10.1 Double integrals over non-rectangular regions

Double integrals over non-rectangular regions let you compute areas, volumes, and other quantities when the domain of integration isn't a simple rectangle. This is where double integration becomes genuinely useful, since most real regions you'll encounter have curved or angled boundaries.

The core skill here is translating a geometric region into the correct limits of integration. That means classifying the region, identifying boundary functions, and writing the iterated integral with proper bounds.

Non-rectangular Regions

Region Types

A non-rectangular region is any region in the $xy$-plane that can't be described as a rectangle with sides parallel to the axes. To integrate over these regions, you describe them using functions as boundaries rather than constants.

There are two standard classifications:

• Type I region: The $x$-values range over a fixed interval $a \leq x \leq b$, and for each $x$, the $y$-values are sandwiched between two continuous functions: $g_1(x) \leq y \leq g_2(x)$. Think of this as slicing the region with vertical lines. Each vertical slice starts at $g_1(x)$ (the bottom curve) and ends at $g_2(x)$ (the top curve).
• Type II region: The $y$-values range over a fixed interval $c \leq y \leq d$, and for each $y$, the $x$-values are bounded by $h_1(y) \leq x \leq h_2(y)$. Here you're slicing with horizontal lines. Each horizontal slice starts at $h_1(y)$ (the left curve) and ends at $h_2(y)$ (the right curve).

Some regions can be described as both Type I and Type II. Others fit naturally into only one type, or need to be split into subregions.

Sketching Regions

Sketching is not optional. It's how you figure out which type of region you're dealing with and what the boundary functions are. Here's the process:

1. Plot each boundary curve in the $xy$-plane.
2. Find the intersection points of the boundary curves to determine the limits on $x$ (or $y$).
3. Shade the enclosed region.
4. Decide whether vertical slices (Type I) or horizontal slices (Type II) give cleaner bounds. If one variable's bounds would require splitting into multiple pieces, try the other order.
5. Read off the boundary functions and their domains from the sketch.

Iterated Integrals Over Non-rectangular Regions

Setting Up Iterated Integrals

An iterated integral evaluates a double integral by integrating one variable at a time. The key difference from rectangular regions is that the inner limits of integration are now functions, not constants.

For a Type I region, you integrate $y$ first (inner integral), with limits that depend on $x$, then integrate $x$ (outer integral) over constant limits:

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

For a Type II region, you integrate $x$ first (inner integral), with limits that depend on $y$, then integrate $y$ (outer integral) over constant limits:

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

The outer integral always has constant limits. The inner integral has variable limits given by the boundary functions. Getting this backwards is one of the most common mistakes.

Fubini's Theorem

Fubini's Theorem guarantees that if $f(x,y)$ is continuous over a closed, bounded region $R$, then the double integral $\iint_R f(x,y)\, dA$ can be evaluated as an iterated integral. If the region can be described as both Type I and Type II, you can choose either order of integration and get the same answer:

• Type I: $\iint_R f(x,y)\, dA = \int_a^b \int_{g_1(x)}^{g_2(x)} f(x,y)\, dy\, dx$
• Type II: $\iint_R f(x,y)\, dA = \int_c^d \int_{h_1(y)}^{h_2(y)} f(x,y)\, dx\, dy$

This flexibility matters. Sometimes one order of integration leads to an integral you can evaluate in closed form, while the other order does not. Choosing wisely can save you significant work.

Examples

Example 1 (Type I): Let $R = \{(x,y) \mid 0 \leq x \leq 1,\; x^2 \leq y \leq \sqrt{x}\}$.

The region sits between the parabola $y = x^2$ (below) and the curve $y = \sqrt{x}$ (above), with $x$ running from 0 to 1. These curves intersect at $(0,0)$ and $(1,1)$. Vertical slices give clean bounds, so the iterated integral is:

$\int_0^1 \int_{x^2}^{\sqrt{x}} f(x,y)\, dy\, dx$

Example 2 (Type II): Let $R = \{(x,y) \mid 0 \leq y \leq 1,\; y \leq x \leq y^2 + 1\}$.

For each fixed $y$, $x$ ranges from the line $x = y$ (left boundary) to the parabola $x = y^2 + 1$ (right boundary), with $y$ running from 0 to 1. Horizontal slices work naturally here:

$\int_0^1 \int_y^{y^2+1} f(x,y)\, dx\, dy$

Example 3 (Switching order): Suppose you need to evaluate $\int_0^1 \int_x^1 e^{y^2}\, dy\, dx$. The inner integral $\int_x^1 e^{y^2}\, dy$ has no elementary antiderivative in $y$, so you can't evaluate it directly.

1. Sketch the region: $0 \leq x \leq 1$, $x \leq y \leq 1$. This is the triangle above the line $y = x$ and below $y = 1$.
2. Rewrite as Type II: for each $y$ from 0 to 1, $x$ runs from 0 to $y$.
3. The switched integral is $\int_0^1 \int_0^y e^{y^2}\, dx\, dy$.
4. Now the inner integral is just $\int_0^y e^{y^2}\, dx = y\, e^{y^2}$, which you can integrate in $y$ using substitution ($u = y^2$).

This is exactly the kind of situation where choosing the right order of integration makes the problem solvable.