Evaluating Double Integrals over General Domains Examples 1

# 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:
(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 $\iint_D x^2 + y^2 + 1 \: dA$ 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)
\begin{align} \quad \iint_D x^2 + y^2 + 1 \: dA = \int_0^2 \int_0^{x^3} x^2 + y^2 + 1 \: dy \: dx = \int_0^2 \left [ x^2y + \frac{y^3}{3} + y \right ]_{y=0}^{y=x^3} \: dx = \int_0^1 x^5 + \frac{x^9}{3} + x^3 \: dx \\ = \left [ \frac{x^6}{6} + \frac{x^{10}}{30} + \frac{x^4}{4} \right ]_{x=0}^{x=1} = \frac{64}{6} + \frac{1024}{30} + 4 = \frac{244}{5} \end{align}

## 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)
\begin{align} \quad \iint_D xy \: dA = \int_0^1 \int_0^x xy \: dy \: dx = \int_0^1 \left [ \frac{xy^2}{2} \right ]_{y=0}^{y=x} \: dx = \int_0^1 \frac{x^3}{2} \: dx = \left [ \frac{x^4}{8} \right ]_{x=0}^{x=1} = \frac{1}{8} \end{align}

## 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)
\begin{align} \quad \iint_D 2x + y \: dA = \int_0^2 \int_0^{-\frac{3}{2}x + 3} 2x + y \: dy \: dx = \int_0^1 \left [ 2xy + \frac{y^2}{2} \right ]_{y=0}^{y = -\frac{3}{2}x + 3} \: dx = \int_0^1 -3x^2 + 6x + \frac{9x^2}{8} - \frac{9}{2}x + \frac{9}{2} \: dx \\ = \int_0^1 -\frac{15x^2}{8} + \frac{3x}{2} + \frac{9}{2} \: dx = \left [ -\frac{15x^3}{24} + \frac{3x^2}{4} + \frac{9}{2}x \right ]_{x=0}^{x=2} = -5 + 3 + 9 = 7 \end{align}

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)
\begin{align} \quad \iint_D 2x + y \: dA = \int_0^3 \int_{0}^{-\frac{2}{3}y + 2} 2x + y \: dx \: dy = \int_0^3 \left [ x^2 + xy \right ]_{x=0}^{x = -\frac{2}{3}y + 2} \: dy = \int_0^3 \frac{4y^2}{9} - \frac{8y}{3} + 4 - \frac{2y^2}{3} + 2y \: dy \\ = \int_0^3 -\frac{2y^2}{9} - \frac{2y}{3} + 4 \: dy = \left [ -\frac{2y^3}{27} - \frac{2y^2}{6} + 4y \right ]_{y=0}^{y=3} = -2 -3 + 12 = 7 \end{align}

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