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
The double integral of, $f$ over $D$ is: $$ \iint_D f(x, y)\ dA = \iint_R \tilde{f}(x, y)\ dA $$
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.
Type I and II regions
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: $$ \boxed{\int_a^b \int_{g_1(x)}^{g_2(x)} f(x, y)\ dy\ dx} $$
Type I definition
If $D$ is of type I, then: $$ \iint_D f(x, y)\ dA = \int_a^b \int_{g_1(x)}^{g_2(x)} f(x, y)\ dy\ dx $$
Type II
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 $y$.
Type II definition
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
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
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
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:
$$ -\dfrac{1}{2} cos(y^2) \bigg\rvert_{y = 0}^{y = 1} $$
$$ \left( -\dfrac{1}{2} cos(1)\right) - \left( -\dfrac{1}{2} cos(0)\right) $$
$$ -\dfrac{1}{2} cos(1) + \dfrac{1}{2} $$
$$ \boxed{\dfrac{1}{2} \left(-cos(1) + 1 \right)} $$