geometric interpretation, and it’s the most visual way to think about it.

But double integrals are more general than just volume. The double integral accumulates (adds up) the values of f over the entire region R. What that “accumulation” means depends on context:

  • If f(x, y) is a height, the double integral is volume
  • If f(x, y) is a density (mass per unit area), the double integral is total mass
  • If f(x, y) = 1 everywhere, the double integral just gives the area of R
  • If f(x, y) is a temperature distribution, the double integral (divided by the area) gives the average temperature

Think of it this way: a single integral adds up values along a line. A double integral adds up values over a flat region. The result is whatever those values represent, accumulated over the whole region.

For this lesson, we’ll focus on the volume interpretation because it’s the most intuitive and visual. But keep in mind that the same math works for any quantity you want to add up over a 2D region.

Worked Examples

Section titled “Worked Examples”

Example 1: Evaluating a double integral step by step

Evaluate the double integral of f(x, y) = x + 2y over the rectangle R = [0, 2] x [1, 3].

Start with the inner integral (integrate with respect to y first, treating x as constant)

∫13(x+2y) dy=[xy+y2]13=(3x+9)−(x+1)=2x+8\int_1^3 (x + 2y) \, dy = \left[ xy + y^2 \right]_1^3 = (3x + 9) - (x + 1) = 2x + 8

Now integrate the result with respect to x

∫02(2x+8) dx=[x2+8x]02=(4+16)−0=20\int_0^2 (2x + 8) \, dx = \left[ x^2 + 8x \right]_0^2 = (4 + 16) - 0 = 20

The volume under z = x + 2y over the rectangle [0, 2] x [1, 3] is 20 cubic units.

The blue surface is z = x + 2y (a tilted plane). The green rectangle on the floor is the region R = [0, 2] x [1, 3]. The volume between the surface and the floor over that rectangle is what the double integral computes: 20 cubic units.

Example 2: Switching the order of integration

We can verify by integrating in the opposite order (dx first, then dy)

∫13∫02(x+2y) dx dy=∫13[x22+2xy]02dy=∫13(2+4y) dy\int_1^3 \int_0^2 (x + 2y) \, dx \, dy = \int_1^3 \left[ \frac{x^2}{2} + 2xy \right]_0^2 dy = \int_1^3 (2 + 4y) \, dy =[2y+2y2]13=(6+18)−(2+2)=20= \left[ 2y + 2y^2 \right]_1^3 = (6 + 18) - (2 + 2) = 20

Same answer. For rectangular regions, the order doesn’t matter. Pick whichever makes the inner integral easier.

Example 3: A product function

Evaluate the double integral of f(x, y) = xy over R = [0, 1] x [0, 2].

∫01∫02xy dy dx=∫01[xy22]02dx=∫012x dx=[x2]01=1\int_0^1 \int_0^2 xy \, dy \, dx = \int_0^1 \left[ \frac{xy^2}{2} \right]_0^2 dx = \int_0^1 2x \, dx = \left[ x^2 \right]_0^1 = 1

When f(x, y) is a product g(x)h(y), you can actually separate the integrals

∫01x dx⋅∫02y dy=12⋅2=1\int_0^1 x \, dx \cdot \int_0^2 y \, dy = \frac{1}{2} \cdot 2 = 1

This separation trick only works for products over rectangles, but it’s a nice shortcut when it applies.

The surface z = xy starts flat at the origin and twists upward toward the corner (1, 2), where it reaches its maximum value of 2. The volume under this surface over the green rectangle is 1 cubic unit.

Real-World Application

Section titled “Real-World Application”

Double integrals over rectangular regions show up in:

  • Physics uses them to compute total mass when density varies over a rectangular plate, or total charge over a rectangular surface
  • Game engines use them to compute average terrain height over rectangular chunks for level-of-detail rendering and AI navigation
  • Engineering uses them for heat distribution calculations over rectangular cross-sections
  • Data science uses them when computing expected values over 2D probability distributions on rectangular domains

Quiz

Section titled “Quiz”
A double integral over a rectangular region computes
When evaluating an iterated integral, you first
For rectangular regions, switching the order of integration
The double integral of $f(x,y) = x + 2y$ over $[0,2] \times [1,3]$ equals
If $f(x,y) = g(x) \cdot h(y)$ over a rectangle, the double integral equals