Evaluating Double Integrals over General Domains Examples 5

# Evaluating Double Integrals over General Domains Examples 5

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 $\iint_R e^{-x^2} \: dA$ where $D = \{ (x, y) \in \mathbb{R}^2 : y ≤ x ≤ 1, 0 ≤ y ≤ 1 \}$.

We see that $D$ is a type 1 region, and so:

(3)
\begin{align} \quad \iint_D e^{-x^2} \: dA = \int_0^1 \int_y^1 e^{-x^2} \: dx \: dy \end{align}

We will need to flip the order of integration since $\int_y^1 e^{-x^2} \: dx$ cannot be readily evaluated with elementary functions.

We are given that $y ≤ x ≤ 1$ and $0 ≤ y ≤ 1$. We immediately have that $0 ≤ x ≤ 1$, and since $y ≤ x$, we have that $0 ≤ y ≤ x$ so $D = \{ (x, y) \in \mathbb{R}^2 : 0 ≤ x ≤ 1, 0 ≤ y ≤ x \}$ and so:

(4)
\begin{align} \quad \iint_D e^{-x^2} \: dA = \int_0^1 \int_0^x e^{-x^2} \: dy \: dx = \int_0^1 \left [ ye^{-x^2} \right ]_{y=0}^{y=x} \: dx = \int_0^1 xe^{-x^2} \: dx \end{align}

We will need to use substitution to evaluate this integral. Let $u = -x^2$. Then $du = -2x \: dx$ so $-\frac{1}{2} du = x \: dx$ and thus for some $\alpha, \beta \in \mathbb{R}$ we have that:

(5)
\begin{align} \quad = - \frac{1}{2} \int_{\alpha}^{\beta} e^u \: du = - \frac{1}{2} e^{u} \biggr \rvert_{u = \alpha}^{u = \beta} = - \frac{1}{2} e^{-x^2} \biggr \rvert_{x = 0}^{x=1} = \frac{1 - e^{-1}}{2} \end{align}

## Example 2

Evaluate the double integral $\iint_R x^{\lambda} y^{\lambda - 1} \: dA$ where $\lambda$ is a positive integer, and $D = \{ (x, y) \in \mathbb{R}^2 : 0 ≤ x ≤ 1, 0 ≤ y ≤ \alpha + x \}$ where $\alpha > 0$. (Hint: You will need to use the binomial expansion formula). Use your formula to compute $\iint_R x^3 y^2 \: dA$ where $R = \{ (x, y) \in \mathbb{R}^2 : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 + x \}$.

This is a type 1 integral, and we have that:

(6)
\begin{align} \quad \iint_R x^{\lambda} y^{\lambda - 1} \: dA = \int_0^1 \int_0^{\alpha + x} x^{\lambda} y^{\lambda - 1} \: dy \: dx = \frac{1}{\lambda} \int_0^1 \left [ x^{\lambda} y^{\lambda} \right ]_{y = 0}^{y = \alpha + x} \: dx = \frac{1}{\lambda} \int_0^1 x^{\lambda}(\alpha + x)^{\lambda} \: dx \\ = \frac{1}{\lambda} \int_0^1 x^{\lambda} \left [ \binom{\lambda}{0} \alpha^{\lambda} x^0 + \binom{\lambda}{1} \alpha^{\lambda - 1}x^1 + ... + \binom{\lambda}{\lambda - 1} \alpha^1 x^{\lambda-1} + \binom{\lambda}{\lambda} \alpha^0 x^{\lambda} \right ] \: dx \\ = \frac{1}{\lambda} \int_0^1 \binom{\lambda}{0} \alpha^{\lambda} x^{\lambda} + \binom{\lambda}{1} \alpha^{\lambda - 1}x^{\lambda + 1} + ... + \binom{\lambda}{\lambda - 1} \alpha^1 x^{2\lambda-1} + \binom{\lambda}{\lambda} \alpha^0 x^{2\lambda} \: dx \\ = \frac{1}{\lambda} \left [ \binom{\lambda}{0} \alpha^{\lambda} \frac{x^{\lambda+1}}{\lambda + 1} + \binom{\lambda}{1} \alpha^{\lambda - 1} \frac{x^{\lambda + 2}}{\lambda + 2} + ... + \binom{\lambda}{\lambda - 1} \alpha^1 \frac{x^{2\lambda}}{2 \lambda} + \binom{\lambda}{\lambda} \alpha^0 \frac{x^{2\lambda + 1}}{2\lambda + 1} \right]_{x=0}^{x=1} \\ = \frac{1}{\lambda} \left [ \binom{\lambda}{0} \frac{\alpha^{\lambda}}{\lambda + 1} + \binom{\lambda}{1} \frac{\alpha^{\lambda-1}}{\lambda + 2} + ... + \binom{\lambda}{\lambda - 1} \frac{\alpha^{1}}{2 \lambda} + \binom{\lambda}{\lambda} \frac{\alpha^{0}}{2\lambda + 1} \right ] \end{align}

Now if we want to evaluate $\iint_R x^3 y^2 \: dA$ where $R = \{ (x, y) \in \mathbb{R}^2 : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 + x \}$, then we have that $\lambda = 3$ and $\alpha = 1$, and so our formula yields that:

(7)
\begin{align} \quad \quad \iint_R x^3 y^2 \: dA = \frac{1}{3} \left [ \binom{3}{0} \frac{1^3}{3 + 1} + \binom{3}{1} \frac{1^2}{3 + 2}+ \binom{3}{2} \frac{1^1}{3 + 3} + \binom{3}{3} \frac{1^0}{3 + 4} \right ] = \frac{1}{3} \left [ \frac{1}{4} + \frac{3}{5} + \frac{3}{6} + \frac{1}{7} \right ] = \frac{1254}{2520} \end{align}

Let's also evaluate this double integral the conventional way.

(8)
\begin{align} \quad \iint_R x^3 y^2 \: dA = \int_0^1 \int_0^{1 + x} x^3 y^2 \: dy \: dx = \frac{1}{3} \int_0^1 \left [ x^3 y^3 \right]_{y=0}^{y = 1+x} \: dx = \frac{1}{3} \int_0^1 x^3(1 + x)^3 \: dx \\ = \frac{1}{3} \int_0^1 x^3 (1 + 3x + 3x^2 + x^3) \: dx = \frac{1}{3} \int_0^1 x^3 + 3x^4 + 3x^5 + x^6 \: dx = \frac{1}{3} \left [ \frac{x^4}{4} + \frac{3x^5}{5} + \frac{3x^6}{6} + \frac{x^7}{7} \right ]_{x=0}^{x=1} = \frac{1}{3} \left [ \frac{1}{4} + \frac{3}{5} + \frac{3}{6} + \frac{1}{7} \right ] = \frac{1254}{2520} \end{align}