# integral over plane region

The integrals over a planar region are generalisations of usual Riemann integrals, but special cases of http://planetmath.org/node/6660surface integrals.

## 0.1 Integral over a rectangle

Let $R$ be the rectangle of $xy$-plane defined by

 $\displaystyle a\leqq x\leqq b,\quad c\leqq y\leqq d$ (1)

and the function $f$ be defined and bounded in $R$.  Let

 $\displaystyle D:\begin{cases}x_{0}=a,\,x_{1},\,\ldots,\,x_{m}=b\\ y_{0}=c,\,y_{1},\,\ldots,\,y_{n}=d\end{cases}$ (2)

a of $R$ into the rectangular parts $\Delta_{i}$ with areas $\Delta_{i}A$ ($i=1,\,\ldots,\,mn$).  Denote

 $m_{i}\;:=\;\inf_{\Delta_{i}}f(x,\,y),\qquad M_{i}\;:=\;\sup_{\Delta_{i}}f(x,\,y)$

and

 $s_{D}\;:=\;\sum_{D}m_{i}\Delta_{i}A,\qquad S_{D}\;:=\;\sum_{D}M_{i}\Delta_{i}A.$

Definition 1.  If  $\displaystyle\sup_{D}\{s_{D}\}\,=\,\inf_{D}\{S_{D}\}$,  then we say that $f$ is integrable over $R$ and call the common value the (Riemann) integral of $f$ over the rectangle $R$ and denote it by

 $\int_{R}f,\quad\int_{R}f(x,\,y)\,dx\,dy\quad\mbox{or}\quad\iint_{R}f(x,\,y)\,% dx\,dy.$

Let then $f$ be defined in a region $A$ of $xy$-plane such that that it can be enclosed in a rectangle $R$ defined by (1).  Define the new function $f_{1}$ through

 $\displaystyle f_{1}(x,\,y)\;:=\;\begin{cases}f(x,\,y)\quad\mbox{when\;\;}(x,\,% y)\in A,\\ 0\qquad\mbox{otherwise.}\end{cases}$ (3)

Definition 2.  If $f_{1}$ is integrable over the rectangle $R$, we say that $f$ is integrable over $A$ and define

 $\displaystyle\int_{A}f\;:=\;\int_{R}f_{1}.$ (4)

It’s apparent that (4) is on the choice of $R$ since the points of $\mathbb{R}^{2}\!\smallsetminus\!A$ give zero-terms to the lower and upper sums.

## 0.2 Double integrals

Definition 3.  Let $f$ be bounded in $R$ as before.  Suppose that

 $\varphi(x)\;:=\;\int_{c}^{d}f(x,\,y)\,dy$

is defined on  $[a,\,b]$.  If also the integral

 $\displaystyle\int_{a}^{b}\varphi(x)\,dx\;=\;\int_{a}^{b}\left[\int_{c}^{d}f(x,% \,y)\,dy\right]dx$ (5)

exists, it is called a double integral or iterated integral and denoted by

 $\int_{a}^{b}dx\int_{c}^{d}f(x,\,y)\,dy.$

One may prove the

Theorem.$\displaystyle\int_{R}f(x,\,y)\,dx\,dy\,=\,\int_{a}^{b}dx\int_{c}^{d}f(x,\,y)\,dy$,  provided that the integral of the left side exists and that the inner integral $\int_{c}^{d}f(x,\,y)\,dy$ of the right side exists for every $x$ in  $[a,\,b]$.

It’s clear that  $\displaystyle\int_{a}^{b}dx\int_{c}^{d}f(x,\,y)\,dy\,=\,\int_{c}^{d}dy\int_{a}% ^{b}f(x,\,y)\,dx$  if also the integral $\int_{a}^{b}f(x,\,y)\,dx$ exists for every $y$ in  $[c,\,d]$.  If especially the function $f$ is continuous in the rectangle $R$, then surely

 $\int_{R}f(x,\,y)\,dx\,dy\;=\;\int_{a}^{b}dx\int_{c}^{d}f(x,\,y)\,dy\;=\;\int_{% c}^{d}dy\int_{a}^{b}f(x,\,y)\,dx.$

Assume now, that $f$ is defined and bounded in the region

 $A\;:=\;\{(x,\,y)\in\mathbb{R}^{2}\vdots\;a\leqq x\leqq b,\;h_{1}(x)\leqq y% \leqq h_{2}(x)\}$

where $A$ is contained in the rectangle $R$ determined by (1).  Then the planar integral $\displaystyle\int_{A}f$ is defined as

 $\int_{A}f\;=\;\int_{R}f$

if the integral of right side exists.  For this, he continuity of $f$ in $A$ does not necessarily suffice, because $f_{1}$ may have a jump discontinuity on the border of $A$ whence the integrability of $f_{1}$ needs not be guaranteed.  One case where the integrability is true is that the graphs of the functions $h_{1}$ and $h_{2}$ are rectifiable (i.e. the functions have continuous derivatives).  For a continuous $f$, we then have

 $\int_{c}^{d}f_{1}(x,\,y)\,dy\;=\;\int_{c}^{h_{1}(x)}0\,dy+\int_{h_{1}(x)}^{h_{% 2}(x)}f(x,\,y)\,dy+\int_{h_{2}(x)}^{d}0\,dy\;=\;\int_{h_{1}(x)}^{h_{2}(x)}f(x,% \,y)\,dy.$

Thus

 $\displaystyle\int_{A}f\;=\;\int_{R}f_{1}\;=\;\int_{a}^{b}dx\int_{h_{1}(x)}^{h_% {2}(x)}f(x,\,y)\,dy,$ (6)

i.e. the planar integral has been expressed as a double integral.