Iterated integral on a bounded closed set

We saw the method of iterative integral for a rectangular region. We now extend the method to the case of non-rectangular regions.

See also: Iterated integral on a rectangular region

Consider two continuous functions $y=\phi (x)$ and  $y=\psi (x)$ on $[a,b]$ such that $\psi (x)<\phi (x)$ for all $x\in [a,b]$. We can define the following region $D$ in ${\mathbb{R}}^2$ sandwiched by these functions:

$$\begin{array}{}\text{(Eq:Dsand)}& D=\{(x,y)\mid a\le x\le b,\psi (x)\le y\le \phi (x)\}.\end{array}$$

Clearly, $D$ is a closed Jordan region. Let $f(x,y)$ be a continuous function on $D$. Then, $f(x,y)$ is integrable on $D$. We can calculate the integral of $f(x,y)$ on $D$ by the method of iterated integral.

See also: Multiple integral on a bounded closed set for the explanation of closed Jordan regions.

First, fix an $x\in [a,b]$, and define the univariate function $F_1(x)$ by 

$$\begin{array}{}\text{(Eq:F1)}& F_1(x)={\int }_{\psi (x)}^{\phi (x)}f(x,y)dy.\end{array}$$

Lemma 

The function $F_1(x)$ defined by (Eq:F1) is continuous on $[a,b]$.

Proof. We show that $F_1(x)$ is continuous at an arbitrary $x_0\in [a,b]$. Let $\epsilon$ be any positive real number. Since $\phi (x)$ and $\psi (x)$ are continuous, there exists a $\delta >0$ such that, if $|x-x_0|<\delta$, then $|\psi (x)-\psi (x_0)|$ and $|\phi (x)-\phi (x_0)|$ are arbitrarily small. Therefore, with $c=max\{\psi (x),\psi (x_0)\}$ and $d=min\{\phi (x),\phi (x_0)\}$, we may assume that the following inequalities hold:

$$\begin{array}{rcl}|F_1(x)-{\int }_c^{d}f(x,y)dy|& <& \frac{\epsilon }{3},\\ |F_1(x_0)-{\int }_c^{d}f(x_0,y)dy|& <& \frac{\epsilon }{3}.\end{array}$$

Noting that $f(x,y)$ is uniformly continuous on $D$, we can redefine $\delta$ to be sufficiently small so that, for all $(x,y),(x^{\prime },y^{\prime })\in D$, if $|x-x^{\prime }|<\delta$ and $|y-y^{\prime }|<\delta$, then

$$|f(x,y)-f(x^{\prime },y^{\prime })|<\frac{\epsilon }{3(d-c)}.$$

Then, for all $x\in [a,b]$, if $|x-x_0|<\delta$, then

$$\begin{array}{rcl}|F_1(x)-F_1(x_0)|& =& |F_1(x)-{\int }_c^{d}f(x,y)dy+{\int }_c^{d}f(x,y)dy\\ & & -{\int }_c^{d}f(x_0,y)dy+{\int }_c^{d}f(x_0,y)dy-F_1(x_0)|\\ & \le & |F_1(x)-{\int }_c^{d}f(x,y)dy|\\ & & +{\int }_c^d|f(x,y)dy-f(x_0,y)|dy+|{\int }_c^{d}f(x_0,y)dy-F_1(x_0)|\\ & <& \frac{\epsilon }{3}+{\int }_c^d\frac{\epsilon }{3(d-c)}dy+\frac{\epsilon }{3}\\ & =& \frac{\epsilon }{3}+\frac{\epsilon }{3}+\frac{\epsilon }{3}=\epsilon .\end{array}$$

Therefore, $F_1(x)$ is continuous at $x_0$. ■

Since $F_1(x)$ is continuous, it is integrable. The integral

$${\int }_a^b{F}_1(x)dx={\int }_a^b({\int }_{\psi (x)}^{\phi (x)}f(x,y)dy)dx$$

is equal to the double integral ${\iint }_{D}f(x,y)dxdy$. That is,

Theorem

With the region $D$ as defined in (Eq:Dsand) and $f(x,y)$ a continuous function on $D$, we have

$${\iint }_{D}f(x,y)dxdy={\int }_a^b({\int }_{\psi (x)}^{\phi (x)}f(x,y)dy)dx.$$

Proof. Let $R=[a,b]\times [c,d]$ be a region such that $D\subset R$. Define the following bounded function on $R$:

$$\tilde{f}(x,y)=\{\begin{array}{cc}f(x,y)& ((x,y)\in D),\\ 0& ((x,y)\notin D).\end{array}$$

