Evaluating Double Integrals over General Domains Examples 2

# 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:
(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 $\iint_D y^2 e^{xy} \: dA$ 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)
\begin{align} \quad \iint_D y^2 e^{xy} \: dA = \int_0^2 \int_0^y y^2 e^{xy} \: dx \: dy = \int_0^2 \left [ y e^{xy} \right ]_{x = 0}^{x = y} \: dy = \int_0^2 y e^{y^2} - y \: dy = \int_0^2 ye^{y^2} \: dy - \int y \: dy \end{align}

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)
\begin{align} \quad = \frac{1}{2} \int_{\alpha}^{\beta} e^u \: du - \int y \: dy = \frac{1}{2} e^u \biggr \rvert_{u = \alpha}^{u = \beta} - \frac{1}{2} [y^2]_{y=0}^{y=2} = \frac{1}{2} e^{y^2} \biggr \rvert_{y = 0}^{y = 2} - \frac{1}{2} [4 - 0 ] = \frac{1}{2} (e^4 - 1) - 2 = \frac{e^4 - 5}{2} \end{align}

## 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)
\begin{align} \quad \iint_D e^{y^3} \: dA= \int_0^1 \int_{\sqrt{y}}^{1} e^{y^3} \: dy \: dx \end{align}

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

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

## Example 3

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

(8)
\begin{align} \quad \iint_D x y^2 e^x \: dA = \int_0^1 \int_0^{x^2} xy^2 e^x \: dy \: dx = \int_0^1 \left [ \frac{xy^3e^x}{3} \right ]_{y=0}^{y = x^2} \: dx = \int_0^1 \frac{x^7 e^{x}}{3} \: dx \end{align}

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)
\begin{align} \quad \int_0^1 \frac{x^9 e^x}{3} \: dx = \frac{1}{3} \left [ (x^7 - 7x^6 + 42x^5 - 210x^4 + 840x^3 - 2520x^2 + 5040x - 5040)e^x \right]_{x=0}^{x=1} = \frac{-1854e + 5040}{3} \end{align}