Double Integrals

Let $z = f(x,y)$ be a bounded function on the rectangular region $W$ over the $xy$ -plane. Divide the rectangular region $W$ by the straight lines parallel to $x$ -axis and $y$ -axis and denote the partitioned small rectangles by $W_{1,1},W_{1,2},\ldots,W_{m,n}$ . We denote this partition by $\Delta$ . Now for each $W_{ij}$ , take an arbitrary point $(x_i,y_j)$ and consider the sum of small rectangular parallelpiped $S(\Delta)$ . Let

$\displaystyle S(\Delta) = \sum_{i=1}^{m}\sum_{j=1}^{n}f(x_i,y_j)\vert W_{ij}\vert$

where $\vert W_{ij}\vert$ is the area of $W_{ij}$ and $\vert\Delta\vert$ is the longest diagonal of $W_{ij}$ . If $S(\Delta)$ approaches the same value as $\vert\Delta\vert$ approaches 0 independent of the partition and the choice of $(x_i,y_j)$

, then

is called Double Integrable on $W$

$\displaystyle \lim_{\Delta \to 0}S(\Delta) = \lim_{\stackrel{m \to \infty}{ n \... ...sum_{i=1}^{m}\sum_{j - 1}^{n}f(x_i,y_j)\vert W_{ij}\vert = \iint_{W}f(x,y)dxdy$

**Figure 5.1:** small parallelpiped
$\includegraphics[width=5cm]{SOFTFIG-5/doubleint.eps}$

**Figure 5.2:** small rectangle
$\includegraphics[width=4cm]{SOFTFIG-5/rectangleregion}$

Sum
$S(\Delta)$ is written as $\sum_{j=1}^{n}\sum_{i=1}^{m}f(x_i,y_j)\vert W_{ij}\vert$ . Thus add the small rectangular parallelpiped in the direction of -axis, then add in the direction of -axis is the same as add the small rectangular parallelpiped in the direction of -axis, then add in the direction of -axis. This is the basic concept of the repeated integrals.

Theorem 5..1 (Fubini Theorem) Let $W = \{(x,y):a\leq x \leq b, c\leq y \leq d\}$ . If $f(x,y)$

is continuous on $W$

, then

$\displaystyle \iint_W f(x,y)dxdy$

$\displaystyle =$

$\displaystyle \int_a^b\int_c^d f(x,y)dydx = \int_c^d \int_a^b f(x,y)dxdy$

NOTE Fix $y$ and conside the integration of $f(x,y)$ from $a$ to $b$ with respect to $x$ . Then we have

$\displaystyle \int_{a}^{b}f(x,y)dx$

Now integrate this from $c$

with respect to $y$

to obtain

$\displaystyle \int_c^d \left(\int_a^b f(x,y)dx\right)dy {\mbox{or} \int_c^d \int_a^b f(x,y)dxdy}$

This is called repeated integral.

Example 5..1 Let $W = \{(x,y) : 0 \leq x \leq 1, 0 \leq y \leq 2\}$ . Evaluate the following repeated integral $\iint_{W}x^2 y dx dy$ .

SOLUTION

$\displaystyle \iint_W f(x,y)dxdy$	$\displaystyle =$	$\displaystyle \int_0^1 \big(\int_0^2 x^2ydy\big)dx = \int_0^1 x^2 \big(\int_0^2 ydy\big)dx$
	$\displaystyle =$	$\displaystyle \int_0^1 x^2 \big(\big[\frac{y^2}{2}\big]_{0}^{2}\big)dx$
	$\displaystyle =$	$\displaystyle \int_0^1 2x^2 dx = \big[\frac{2}{3}x^3\big]_0^1 = \frac{2}{3}$

We evaluate $\int_0^2 x^2ydy$ by keeping $x$ as a constant.

