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Integration Over the Rectangle

In the previous blog, we learned how to use single integral to calculate the volume and center of mass. For single integrals, the region over which we integrate is always an interval. (i.e. a one-dimensional space)

In this blog we are intereted in integrating a function of two variables \(f(x,y)\). With functions of two variable we integrated over a region of \(\mathbb{R}^2\) (i.e. a two-dimensional space). We will start out by assuming that the region in \(\mathbb{R}^2\) is a rectangle which we will denote as \(R=[a,b]\times [c,d]\). This means that the ranges for \(x\) and \(y\) are \(a\leq x\leq b\), and \(c \leq y\leq d\). Also let's assume \(f(x,y) \geq 0\). The graph of \(f\) is a aurface with equation \(z=f(x,y)\)
Our goal is to find the volume between surface and \(R\). First, we want to approximate it. We divide up \(R\) into a series of smaller rectangles, such that divide up \(a\leq x\leq b\) into \(n\)subintervals and divide up \(c \leq y\leq d\) into \(m\) subintervals, we denote each rectangle as \((x_i, y_j)\). Then, we will construct a box whose height is given by \(f(x_i, y_j)\) over each of these smaller rectangles. Then, each of rectangle has a base \(\Delta A\) and a height of \(f(x_i, y_j)\), so we have the volume of each boxes will be \(f(x_i, y_j) \Delta A\). The volume under \(S\) will be approximately equal to the following: \[ V \approx \sum^n_{i=1} \sum^m_{j=1} f(x_i, y_j) \Delta A\] Our intuition tells us that if the approximation becomes better as \(m\) and \(n\) become larger, we would expect that : \[ V = \lim _{m, n \to \infty} \sum^n_{i=1} \sum^m_{j=1} f(x_i, y_j) \Delta A\] It looks like the function of single integral's definition. In fact, it is the definition of double integrals. It can be written as: The double integral of \(f\) over the rectangle \(R\) is \[V= \int \int_Rf(x,y)dA = \lim _{m, n \to \infty} \sum^n_{i=1} \sum^m_{j=1} f(x_i, y_j) \Delta A\] if this limit exists. Let's use an example to learn how we use double integral to find the volume below the function and above the rectangle and how we compute double integrals.
Calculate the solid volume that lies above the square \(R=[0,2]\times[0,2]\) and below the paraboloid \(z=16-x^2-2y^2\).
First, Let's sketch this paraboloid and the square.
We want to approximate the volume of the solid. First, let's divide 100 equal squares, which means \(n=m=10\).Then we know the \(\Delta A = 0.2\times 0.2 = 0.04\) and the height of boxes will be following:
We could use 3D printer to print the model as follwoing:
Now, we know the base and height of each box, we could approximate the total volume of this solid as following: \[V \approx \sum^{10}_{i=1} \sum^{10}_{j=1} f(x_i, y_j) \Delta A = 45.52\] Let's use integral to calculate it to get an accurate result. \[\begin{split} V &=\int\int_R f(x,y) dA = \int^2_0 \int^2_0 16-x^2-2y^2 dx dy\\ &= \int^2_0 \left[16x-\frac{x^3}{3} -2xy \bigg|^{x=2}_{x=0}\right]dy = \int^2_0 32-\frac{8}{3}-4y dy \\ &= 32y-\frac{8}{3}y-4\frac{y^3}{3} \bigg |^{y=2}_{y=0} = 64-\frac{16}{3}-\frac{32}{3} \\ &= 48 \end{split} \] Comparing the approximation result with the actual result, they are closed.

I choose this function because it is an easy example for beginners to learn how to integrate over the rectangle. After they know how to integrate over the rectangle, they could continue learning how to integrate over the general regions in two dimensions.

Fig. 1 Double Integrals over Rectangles. (2020, November 10). Retrieved September 27, 2021, from https://math.libretexts.org/@go/page/4545

Author: Yueqi Li (Nicole)

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