Evaluating Double Integrals over General Domains Examples 3

# 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:
(1)
\begin{align} \quad \iint_D f(x, y) \: dA = \int_a^b \int_{g_1(x)}^{g_2(x)} f(x, y) \: dy \:dx \end{align}
• 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:
(2)
\begin{align} \quad \iint_D f(x, y) \: dA = \int_c^d \int_{h_1(y)}^{h_2(y)} f(x, y) \: dx \: dy \end{align}

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

## 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)
\begin{align} \quad \iint_D c= \int_0^{\sqrt{2}} \int_{y^2}^2 y^3 e^{x^3} \: dx \: dy \end{align}

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)
\begin{align} \quad \iint_D \int_{y^2}^2 y^3 e^{x^3} \: dx = \int_0^2 \int_0^{\sqrt{x}} y^3 e^{x^3} \: dy \: dx = \int_0^2 \left [ \frac{1}{4} y^4 e^{x^3} \right ]_{y=0}^{y = \sqrt{x}}\: dx = \int_0^2 \frac{1}{4} x^2 e^{x^3} \: dx = \frac{1}{4} \int_0^2 x^2 e^{x^3} \: dx \end{align}

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)
\begin{align} \quad = \frac{1}{12} \int_{\alpha}^{\beta} e^u \: du = \frac{1}{12} \left [ e^u \right ]_{u = \alpha}^{u = \beta} = \frac{1}{12} \left [ e^{y^3} \right ]_{y = 0}^{y=2} = \frac{1}{12} (e^8 - 1) = \frac{e^8 - 1}{12} \end{align}

## 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)
\begin{align} \quad \iint_D xy \: dA = \int_0^1 \int_{x - x^2}^{12x - 12x^2} xy \: dy \: dx =\int_0^1 \left [ \frac{1}{2} xy^2 \right ]_{y = x - x^2}^{y = 12x - 12x^2} \: dx = \frac{1}{2} \int_0^1 x(12x - 12x^2)^2 - x(x - x^2)^2 \: dx \\ = \frac{1}{2} \int_0^1 x [ (144x^2 - 288x^3 + 144x^4) - (x^2 - 2x^3 + x^4)] = \frac{1}{2} \int_0^1 x [ 143x^2 -286x^3 + 143x^4 ] \: dx = \frac{143}{2} \int_0^1 x^3 - 2x^4 + x^5 \: dx \\ = \frac{143}{2} \left [ \frac{x^4}{4} - \frac{2x^5}{5} + \frac{x^6}{6} \right ]_{x=0}^{x=1} = \frac{143}{2} \left [ \frac{1}{4} - \frac{2}{5} + \frac{1}{6} \right ] = \frac{143}{2} \left [ \frac{1}{60} \right ] = \frac{142}{120} \end{align}

## Example 3

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

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

(7)
\begin{align} \quad \iint_D \sqrt{x^4 + 1} \: dA = \int_0^8 \int_{\sqrt{y}}^{2} \sqrt{x^4 + 1} \: dx \: dy \end{align}

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

We are given that $\sqrt{y} ≤ x ≤ 2$ and $0 ≤ y ≤ 8$ and so we immediately have that $0 ≤ x ≤ 2$. Now since $\sqrt{y} ≤ x$ 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)
\begin{align} \quad \iint_D \sqrt{x^4 + 1} \: dA = \int_0^2 \int_0^{x^3} \sqrt{x^4 + 1} \: dy \: dx = \int_0^2 \left [ y \sqrt{x^4 + 1}\right ]_{y=0}^{y=x^3} \: dx = \int_0^2 x^3 \sqrt{x^4 + 1} \: dx \end{align}

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)
\begin{align} \quad = \frac{1}{4} \int_{\alpha}^{\beta} u^{1/2} \: du = \frac{2}{12} \left [ (u)^{3/2} \right ]_{u = \alpha}^{u = \beta} = \frac{1}{6} \left [ (x^4 + 1)^{3/2} \right ]_{y=0}^{y=2} = \frac{ 17^{3/2} - 1}{6} \end{align}