Double integrals over rectangles

Section 1.1 Double integrals over rectangles

Definition 1.1.1. Rectangles, partitions, pointed partitions.

Consider the rectangle \(R=[a,b]\times [c,d]\subseteq \R^2\text{:}\) i.e.,

\begin{equation*} R=\{(x,y)\in \R^2\colon a\leq x\leq b, c\leq y\leq d\}\text{.} \end{equation*}

A partition \(P\) of the rectangle \(R\) is a subdivision \(R=R_1\cup R_2\cdots R_n\) of \(R\) into smaller subrectangles \(R_k\) using a grid of lines parallel to the \(x\)- and \(y\)-axes. The norm of \(P\text{,}\) denoted \(\norm{P}\text{,}\) is defined as the maximum width or length of any of the subrectangles of \(P\text{.}\) A pointed partition is a partition \(P\) along with a selection of sample points \((x_k, y_k)\in R_k\) from each subrectangle.

Definition 1.1.2. Riemann sums in two variables.

Let \(f\) be a function of two variables defined on the rectangle \(R=[a,b]\times [c,d]\text{.}\) Let \(P\) be a pointed partition of \(R\) into subrectangles \(R_1,R_2,\dots, R_n\) with sample points \((x_k, y_k)\in R_k\text{,}\) and let \(\Delta A_k\) be the area of \(R_k\) for each \(1\leq k\leq n\text{.}\) The Riemann sum \(S_P\) associated to this pointed partition is defined as

\begin{equation} S_P=\sum_{k=1}^nf(x_k,y_k)\Delta A_k\text{.}\tag{1.1.1} \end{equation}

Definition 1.1.3. Double integral over a rectangle.

A function \(f\) is said to be integrable over the rectangle \(R=[a,b]\times [c,d]\) if there is a number \(I\) such that given any sequence of pointed partitions \(P_i\) satisfying \(\norm{P_i}\rightarrow 0\text{,}\) we have

\begin{equation} I=\lim_{i\rightarrow \infty}S_i=\lim_{i\rightarrow\infty}\sum_{k=1}^{n_i}f(x_{ki},y_{ki})\Delta A_{ki}\text{.}\tag{1.1.2} \end{equation}

In this case \(I\) is called the integral of \(f\) over \(R\text{,}\) denoted \(I=\iint\limits_Rf(x,y)\, dA\text{.}\) As an abbreviated form of (1.1.2), we will write

\begin{equation} \iint\limits_Rf(x,y)\, dA =\lim_{\norm{P}\rightarrow 0}\sum_{k=1}^nf(x_k, y_k)\Delta A_k\text{.}\tag{1.1.3} \end{equation}

Example 1.1.4. Nonintegrable functions.

The following nonintegrable example highlights the importance of understanding what exactly the “for all” in Definition 1.1.3 quantifies over. Define \(f\colon \R^2\rightarrow \R\) as

\begin{equation*} f(x,y)=\begin{cases}1\amp \text{if } x,y \text{ are both rational}\\ 0 \amp \text{otherwise} \end{cases}\text{.} \end{equation*}

Let \(R=[0,1]\times [0,1]\text{,}\) and for each \(i\geq 1\) let \(P_i\) be the pointed partition obtained by subdividing the base and height of \(R\) equally into \(i\) subintervals, with sample points \((x_k, y_k)\) chosen to lie on the top-right of the \(k\)-th subrectangle. It is easy to see that \(x_k, y_k\) are both rational for all \(k\) (in fact we have \((x_k, y_k)=(m/i, n/i)\) for some integers \(1\leq m,n\leq i\)), and hence that \(f(x_k,y_k)=1\) for all \(k\text{.}\) The corresponding Riemann sum for this pointed partition \(P_i\) is

\begin{equation*} S_i=\sum_{k=1}^{i^2}f(x_k, y_k)\Delta A_k=\sum_{k=1}^{i^2} 1\cdot\Delta A_k=1\text{,} \end{equation*}

since the sum of the areas of the subrectangles is equal to one. We have \(\norm{P_n}=\frac{1}{n}\rightarrow 0\) and

\begin{equation*} \lim_{i\rightarrow \infty}S_i=\lim_{i\rightarrow \infty} 1=1\text{.} \end{equation*}

Now consider the sequence of pointed partitions \(P'_i\) constructed in exactly the same way, except now the sample points \((x'_k, y'_k)\) are now chosen so that \(x'_k\) is irrational. (Yes, this can be done!) Since now \(f(x'_k,y'_k)=0\) for all \(k\) it follows that we have \(S'_i=0\) for the corresponding Riemann sum, and hence

\begin{equation*} \lim_{i\rightarrow \infty}S'_i=\lim_{i\rightarrow \infty} 0=0\text{.} \end{equation*}

We have found two sequences of pointed partitions with norm tending to 0 that give rise to two different limits of the corresponding Riemann sums for \(f\) over \(R\text{.}\) We conclude that \(f\) is not integrable over \(R\text{!}\)

Theorem 1.1.5. Integrable over a rectangle.

In each of the following cases the function \(f\) is integrable over the rectangle \(R\text{:}\)

\(f\) is continuous on \(R\text{;}\)
\(f\) is bounded, and \(f\) is continuous at all but finitely many points of \(R\text{;}\)
\(f\) is bounded, and the set of points in \(R\) where \(f\) is not continuous is a finite union of smooth curves.

