# Evaluating Double Integrals over General Domains Examples 1

Recall from the Evaluating Double Integrals over General Domains page that we can evaluate double integrals over general domains by breaking them up into their respective iterated integrals.

- For type 1 regions ($y$-simple domains) $D = \{ (x, y) : a ≤ x ≤ b , g_1 (x) ≤ y ≤ g_2(x) \}$ where $g_1$ and $g_2$ are continuous, then:

- For type 2 regions ($x$-simple domains) $D = \{ (x, y) : h_1(y) ≤ x ≤ h_2(y), c ≤ y ≤ d \}$ where $h_1(y)$ and $h_2(y)$ are continuous, then:

We will now look at some more examples of evaluating double integrals over general domains. More examples can be found on the following pages:

**Evaluating Double Integrals over General Domains Examples 1**- Evaluating Double Integrals over General Domains Examples 2
- Evaluating Double Integrals over General Domains Examples 3
- Evaluating Double Integrals over General Domains Examples 4
- Evaluating Double Integrals over General Domains Examples 5
- Evaluating Double Integrals over General Domains Examples 6

## Example 1

**Evaluate the double integral $\iint_D x^2 + y^2 + 1 \: dA$ where $D$ is the region trapped between the lines $x = 2$, $y = 0$, and $y = x^3$.**

We will first sketch the domain that we are integrating over:

Therefore we want $0 ≤ x ≤ 2$ and $0 ≤ y ≤ x^3$. Thus $D = \{ (x, y) : 0 ≤ x ≤ 2 , 0 ≤ y ≤ x^3 \}$ (here we have that $g_1(x) = 0$ and $g_2(x) = x^3$). Therefore:

(3)## Example 2

**Evaluate the double integral $\iint_D xy \: dA$ where $D$ is the region of points $(x, y)$ contained in the triangle with vertices $(0, 0)$, $(1, 0)$, and $(1, 1)$.**

Let's first sketch this domain to see if we can represent it in terms of some functions:

Note that $D$ can be described as $D = \{ (x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ x \}$ (here we have that $g_1(x) = 0$ and $g_2(x) = x$, so we want $y$ between the lines $g_1(x) = 0$ and $g_2(x) = x$). Thus we have that:

(4)## Example 3

**Evaluate the double integral $\iint_D 2x + y \: dA$ where $D$ is the region of points $(x, y)$ contained in the triangle with vertices $(0, 0)$, $(0, 3)$, and $(2,0)$.**

Once again, let's first sketch this domain to see if we can represent it in terms of some functions:

Note that $D$ can be described as $D = \{ (x, y) : 0 ≤ x ≤ 2, 0 ≤ y ≤ -\frac{3}{2}x + 3 \}$ (here we have that $g_1(x) = 0$ and $g_2(x) = -\frac{3}{2}x + 3$). Thus:

(5)Alternatively, we could describe $D$ as $D = \{ (x, y) : 0 ≤ x ≤ -\frac{2}{3}y + 2, 0 ≤ y ≤ 3 \}$ (here we have that $h_1(y) = 0$ and $h_2(y) = -\frac{2}{3}y + 2$ which also represents the line defining out triangle). Thus:

(6)No matter which way we interpreted $D$ and evaluated the double integral, we obtained the same result.