Part 10 - General regions (1)

Introduction

In this part we’ll cover how to compute the double integral over general regions.

Double integrals over general regions

Let $f(x, y)$ be defined on the domain, $D$ , where $D$ is a bounded region. $D$ is a general region, meaning it can have any geometrical shape.

However, if we reduce our problem into something we know, a rectangular region, we know how to compute the double integral.

Therefore, let’s enclose $D$ , with a rectangle, let’s call this rectangle for $R$ .

We define a function, $\tilde{f}$ , on $R$ as the following:

\tilde{f}(x, y) = \begin{cases} f(x, y) & (x, y) \in D \newline 0 & (x, y) \notin D \end{cases}

Definition 1 (Double integral over a general region)

The double integral of, $f$ over $D$ is:

\iint_D f(x, y)\ dA = \iint_R \tilde{f}(x, y)\ dA

Definition 2 (Equivalent definition)

Recall our original definition of the double integral, we can also divide the domain, $D$ , into smaller pieces.

Therefore, the double integral of, $f$ , over $D$ is also:

\iint_{D} f(x, y)\ dA = \lim_{\substack{\Delta x \to 0 \newline \Delta y \to 0}} \sum_{j = 1}^{m} \sum_{i = 1}^{n} f(x_{i}^{\*}, y_{j}^{\*}) (\Delta A)_{ij}

Since everything outside $D$ , is 0, they contribute nothing to the overall volume. Therefore, the volume under $\tilde{f}$ is the same as the volume under $f$ .

How to compute general double integrals

To compute double integrals over a general region, we firstly need to introduce a new concept.

Definition 3 (Type I region)

A domain, $D$ , is of type I, if it lies between two graphs of two continuous functions of x.

Meaning:

D = \{(x, y) | a \leq x \leq b, g_1(x) \leq y \leq g_2(x)\}

Let’s see how we can compute this now. Let $D$ be of type I. Let’s enclose this region within a rectangle as well:

\iint f(x, y)\ dA = \iint_R \tilde{f}(x, y)\ dA = \int_a^b \int_c^d \tilde{f}(x, y)\ dy\ dx

If we fix $x$ , then $\tilde{f}$ becomes:

\tilde{f}(x, y) = \begin{cases} 0 & c \leq y \leq g_1(x) \newline f(x, y) & g_1(x) \leq y \leq g_2(x) \newline 0 & g_2(x) \leq y \leq d \end{cases}

Which means:

\boxed{\int_a^b \int_{g_1(x)}^{g_2(x)} f(x, y)\ dy\ dx}

Proposition 1 (Integration over a type I region)

If $D$ is of type I, then:

\iint_D f(x, y)\ dA = \int_a^b \int_{g_1(x)}^{g_2(x)} f(x, y)\ dy\ dx

Definition 4 (Type II region)

D is of type II, if it lies between two graphs of two continuous functions of y.

D = \{(x, y) | h_1(y) \leq x \leq h_2(y), c \leq y \leq d\}

We perform the same procedure as type I, instead of fixing $x$ , we fix $y$ .

Proposition 2 (Integration over a type II region)

If $D$ is of type II, then:

\iint_D f(x, y)\ dA = \int_c^d \int_{h_1(y)}^{h_2(y)} f(x, y)\ dx\ dy

Before we head into doing a few examples, it’s a good practice always draw where our $D$ lies. So we easily get a geometrical sense of how it lies.

Example 1

Find $\iint_D y\ dA$ , where $D$ is the region bounded by, $y = 2x^2, y = 1 + x^2$ .

So, our region is bounded by two parabolas, we have two functions of $x$ , therefore, our region is of type I.

Let’s firstly find where these graphs intersect:

2x^2 = 1 + x^2 \newline x^2 = 1 \newline x = \pm 1

Graphing these we also see that, $y = 1 + x^2$ is the upper graph, and the lower is $y = 2x^2$ .

