Double Integrals over General Domains

In the previous note we integrated over rectangles. Most regions of interest in applications are not rectangular. This note explains how to integrate over regions bounded by curves.

1. Type I regions

A Type I region $D$ is bounded on the left and right by $x = a$ and $x = b$, and vertically by two continuous curves:

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

where $g_1(x) \le g_2(x)$ for all $x \in [a,b]$.

Think of a Type I region as one that can be sliced into vertical strips. For each fixed $x$, the strip runs from $y = g_1(x)$ (the lower curve) up to $y = g_2(x)$ (the upper curve).

The interactive graph below shows the Type I region between $y = x^2$ and $y = 2x$ on $[0, 2]$. Adjust the slider to see the vertical strips that make up the region.

2. Iterated integral over a Type I region

The double integral over a Type I region is evaluated by integrating $y$ first (over the strip), then integrating $x$:

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

How to read this: for each fixed $x$, compute the inner integral $\displaystyle A(x) = \int_{g_1(x)}^{g_2(x)} f(x,y)\,dy$ (one slice of volume), then integrate $A(x)$ over $x \in [a,b]$.

Key point: The limits of the inner integral depend on $x$, so they vary as you move the outer integration variable.

3. Type II regions

A Type II region $D$ is bounded above and below by $y = c$ and $y = d$, and horizontally by two continuous curves:

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

where $h_1(y) \le h_2(y)$ for all $y \in [c,d]$.

A Type II region can be sliced into horizontal strips. For each fixed $y$, the strip runs from $x = h_1(y)$ (the left curve) to $x = h_2(y)$ (the right curve).

The double integral over a Type II region is

$\boxed{\iint_D f(x,y)\,dA = \int_c^d \!\int_{h_1(y)}^{h_2(y)} f(x,y)\,dx\,dy.}$

4. Example: Area between $y = 2x^2$ and $y = 1 + x^2$

Find the area of the region $D$ between the parabolas $y = 2x^2$ and $y = 1 + x^2$.

Step 1: Find the intersection.

$2x^2 = 1 + x^2 \implies x^2 = 1 \implies x = \pm 1.$

At $x = 0$: $2x^2 = 0$ and $1+x^2 = 1$, so $2x^2 \le 1+x^2$ on $[-1,1]$. The lower curve is $y = 2x^2$, the upper curve is $y = 1 + x^2$.

Step 2: Set up as Type I.

$\text{Area} = \iint_D dA = \int_{-1}^{1}\!\int_{2x^2}^{1+x^2} dy\,dx.$

Step 3: Evaluate the inner integral.

$\int_{2x^2}^{1+x^2} dy = (1+x^2) - 2x^2 = 1 - x^2.$

Step 4: Evaluate the outer integral.

$\int_{-1}^{1}(1 - x^2)\,dx = \Bigl[x - \tfrac{x^3}{3}\Bigr]_{-1}^{1} = \Bigl(1 - \tfrac{1}{3}\Bigr) - \Bigl(-1 + \tfrac{1}{3}\Bigr) = \frac{4}{3}.$

The area is $\dfrac{4}{3}$.

5. Switching order: Type I vs Type II

Sometimes it is easier to describe a region as Type II and integrate in the order $dx\,dy$, even if the natural first description is Type I.

Example. Evaluate $\displaystyle\int_0^1\!\int_{\sqrt{x}}^{1} \sin(y^3)\,dy\,dx$.

The inner integrand $\sin(y^3)$ has no elementary antiderivative with respect to $y$, so the $dy$-first approach fails.

The region is $\{(x,y): 0 \le x \le 1,\,\sqrt{x} \le y \le 1\}$, which equals $\{(x,y): 0 \le y \le 1,\, 0 \le x \le y^2\}$ as a Type II region.

Switching order:

$\int_0^1\!\int_0^{y^2} \sin(y^3)\,dx\,dy = \int_0^1 y^2 \sin(y^3)\,dy = \Bigl[-\tfrac{\cos(y^3)}{3}\Bigr]_0^1 = \frac{1-\cos 1}{3} \approx 0.153.$

6. Properties of double integrals

Let $D$ be a region in the plane and $f$, $g$ be integrable functions on $D$.

Linearity:

$\iint_D \bigl(c_1 f + c_2 g\bigr)\,dA = c_1\iint_D f\,dA + c_2\iint_D g\,dA.$

Additivity: If $D = D_1 \cup D_2$ with $D_1 \cap D_2$ having zero area, then

$\iint_D f\,dA = \iint_{D_1} f\,dA + \iint_{D_2} f\,dA.$

Comparison: If $f(x,y) \le g(x,y)$ on $D$, then

$\iint_D f\,dA \le \iint_D g\,dA.$

Area as a special case: Setting $f = 1$,

$\iint_D 1\,dA = \text{Area}(D).$

Bound: If $m \le f(x,y) \le M$ on $D$, then

$m \cdot \text{Area}(D) \le \iint_D f\,dA \le M \cdot \text{Area}(D).$

7. Exercises