# Evaluating Double Integrals over General Domains Examples 3

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

We first note that $D$ is a type 1 ($y$-simple) region and so:

(3)Now we cannot evaluate the iterated integral above because $\int_{y^2}^2 y^3 e^{x^3} \: dx$ cannot be evaluated with elementary functions. We need to rewrite $D$ as a type 2 ($x$-simple) region.

We are given that $y^2 ≤ x ≤ 2$ and $0 ≤ y ≤ \sqrt{2}$. We immediately have that $0 ≤ x ≤ 2$ and since $y^2 ≤ x$, we have that $y ≤ \sqrt{x}$, and so $0 ≤y ≤ \sqrt{x}$. Thus $D = \{ (x, y) \in \mathbb{R}^2 : 0 ≤ x ≤ 2, 0 ≤ y ≤ \sqrt{x} \}$ and thus:

(4)Now let $u = x^3$ so that $du = 3x^2 \: dx$ and $\frac{1}{3} du = x^2 \: dx$. Thus for some $\alpha, \beta \in \mathbb{R}$ we have that:

(5)## Example 2

**Evaluate the double integral $\iint_D xy \: dA$ where $D$ is the region in the first quadrant that is bounded between the curves $y = 12x - 12x^2$ and $y = x - x^2$.**

To determine the region $D$, let's sketch the region bounded between the curves $y = 12x - 12x^2$ and $y = x - x^2$. Note that both curves are downwards parabolas with roots $x = 0$ and $x = 1$ as both equations can be factored as $y = 12x(1 - x)$ and $y = x(1 - x)$ and so our region $D$ is:

We can see that this is a type 1 region and can be described as $D = \{ (x, y) \in \mathbb{R}^2 : 0 ≤ x ≤ 1, x - x^2 ≤ y ≤ 12x - 12x^2 \}$, and so:

(6)## Example 3

**Evaluate the double integral $\iint_D \sqrt{x^4 + 1} \: dA$ where $D = \{ (x, y) \in \mathbb{R} : \sqrt[3]{y} ≤ x ≤ 2, 0 ≤ y ≤ 8 \}$.**

We first note that $D$ is a type 2 domain, and so:

(7)Now we cannot evaluate the iterated integral above because $\int_{\sqrt[3]{y}}^{2} \sqrt{x^4 + 1} \: dx$ cannot be evaluated with elementary functions. We will rewrite $D$ as a type 1 domain instead.

We are given that $\sqrt[3]{y} ≤ x ≤ 2$ and $0 ≤ y ≤ 8$ and so we immediately have that $0 ≤ x ≤ 2$. Now since $\sqrt[3]{y} ≤ x$ we have that $y ≤ x^3$ and so $0 ≤ y ≤ x^3$ so $D = \{ (x, y) \in \mathbb{R}^2 : 0 ≤ x ≤ 2, 0 ≤ y ≤ x^3 \}$ and so:

(8)Now let $u = x^4 + 1$ so that $du = 4x^3 \: dx$. Therefore $\frac{1}{4} du = x^3 \: dx$ and so for some $\alpha, \beta \in \mathbb{R}$ we have that:

(9)