Question

Let \(R=[0,1] \times[0,1] \times[0,1], \widetilde{R}=[0,1] \times[0,1],\) and \(f(x, y, z)=\left\\{\begin{array}{ll}2 x y+2 x z & \text { if } y \text { and } z \text { are rational, } \\ y+2 x z & \text { if } y \text { is irrational and } z \text { is rational, } \\ 2 x y+z & \text { if } y \text { is rational and } z \text { is irrational, } \\ y+z & \text { if } y \text { and } z \text { are irrational. }\end{array}\right.\) Calculate (a) \(\int_{R} f(x, y, z) d(x, y, z)\) and \(\int_{R} f(x, y, z) d(x, y, z)\) (b) \(\int_{\widetilde{R}} f(x, y, z) d(x, y)\) and \(\overline{\int_{\widetilde{R}}} f(x, y, z) d(x, y)\) (c) \(\int_{0}^{1} d y \int_{0}^{1} f(x, y, z) d x\) and \(\int_{0}^{1} d z \int_{0}^{1} d y \int_{0}^{1} f(x, y, z) d x\).

Short answer

Question: Calculate the values of \(I_1, I_2, I_3\), and \(I_4\). Answer: For \(I_1\), we have: $$I_1 = \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} (2xy + 2xz)dxdydz$$ $$I_1 = 1$$ For \(I_2\), we have: $$I_2 = \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} (y + 2xz)dxdydz$$ $$I_2 = \frac{3}{2}$$ For \(I_3\), we have: $$I_3 = \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} (2xy + z)dxdydz$$ $$I_3 = \frac{3}{2}$$ For \(I_4\), we have: $$I_4 = \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} (y + z)dxdydz$$ $$I_4 = 1$$ Now we can use the calculated values to answer the questions in the exercise: (a) To calculate \(\int_{R} f(x, y, z) d(x, y, z)\), we add the values of \(I_1, I_2, I_3\), and \(I_4\): $$\int_{R} f(x, y, z) d(x, y, z) = I_1 + I_2 + I_3 + I_4 = 1 + \frac{3}{2} + \frac{3}{2} + 1 = 4$$ (b) For this part, we only need to consider the cases where \(z = 0\) or \(z\) is rational in \(\widetilde{R}\): \(\int_{\widetilde{R}} f(x, y, z) d(x, y)\) = \(I_1(\text{where } z=0) + I_2(\text{where } z=0)\) = $\frac{1}{2}$ \(\overline{\int_{\widetilde{R}}} f(x, y, z) d(x, y)\) = \(I_3(\text{where } z=0) + I_4(\text{where } z=0)\) = $\frac{1}{2}$ (c) We will calculate each integral using the results from the four cases as needed: \(\int_{0}^{1} d y \int_{0}^{1} f(x, y, z) d x = I_1(\text{at } z) + I_2(\text{at } z)\) = 2z + 1 \(\int_{0}^{1} d z \int_{0}^{1} d y \int_{0}^{1} f(x, y, z) d x = \int_{0}^{1} d z (I_1(\text{at } z) + I_2(\text{at } z))\) = $\frac{5}{2}$

Step-by-step solution

Case 1: \(y\) and \(z\) are rational

In this case, the function becomes \(f(x, y, z) = 2xy + 2xz\). Integrating over x, y, and z in the region R, we get: $$I_1 = \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} (2xy + 2xz)dxdydz$$

Case 2: \(y\) is irrational and \(z\) is rational

In this case, the function becomes \(f(x, y, z) = y + 2xz\). Integrating over x, y, and z in the region R, we get: $$I_2 = \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} (y + 2xz)dxdydz$$

Case 3: \(y\) is rational and \(z\) is irrational

In this case, the function becomes \(f(x, y, z) = 2xy + z\). Integrating over x, y, and z in the region R, we get: $$I_3 = \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} (2xy + z)dxdydz$$

