5.5 Area of regions in the plane  (Page 3/3)

The next theorem gives the connection between area (geometry) and integration (analysis).In fact, this theorem is what most calculus students think integration is all about.

Let $S$ be a geometric set, i.e., a subset of $R^2$ that is determined in the above manner by a closed bounded interval $[a,b]$ and two bounding functions $l$ and $u.$ Then

Let $P=\{x_0<x_1<...<x_n\}$ be a partition of $[a,b],$ and let $c_i$ and $d_i$ be defined as above. Let $h$ be a step function that equals $d_i$ on the open interval $(x_{i-1},x_i),$ and let $k$ be a step function that equals $c_i$ on the open interval $(x_{i-1},x_i).$ Then on each open interval $(x_{i-1},x_i)$ we have $h\left(x\right)\le u\left(x\right)$ and $k\left(x\right)\ge l\left(x\right).$ Complete the definitions of $h$ and $k$ by defining them at the partition points so that $h\left(x_i\right)=k\left(x_i\right)$ for all $i.$ Then we have that $h\left(x\right)-k\left(x\right)\le u\left(x\right)-l\left(x\right)$ for all $x\in [a,b].$ Hence,

Since this is true for every partition $P$ of $[a,b],$ it follows by taking the supremum over all partitions $P$ that

which proves half of the theorem; i.e., that $A\left(S\right)\le {\int }_a^{b}u-l.$

To see the other inequality, let $h$ be any step function on $[a,b]$ for which $h\left(x\right)\le u\left(x\right)$ for all $x,$ and let $k$ be any step function for which $k\left(x\right)\ge l\left(x\right)$ for all $x.$ Let $P=\{x_0<x_1<...<x_n\}$ be a partition of $[a,b]$ for which both $h$ and $k$ are constant on the open subintervals $(x_{i-1},x_i)$ of $P.$ Let $a_1,a_2,...,a_n$ and $b_1,b_2,...,b_n$ be the numbers such that $h\left(x\right)=a_i$ on $(x_{i-1},x_i)$ and $k\left(x\right)=b_i$ on $(x_{i-1},x_i).$ It follows, since $h\left(x\right)\le u\left(x\right)$ for all $x,$ that $a_i\le d_i.$ Also, it follows that $b_i\ge c_i.$ Therefore,

Finally, let $\left\{h_m\right\}$ be a nondecreasing sequence of step functions that converges uniformly to $u,$ and let $\left\{k_m\right\}$ be a nonincreasing sequence of step functions that converges uniformly to $l.$ See part (d) of [link] . Then

which proves the other half of the theorem.

OK! Trumpet fanfares, please!

( $A=\pi r^2.$ ) If $S$ is a circle in the plane having radius $r,$ then the area $A\left(S\right)$ of $S$ is $\pi r^2.$

Suppose the center of the circle $S$ is the point $(h,k).$ This circle is a geometric set. In fact, we may describe the circle with center $(h,k)$ and radius $r$ as the subset $S$ of $R^2$ determined by the closed bounded interval $[h-r,h+r]$ and the functions

and

By the preceding theorem, we then have that

We leave the verification of the last equality to the following exercise.

REMARK There is another formula for the area of a geometric set that is sometimes very useful. This formula gives the area in terms of a “double integral.”There is really nothing new to this formula; it simply makes use of the fact that the number (length) $u\left(x\right)-l\left(x\right)$ can be represented as the integral from $l\left(x\right)$ to $u\left(x\right)$ of the constant 1. Here's the formula:

The next theorem is a result that justifies our definition of area by verifying that the whole is equal to the sum of its parts, something that any good definition of area should satisfy.

Let $S$ be a closed geometric set, and suppose $S={\cup }_{i=1}^n{S}_i,$ where the sets $\left\{S_i\right\}$ are closed geometric sets for which $S_i^0\cap S_j^0=\varnothing$ if $i\ne j.$ Then

Suppose $S$ is determined by the interval $[a,b]$ and the two bounding functions $l$ and $u,$ and suppose $S_i$ is determined by the interval $[a_i,b_i]$ and the two bounding functions $l_i$ and $u_i.$ Because $S_i\subseteq S,$ it must be that the interval $[a_i,b_i]$ is contained in the interval $[a,b].$ Initially, the bounding functions $l_i$ and $u_i$ are defined and continuous on $[a_i,b_i],$ and we extend their domain to all of $[a,b]$ by defining $l_i\left(x\right)=u_i\left(x\right)=0$ for all $x\in [a,b]$ that are not in $[a_i,b_i].$ The extended functions $l_i$ and $u_i$ may not be continuous on all of $[a,b],$ but they are still integrable on $[a,b].$ (Why?) Notice that we now have the formula

Next, fix an $x$ in the open interval $(a,b).$ We must have that the vertical intervals $(l_i\left(x\right),u_i\left(x\right))$ and $(l_j\left(x\right),u_j\left(x\right))$ are disjoint if $i\ne j.$ Otherwise, there would exist a point $y$ in both intervals, and this would mean that the point $(x,y)$ would belong to both $S_i^0$ and $S_j^0,$ which is impossible by hypothesis. Therefore, for each $x\in (a,b),$ the intervals $\{\left(l_i\left(\right)x\right),u_i\left(x\right))\}$ are pairwise disjoint open intervals, and they are all contained in the interval $\left(l\right(x),u(x\left)\right),$ because the $S_i$ 's are subsets of $S.$ Hence, the sum of the lengths of the open intervals $\left\{(l_i\left(x\right),u_i\left(x\right))\right\}$ is less than or equal to the length of $\left(l\right(x),u(x\left)\right).$ Also, for any point $y$ in the closed interval $\left[l\right(x),u(x\left)\right],$ the point $(x,y)$ must belong to one of the $S_i$ 's, implying that $y$ is in the closed interval $[l_i\left(x\right),u_i\left(x\right)]$ for some $i.$ But this means that the sum of the lengths of the closed intervals $[l_i\left(x\right),u_i\left(x\right)]$ is greater than or equal to the length of the interval $\left[l\right(x),u(x\left)\right].$ Since open intervals and closed intervals have the same length, we then see that $(u\left(x\right)-l\left(x\right)={\sum }_{i=1}^n(u_i\left(x\right)-l_i\left(x\right)).$

We now have the following calculation:

which completes the proof.