# Evaluating Double Integrals over General Domains Examples 2

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 $\iint_D y^2 e^{xy} \: dA$ where $D = \{ (x, y) \in \mathbb{R}^2 : 0 ≤ x ≤ y, 0 ≤ y ≤ 2 \}$.**

We note that $D$ is a type 2 ($x$-simple) region and thus we have that:

(3)Let $u = y^2$ so that $du = 2y \: dy$ and $\frac{1}{2} du = y \: du$. So for some $\alpha, \beta \in \mathbb{R}$ we have that:

(4)## Example 2

**Evaluate the double integral $\iint_D e^{y^3} \: dA$ where $D = \{ (x, y) \in \mathbb{R}^2 : 0 ≤ x ≤ 1 , \sqrt{x} ≤ y ≤ 1 \}$.**

First note that if we try to evaluate this double integral as an iterated integral by integrating with respect to $y$ first, we get that:

(5)However, we cannot compute $\int_{\sqrt{y}}^{1} e^{y^3} \: dy$ since we cannot anti-differentiate $e^{y^3}$ with elementary functions. Let's try to rewrite our type 1 domain into a type 2 domain.

We are given that $0 ≤ x ≤ 1$ (*) and $\sqrt{x} ≤ y ≤ 1$ (x). Since (*) holds, we see that $0 ≤ y ≤ 1$ from (x). Now we need to find bounds for $x$. If $\sqrt{x} ≤ y$ then $x ≤ y^2$. But from (*) we have that $0 ≤ x$ and so $0 ≤ x ≤ y^2$. Thus $D = \{ (x, y) \in \mathbb{R}^2 : 0 ≤ x ≤ y^2, 0 ≤ y ≤ 1 \}$ and so:

(6)We now need to use substitution to evaluate this integral. Let $u = y^3$ so that $du = 3y^2 \: dy$ and $\frac{1}{3} du = y^2 \: dy$. Then for some $\alpha, \beta \in \mathbb{R}$ we have that:

(7)## Example 3

**Evaluate the double integral $\iint_D x y^2 e^x \: dA$ where $D = \{ (x, y) \in \mathbb{R}^2 : 0 ≤ x ≤ 1, 0 ≤ y ≤ x^2 \}$.**

Evaluating the integral above is a lengthly process but can be done by using integration by parts or tabular integration which is preferred. We get that:

(9)