Double integrals approximated with a Riemann sum

Multiple integrals: Double integrals

Double integrals approximated with a Riemann sum

How do you calculate a double integral numerically? We are going to approach this in the same way as numerically integrating functions of one variable. We do it again only for a positive continuous function $f(x,y)$ defined on a rectangle $R=[a,b]\times [c,d]$ .

We always divide the interval $[a,b]$ on the $x$ -axis into $m$ pieces by designating fixed points $x_0, x_1, x_2,\ldots, x_m$ with $a = x_0 < x_1 < x_2 < \cdots < x_m = b$ in that interval. Similarly, we divide the interval $[c,d]$ on the $y$ -axis into $n$ pieces by designating fixed points $y_0, y_1, y_2,\ldots, y_n$ with $c = y_0 < y_1 < y_2 < \cdots < y_n = d$ in that interval. This way we get $m\times n$ sub-rectangles $R_{ij}= [x_{i-1},x_i]\times [y_{i-1},y_i]$ with $i=1,2,\ldots m$ and $j=1,2,\ldots m$ . Such a set is called a partition $\mathcal{V}$ of $[a,b]\times [c,d]$ . We then choose a random point $s_{ij}$ in each of the sub-rectangles $R_{ij}$ ; we call such point a tag. So $x_{i-1}\le s_{ij}\le x_i$ for $i = 1, 2, \ldots m$ and $y_{j-1}\le s_{ij}\le y_j$ for $j = 1, 2, \ldots n$ . We denote the area of each sub-rectangle $R_{ij}$ with ${\vartriangle}_{ij}={\vartriangle}_{i}\cdot {\vartriangle}_{j}$ , where ${\vartriangle}_{i}=x_i-x_{i-1})$ and ${\vartriangle}_{j}=x_j-x_{j-1})$ are the width and height of each sub-rectangle $R_{ij}$ . We call the maximum of the lengths of the sides of the partial rectangles the mesh size. The expression $S(\mathcal{V}; s_{11}, \ldots, s_{mn})=\sum_{i=1}^m\sum_{j=1}^{n}f(s_{ij})\cdot {\vartriangle}_{ij}$ is called a Riemann sum corresponding with the function $f$ . Note: for a given partition $\mathcal{V}$ of $[a, b]\times [c,d]$ there are many different Riemann sums: with each choice of tags you always get one. Such a Riemann sum is therefore the sum of the volume of rectangular pillars with the sub-rectangles used in the partition as bases. The volume of these rectangular pillars (i.e, cuboids) add up to an approximation of the volume under a graph that we want to calculate. The requested volume is, in a sense, a limit of such Riemann sums. See example below.

dubbelintegraal benaderd met een Riemannsom

In the figure above, for the function $f(x,y)=\frac{1}{6}(x^2y+2\sqrt{x+1}-y^3+6)$ , we have the square $[0,2]\times[0,2]$ as integration region $R$ , an evenly divided partition of $R$ with $10^2=100$ sub-squares $m=n=10$ , $\vartriangle_{i}=\vartriangle_{j}=0.2$ for all $i,j=1,2,\ldots 10$ (the mesh size is equal to $0.2$ ), and in each sub-square $R_{ij}$ we have choen the centre point for $s_{ij}$ . Then, in eight significant figures, we get the Riemann sum $\sum_{i=1}^{10}\sum_{j-1}^{10}f(s_{ij})\cdot {\vartriangle}_{ij} \approx 5.4251910$ The smaller the mesh size of the partition, the better the approximation. For example, if you divide the square evenly into $10^4=1000$ sub-squares and the mesh size is therefore equal to $0.02$ , you get $\sum_{i=1}^{100}\sum_{j-1}^{100}f(s_{ij})\cdot {\vartriangle}_{ij} \approx 5.4205590$ If you do the same with a mesh size of $0.002$ , you get $\sum_{i=1}^{1000}\sum_{j-1}^{1000}f(s_{ij})\cdot {\vartriangle}_{ij} \approx 5.4205127$ and the first six decimal places are already good compared to the exact result $\frac{29}9+4/3\sqrt{3}\approx 5.420512188$ for the double integral: $\begin{aligned}\iint_R f(x,y)\,\dd(x,y) &= \lim_{n\to\infty}\left(\lim_{m\to\infty}f(s_{ij})\cdot {\vartriangle}_{i}\right)\cdot {\vartriangle}_{j}=\int_{y=0}^{y=2}\left(\int_{x=0}^{x=2}f(x,y)\,\dd x\right)\dd y\\[0.25cm] &= \lim_{m\to\infty}\left(\lim_{n\to\infty}f(s_{ij})\cdot {\vartriangle}_{j}\right)\cdot {\vartriangle}_{i}=\int_{x=0}^{x=2}\left(\int_{y=0}^{y=2}f(x,y)\,\dd y\right)\dd x\end{aligned}$

Step-by-step approach for a numerical calculation of a double integral Even though there are much better methods to calculate a double integral numerically, the Riemann integration method described above does give a good idea of what a double integral actually is. We can draw up a step-by-step approach that can be applied more broadly:

Divide the integration region $R$ into small sub-regions $\dd R$ , which are also called area elements.
Choose the area elements so small that the function above them is almost constant.
Choose any point $(x,y)$ in each area element $\dd R$ and calculate $f(x,y)$ . Multiply this by the area $|\dd R|$ of $\dd R$ .
Add up all the results obtained in step 3. The sum is then approximately equal to the double integral. The approximation is better when the area elements are smaller.

Negative function values It is clear from the step-by-step approach that area elements in which the function $f(x,y)$ is negative make a negative contribution to the Riemann sum. After all, then the term $f(x,y)\cdot |\dd R|$ is negative.

Non-rectangular integration areas It is clear from the step-by-step approach that it also works when the integration area is not a rectangle. The only thing that matters is that the surface elements cover the entire area $R$ and that the area $|\dd R|$ of each $\dd R$ can be determined.