Evaluating Double Integrals over General Domains

We recently saw that if $z = f(x, y)$ was a two variable real-valued function, then we could evaluate a double integral over a rectangular region $R = [a, b] \times [c, d]$ with iterated integrals, that is $\iint_R f(x, y) \: dA = \int_a^b \int_c^d f(x, y) \: dy \: dx = \int_c^d \int_a^b f(x, y) \: dx \: dy$.

However, we now need to develop a method to evaluate double integrals over general domains. There are two different types of regions for which we will need to integrate upon.

• Type 1 Region: The first type of region that we may need to integrate over would be in the form $D = \{ (x, y) : a ≤ x ≤ b , g_1(x) ≤ y ≤ g_2(x) \}$. These types of domains are sometimes called y-simple domains/regions. Note that $D$ represents the area trapped between the continuous curves $y = g_1(x)$ and $y = g_2 (x)$ for $a ≤ x ≤ b$.

• Type 2 Region: The second type of region that we may need to integrate over would be in the form $D = \{ (x, y) : h_1(y) ≤ x ≤ h_2(y) , c ≤ y ≤ d \}$. These types of domains are sometimes called x-simple domains/regions. Note that here, $D$ represents the area trapped between the continuous curves $x = h_1(y)$ and $x = h_2(x)$ for $c ≤ x ≤ d$.

Now recall that if $D \subseteq D(f)$ and if $R = [a, b] \times [c, d]$ is a rectangle that contains $D$ then for $\hat{f} = \left\{\begin{matrix} f(x,y) & \mathrm{if} (x,y) \in D \\ 0 & \mathrm{if} (x,y) \not \in D \end{matrix}\right.$ we have that:

(1)

\begin{align} \quad \iint_D f(x, y) \: dA =\iint_R \hat{f} (x, y) \: dA = \int_a^b \int_c^d \hat{f} (x, y) \: dy \: dx = \int_c^d \int_a^b \hat{f} (x, y) \: dx \: dy \end{align}

Let's first consider Type 1 Regions. We note that if $(x, y)$ is such that $y < g_1(x)$ or if $y > g_2(x)$, then $(x, y) \not \in D$ and so $\hat{f} (x, y) = 0$. Since $f(x, y) = \hat{f}(x,y)$ when $g_1(x) ≤ y ≤ g_2(x)$, we have that:

(2)

\begin{align} \quad \int_c^d \hat{f} (x, y) \: dy = \int_{g_1(x)}^{g_2(x)} \hat{f} (x, y) \: dy = \int_{g_1(x)}^{g_2(x)} f(x, y) \: dy \end{align}

Therefore we can evaluate the double integral over $D = \{ (x, y) : a ≤ x ≤ b , g_1(x) ≤ y ≤ g_2(x) \}$ with the following formula:

(3)

\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}

Now let's consider Type 2 Regions. We note that if $(x, y)$ is such that $x < h_1(y)$ or if $x > h_2(y)$, then $(x, y) \not \in D$ and so $\hat{f} (x, y) = 0$. Since $f(x, y) = \hat{f}(x, y)$ when $h_1(y) ≤ x ≤ h_2(y)$, we have that:

(4)

\begin{align} \quad \int_a^b \hat{f} (x, y) \: dx = \int_{h_1(y)}^{h_2(y)} \hat{f} (x, y) \: dx = \int_{h_1(x)}^{h_2(x)} f(x,y) \: dx \end{align}

Therefore we can evaluate the double integral over $D = \{ (x, y) : h_1(y) ≤ x ≤ h_2(y) , c ≤ y ≤ d \}$ with the following formula:

(5)

\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}

Remark 1: For type 1 regions an alternative notation for the double integral over $D$ is $\iint_D f(x, y) \: dA = \int_a^b \: dx \int_{g_1(x)}^{g_2(x)} f(x, y) \: dy$, and similarly for type 2 regions we have $\iint_D f(x, y) \: dA = \int_c^d \: dy \int_{h_1(y)}^{h_2(y)} f(x, y) \: dx$.

We will now look at an example of evaluating double integrals over general domains. Once again, it is important to note the following techniques of integration from single variable calculus that we may need to apply:

• Substitution.

• Integration by Parts $\int u \: dv = uv - \int v \: du$.

• Tabular Integration

• Advanced Trigonometric Function Integration

• Integration by Trigonometric Substitution

• Integration with Partial Fractions

Example 1

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

We note that example 1 requires double integration over a type 1 region. The region $D$ is depicted in the following image:

Using the formula above, we have that:

(6)

\begin{align} \quad \iint_D xe^y \: dA = \int_0^1 \int_{0}^{x} xe^y \: dy \: dx \end{align}

We will first evaluate the inner integral $\int_{0}^{x} xe^y \: dy$ as follows, holding $x$ as fixed.

(7)

\begin{align} \quad \int_{0}^{x} xe^y \: dy = \left [ xe^y \right ]_{0}^{x} = xe^{x} - x \end{align}

Therefore we have that:

(8)

\begin{align} \quad \iint_D xe^y \: dA = \int_0^1 xe^{x} - x \: dx = \int_0^1 xe^x - \int_0^1 x \: dx \end{align}

The integral $\int_0^1 xe^x \: dx$ can be solved with either Integration by Parts or Tabular Integration. We get that $\int_0^1 xe^x \: dx = \left [ xe^x - e^x \right]_0^1 = 1$. Furthermore, $\int_0^1 x \: dx = \left [ \frac{x^2}{2} \right ]_0^1 = \frac{1}{2}$. Therefore:

(9)

\begin{align} \quad \iint_D xe^y \: dA = \frac{1}{2} \end{align}