\int_{-1}^1 \int_{2x^2}^{1 + x^2} y\ dy\ dx

\dfrac{1}{2} \int_{-1}^1 y^2 \bigg\rvert_{y = 2x^2}^{y = 1 + x^2}\ dx

\dfrac{1}{2} \int_{-1}^1 (1 + x^2)^2 - (2x^2)^2\ dx

\dfrac{1}{2} \int_{-1}^1 1 + 2x^2 + x^4 - 4x^4 \ dx

\dfrac{1}{2} \int_{-1}^1 1 + 2x^2 - 3x^4\ dx

\dfrac{1}{2} \left[x + \dfrac{2}{3} x^3 - \dfrac{3}{5} x^5 \bigg\rvert_{x = -1}^{x = 1} \right]

\dfrac{1}{2} \left[\left(1 + \dfrac{2}{3} - \dfrac{3}{5}\right) - \left(-1 - \dfrac{2}{3} + \dfrac{3}{5}\right) \right]

\dfrac{1}{2} \left[1 + \dfrac{2}{3} - \dfrac{3}{5} + 1 + \dfrac{2}{3} - \dfrac{3}{5} \right]

\boxed{1 + \dfrac{2}{3} - \dfrac{3}{5}}

Example 2

Find $\iint_D xy\ dA$ , where $D$ is the region bounded by, $y = x -1, y^2 = x+ 1$ .

Now, this region is technically both of type I and II. However, one of the choices will make our lives easier.

Let’s rewrite our bounds as, $x = y + 1, x = y^2 -1$ .

Let’s find where these intersect:

y + 1 = y^2 - 1 \newline y^2 -y -2 = 0 \newline y_1 = \ldots = -1 \newline y_2 = \ldots = 2 \newline

Which means:

\int_{-1}^{2} \int_{y^2 - 1}^{y + 1} xy\ dx\ dy

\int_{-1}^{2} \dfrac{y}{2} \left[x^2\right] \bigg\rvert_{x = y^2 - 1}^{x = y + 1}\ dy

\int_{-1}^{2} \dfrac{y}{2} \left[(y + 1)^2 - (y^2 - 1)^2\right]\ dy

\int_{-1}^{2} \dfrac{y}{2} \left[y^2 + 2y + 1 - y^4 + 2y^2 - 1 \right]\ dy

\int_{-1}^{2} \dfrac{y}{2} \left[3y^2 + 2y - y^4 \right]\ dy

\int_{-1}^{2} \dfrac{3}{2} y^3 + y^2 - \dfrac{y^5}{2}\ dy

\dfrac{3}{8} y^4 + \dfrac{y^3}{3} - \dfrac{y^6}{12} \bigg\rvert_{y = -1}^{y = 2} = \ldots = \boxed{\dfrac{27}{8}}

Changing order of integral

One might now think that we can change the order of integration, using Fubini’s Theorem.

That is not the case.

Say we have:

\int_{g_1(x)}^{g_2(x)} \int_a^b f(x, y)\ dx\ dy

Computing this entire integral will not yield a singular number, it will yield a function of $x$ instead.

However, if $D$ is both type I and II, we can change order of integration, but we need to change our limits accordingly.

Example 3

Find $\iint_D sin(y^2)\ dA$ , where $D$ is the region bounded by, $x = 0, y = 1, y = x$

Graphing this we can see that this a region of both types.

For, type I:

\iint_D f(x, y)\ dA = \int_0^1 \int_x^y sin(y^2)\ dy\ dx

Type II:

\iint_D f(x, y)\ dA = \int_0^1 \int_0^y sin(y^2)\ dx\ dy

In this case, one of these isn’t even integrable in terms of elementary functions. So yeah…

Solving the second integral:

\int_0^1 \int_0^y sin(y^2)\ dx\ dy

\int_0^1 x \cdot sin(y^2) \bigg\rvert_{x = 0}^{x = y}\ dy

\int_0^1 y \cdot sin(y^2) \ dy

Using normal u-sub, we obtain:

-\dfrac{1}{2} cos(y^2) \bigg\rvert_{y = 0}^{y = 1}

\left( -\dfrac{1}{2} cos(1)\right) - \left( -\dfrac{1}{2} cos(0)\right)

-\dfrac{1}{2} cos(1) + \dfrac{1}{2}

\boxed{\dfrac{1}{2} \left(-cos(1) + 1 \right)}