Case 4: \(y\) and \(z\) are irrational

In this case, the function becomes \(f(x, y, z) = y + z\). Integrating over x, y, and z in the region R, we get: $$I_4 = \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} (y + z)dxdydz$$ Now we have four integrals corresponding to the four cases mentioned above. We can use these integrals to answer each question in the exercise. (a) To calculate \(\int_{R} f(x, y, z) d(x, y, z)\), we need to add the values of \(I_1, I_2, I_3\), and \(I_4\). (b) For this part, we only need to consider the cases where \(z = 0\) or \(z\) is rational in \(\widetilde{R}\): \(\int_{\widetilde{R}} f(x, y, z) d(x, y)\) = \(I_1(\text{where } z=0) + I_2(\text{where } z=0)\) \(\overline{\int_{\widetilde{R}}} f(x, y, z) d(x, y)\) = \(I_3(\text{where } z=0) + I_4(\text{where } z=0)\) (c) We will calculate each integral using the results from the four cases as needed. \(\int_{0}^{1} d y \int_{0}^{1} f(x, y, z) d x = I_1(\text{at } z) + I_2(\text{at } z)\) \(\int_{0}^{1} d z \int_{0}^{1} d y \int_{0}^{1} f(x, y, z) d x = \int_{0}^{1} d z (I_1(\text{at } z) + I_2(\text{at } z))\) By calculating the values of \(I_1, I_2, I_3\), and \(I_4\), and using the results in the respective questions, we can provide the answers to all the exercise's questions.

Key concepts

Multiple Integrals

When students venture into the realm of calculus, one of the concepts they meet is the multiple integral. Multiple integrals extend the idea of a single integral to functions of several variables. For instance, double and triple integrals are used to calculate areas and volumes, respectively, in two and three dimensions. The function f(x, y, z) in the given exercise is evaluated over a three-dimensional region, R, which necessitates the use of a triple integral. Calculating the triple integral means integrating f(x, y, z) with respect to x, y, and z over the specified ranges.

Overall, the computation of multiple integrals allows for the evaluation of more complex functions and the ability to address multidimensional problems in fields such as physics, engineering, and economics.

Rational and Irrational Numbers

In mathematics, numbers are broadly categorized into rational and irrational. Rational numbers are those that can be expressed as the quotient of two integers, while irrational numbers cannot be represented as a simple fraction. The distinction between these two types of numbers is crucial in the function f(x, y, z) provided in the exercise. Here, the function's behavior changes depending on whether y and z are rational or irrational. When dealing with integrals, particularly the Lebesgue integral, this categorization becomes significant as it informs the measure of the set over which we are integrating. In many real-world applications and complex mathematical models, recognizing the difference between rational and irrational numbers aids in providing more accurate descriptions and predictions.

Lebesgue Integral

When considering integrals of functions involving both rational and irrational numbers, traditional Riemann integration might face some limitations. Enter the Lebesgue integral, a mathematical construct which vastly extends the ability to integrate functions. Unlike Riemann integrals, which slice the domain of the function, the Lebesgue integral slices up the function's range. This makes it particularly adept at dealing with intricate functions, such as those defined differently on rational and irrational numbers. Although the exercise initially presents a function more suited for the Riemann integral, understanding the benefits and applications of the Lebesgue integral (particularly in dealing with complex or irregular functions) is essential for students making inroads into higher-level calculus and real analysis.

Measure Theory

Behind every Lebesgue integral is the foundation of measure theory. Measure theory provides a systematic way to assign a 'size' or 'measure' to subsets of a given space, much like how length is assigned to intervals on the real line. This framework is key to understanding how the Lebesgue integral calculates the 'area under the curve' when that 'area' can be quite abstract. In the context of the exercise, measure theory allows us to rigorously define and handle the integration over regions that contain both rational and irrational numbers. With this tool, mathematicians and scientists can resolve problems that involve complicated structures, such as fractals, and other phenomena where traditional notions of 'size' and 'distance' don't apply.