Let $\mathrm{\Delta }$ be a partition

$$\mathrm{\Delta }:\{\begin{array}{ccc}a=& a_0<a_1<\cdots <a_{n-1}<a_n& =b,\\ c=& c_0<c_1<\cdots <c_{m-1}<c_m& =d.\end{array}$$

Consider the $nm$ rectangular regions $D_{ij}=[a_i,a_{i+1}]\times [c_j,c_{j+1}]$ $(i=0,1,\cdots ,n-1;j=0,1,\cdots ,m-1)$ and let

$$\begin{array}{rcl}{M}_{ij}& =& sup\{\tilde{f}(x,y)\mid (x,y)\in D_{ij}\},\\ m_{ij}& =& inf\{\tilde{f}(x,y)\mid (x,y)\in D_{ij}\}.\end{array}$$

Thus, if $(x,y)\in D_{ij}$, then $m_{ij}\le \tilde{f}(x,y)\le M_{ij}$. Integrating these by $y$ from $c_j$ to $c_{j+1}$, we have

$$m_{ij}(c_{j+1}-c_j)\le {\int }_{c_j}^{c_{j+1}}\tilde{f}(x,y)dy\le M_{ij}(c_{j+1}-c_j).$$

Integrating these by $x$ from $a_i$ to $a_{i+1}$, we have

$$m_{ij}(a_{i+1}-a_i)(c_{j+1}-c_j)\le {\int }_{a_i}^{a_{i+1}}({\int }_{c_j}^{c_{j+1}}\tilde{f}(x,y)dy)dx\le M_{ij}(a_{i+1}-a_i)(c_{j+1}-c_j).$$

Summing these over $i$ and $j$, we have

$$\begin{array}{rcl}\sum _{i=0}^{n-1}\sum _{j=0}^{m-1}{m}_{ij}(a_{i+1}-a_i)(c_{j+1}-c_j)& \le & {\int }_a^b({\int }_c^d\tilde{f}(x,y)dy)dx\\ \text{(Eq:mMij)}& & \le & \sum _{i=0}^{n-1}\sum _{j=0}^{m-1}{M}_{ij}(a_{i+1}-a_i)(c_{j+1}-c_j).\end{array}$$

Since $f(x,y)$ is integrable on $D$, the left-most-hand side (lower Riemann sum) and the right-most-hand side (upper Riemann sum) of (Eq:mMij) converge to the same value, namely, ${\iint }_{D}f(x,y)dxdy$. On the other hand, by the definition of $\tilde{f}(x,y)$, we have

$$\begin{array}{}\text{(Eq:iinty)}& {\int }_c^d\tilde{f}(x,y)dy={\int }_{\psi (x)}^{\phi (x)}f(x,y)dy\end{array}$$

for all $x\in [a,b]$. Substituting this into (Eq:mMij) and applying the Squeeze Theorem, we have

$${\iint }_{D}f(x,y)dxdy={\int }_a^b({\int }_c^d\tilde{f}(x,y)dy)dx={\int }_a^b({\int }_{\psi (x)}^{\phi (x)}f(x,y)dy)dx.$$

■

Example. Let us calculate ${\iint }_D(x+y)dxdy$ where $D=\{(x,y)\mid x\ge 0,y\ge 0,x+y\le 2\}$. (Draw a picture!)

$$\begin{array}{rcl}{\iint }_D(x+y)dxdy& =& {\int }_0^2({\int }_0^{2-x}(x+y)dy)dx\\ & =& {\int }_0^2{[xy+\frac{y^2}{2}]}_{y=0}^{y=2-x}dx\\ & =& {\int }_0^2(2-\frac{x^2}{2})dx\\ & =& {[2x-\frac{x^3}{6}]}_{x=0}^{x=2}=\frac{8}{3}.\end{array}$$

□

Example. Let $D=\{(x,y)\mid 0\le x\le 1,0\le y\le x^2\}$ (Draw a picture!) and $f(x,y)$ be a continuous function on $D$. Then,

$$\begin{array}{rcl}{\iint }_{D}f(x,y)dxdy& =& {\int }_0^1({\int }_0^{x^2}f(x,y)dy)dx\\ & =& {\int }_0^1({\int }_{\sqrt{y}}^{1}f(x,y)dx)dy.\end{array}$$

□