$\displaystyle \int_0^2 x^2y dy$	$\displaystyle =$	$\displaystyle x^2 \int_0^2 ydy$
	$\displaystyle =$	$\displaystyle x^2 [\frac{y^2}{2}]_0^2$
	$\displaystyle =$	$\displaystyle 2x^2\ensuremath{ \blacksquare}$

Exercise 5..1 Let $W = \{(x,y) : 0 \leq x \leq 2, 0 \leq y \leq 1\}$ . Evaluate the following repeated integrals $\iint_{W}x\sin{\pi y} dx dy$ .

Exercise5-1
It is possible to evaluate $\int_0^1 \big(\int_0^2 x\sin{\pi y}dx\big)dy$ .

SOLUTION

$\displaystyle \iint_W x\sin{\pi y}dx dy$	$\displaystyle =$	$\displaystyle \int_0^2 \big(\int_0^1 x\sin{\pi y}dy\big)dx$
	$\displaystyle =$	$\displaystyle \int_0^2 x \big(\int_0^1 \sin{\pi y}dy\big)dx$
	$\displaystyle =$	$\displaystyle \int_0^2 x \big(\big[-\frac{1}{\pi}\cos{\pi y}\big]_0^1\big)dx$
	$\displaystyle =$	$\displaystyle \int_0^2 x \big(-\frac{1}{\pi}(-1 - 1)\big)dx$
	$\displaystyle =$	$\displaystyle \int_0^2 \frac{2}{\pi}x dx = \big[\frac{2}{\pi}(\frac{x^2}{2})\big]_0^2 = \frac{2}{\pi}(\frac{4}{2}) = \frac{4}{\pi} \ensuremath{ \blacksquare}$

Let $z = f(x,y)$ be a function defined on the closed bounded region $\Omega$ on $xy$ -plane. Let $W$ be a rectangular region containing $\Omega$ . Now divide the rectangular region $W$ by the straight lines parallel to $x$ -axis and $y$ -axis and denote the partitioned small rectangles by $W_{1,1},W_{1,2},\ldots,W_{m,n}$ .

**Figure 5.3:** Partition
$\includegraphics[width=7cm]{SOFTFIG-5/doubleintoveromega.eps}$

Let $\hat{f}(x,y)$ be the function on $W$

defined by

$\displaystyle \hat{f}(x,y) = \left\{\begin{array}{cl} f(x,y) & (x,y) \in \Omega\\ 0 & (x,y) \in W - \Omega \end{array} \right.$

If $\hat{f}(x,y)$ is integrable on $W$

, then we say $f(x,y)$

is integrable on $\Omega$ and the integration of $f(x,y)$

on $\Omega$ is expressed as follows:

$\displaystyle \iint_{W}\hat{f}(x,y) dxdy = \iint_{\Omega}f(x,y)dxdy$

Theorem 5..2 Suppose that $f(x,y),g(x,y)$

are continuous on $\Omega$ . Then we have the followings.

1. Let $a,b$ be constants. Then

$\displaystyle \iint_{\Omega}\{af(x,y) + bg(x,y)\}dxdy = a\iint_{\Omega}f(x,y)dxdy + b\iint_{\Omega}g(x,y)dxdy$

2. If $\Omega = \Omega_{1} \cup \Omega_{2}$ , then

$\displaystyle \iint_{\Omega}f(x,y)dxdy = \iint_{\Omega_{1}}f(x,y)dxdy + \iint_{\Omega_{2}}f(x,y)dxdy$

3. $\displaystyle{\vert\iint_{\Omega}f(x,y)dxdy\vert \leq \iint_{\Omega}\vert f(x,y)\vert dxdy }$

Understanding
If the region $\Omega$ is not a rectangular region, then consider the rectangle containing $\Omega$ . For the rectangle inside of $\Omega$ , use as it is given. For the rectangular region outside of $\Omega$ , we set . This way we can define repeated integral over $\Omega$ .

Linearity
Theorem5-2-1. is called linearity of double integral.

**Figure 5.4:** Theorem5-2-2
$\includegraphics[width=3.5cm]{SOFTFIG-5/dividedregion.eps}$