Once we know a function \(f\) is integrable over a rectangle \(R\) (for example by using Theorem 1.1.5), we know that we can compute this as the limit of any sequence of pointed partitions of \(R\) whose norms tend to zero. Procedure 1.1.6 outlines a direct way of computing integrals where the intervals defining \(R\) are equally partitioned.

Procedure 1.1.6. Limit definition of double integral (equal partition).

If \(f\) is integrable over the rectangle \(R=[a,b]\times [c,d]\text{,}\) one way of computing \(\iint\limits_R f(x,y)\, dA\) is as follows.

For each \(n\geq 1\) let \(P_n\) be the partition obtained by dividing \([a,b]\) and \([c,d]\) into \(n\) equal parts. This partitions \(R\) into \(n^2\) subrectangles \(R_k\text{,}\) each of area \(\Delta A_k=\frac{(b-a)(d-c)}{n^2}\text{.}\)
For each \(n\geq 1\) make any convenient choice of sample points \((x_k, y_k)\) lying in the subrectangles \(R_k\) of \(P_n\text{.}\) As an example, choosing the top right of each rectangle would yield sample points of the form \((i(b-a)/n, j(d-c)/n)\text{,}\) where \(1\leq i,j\leq n\text{.}\)
Compute the Riemann sum

\begin{equation*} S_n=\sum_{k=1}^{n^2} f(x_k, y_k)\Delta A_k=(b-a)(d-c)\sum_{k=1}^{n^2} \frac{f(x_k, y_k)}{n^2}\text{,} \end{equation*}

and attempt to simplify the summation expression as much as possible.
Compute \(\lim\limits_{n\rightarrow\infty}S_n\text{.}\)

Definition 1.1.7. Volume between graph and \(xy\)-plane (rectangular base).

Let \(f\) be nonnegative and integrable on the rectangle \(R=[a,b]\times[c,d]\) (i.e., \(f(x,y)\geq 0\) for all \((x,y)\in R\)), and let \(\mathcal{S}\) be the region consisting of all points lying on or above \(R\) and below the graph of \(f\text{:}\) i.e.,

\begin{equation*} \mathcal{S}=\{(x,y,z)\colon \R^3\colon a\leq x\leq b, c\leq y\leq d, 0\leq z\leq f(x,y)\}\text{.} \end{equation*}

We define the volume of \(\mathcal{S}\text{,}\) denoted \(\operatorname{vol} S\text{,}\) as

\begin{equation*} \operatorname{vol} \mathcal{S}=\iint_R f(x,y)\, dA\text{.} \end{equation*}

Example 1.1.8. Volume below graph.

Let \(f(x,y)=x+y^2\text{,}\) and let \(\mathcal{S}\) be the region lying above the rectangle \(R=[0,1]\times[0,1]\) and below the graph of \(f\text{.}\) Compute \(\operatorname{vol} \mathcal{S}\) using the limit definition of the double integral.

Solution.

First observe that since \(f(x,y)\) is continuous, it is integrable over \(R\text{,}\) and we can choose any sequence of pointed partitions with norm tending to 0 to compute its integral.

Next, for each \(n\) let \(P_n\) be the pointed partition obtained by dividing both edges of \(R\) into \(n\) equal subintervals and choosing as sample points the top-right corner of each rectangle. We have \(n^2\) rectangles in all, each with area \(\Delta A_k=\frac{1}{n^2}\text{.}\) The sample points are given as

\begin{equation*} (i/n, j/n), 1\leq i,j\leq n\text{.} \end{equation*}

Note that \(\norm{P_n}=\frac{1}{n}\rightarrow 0\text{,}\) as needed.

The Riemann sum \(S_n\) corresponding to \(P_n\) is given by

\begin{align*} S_n \amp=\sum_i^n\sum_j^nf(i/n,j/n)\cdot \frac{1}{n^2} \\ \amp=\frac{1}{n^2}\sum_i^n\sum_j^n \frac{i}{n}+\frac{j^2}{n^2} \amp (f(x,y)=x+y^2)\\ \amp =\frac{1}{n^2}\sum_{i=1}\frac{n\, i}{n}+\frac{1}{n^2}\left(\frac{n(n+1)(2n+1)}{6}\right) \amp \left(\sum_{j=1}^nj^2=\frac{n(n+1)(2n+1)}{6}\right)\\ \amp = \frac{1}{n^2}\left(\frac{n(n+1)}{2}+\frac{n(n+1)(2n+1)}{6n}\right) \amp \left(\sum_{i=1}^n i=\frac{n(n+1)}{2}\right)\\ \amp = \frac{1+1/n^2}{2}+\frac{1(1+1/n)(2+1/n)}{6} \amp (\text{algebra})\text{.} \end{align*}

Lastly, we compute

\begin{align*} \operatorname{vol} S \amp =\iint\limits_Rf(x,y)\, dA \amp (\text{def. of vol})\\ \amp=\lim_{n\to\infty}S_n \amp (\text{def. of integral}) \\ \amp = \lim_{n\to\infty} \frac{1+1/n^2}{2}+\frac{1(1+1/n)(2+1/n)}{6} \\ \amp = \frac{1+0}{2}+\frac{1(1+0)(2+0)}{6}\\ \amp = \frac{1}{2}+\frac{2}{6}=\frac{5}{6}\text{.} \end{align*}

Math 230-2: Kursobjekt

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