Question

Let \(\mathcal{C}\) be the set of points interior to or on the boundary of a cube with edge of length \(1 .\) Moreover, say that the cube is in the first octant with one vertex at the point \((0,0,0)\) and an opposite vertex at the point \((1,1,1) .\) Let \(Q(C)=\) \(\iiint_{C} d x d y d z\) (a) If \(C \subset \mathcal{C}\) is the set \(\\{(x, y, z): 0

Short answer

The value of \(Q(C)\) for the given subsets of the cube are \(\frac{1}{6}\) and 0 respectively.

Step-by-step solution

Setting up the first volume integral

For part (a), the given inequalities define a region within the unit cube. These can be directly converted into limits for the triple integral: \[Q(C) = \int_{0}^{1} \int_{x}^{1} \int_{y}^{1} dz \, dy \, dx\]

Evaluating the first integral

Evaluating this integral step-by-step gives: \[\int_{0}^{1} \int_{x}^{1} (1-y) \, dy \, dx = \int_{0}^{1} \left[ y - \frac{y^2}{2} \right]_x^1 \, dx\] Simplifying this gives: \[\int_{0}^{1} \left[ 1-x - \frac{(1-x)^2}{2} \right] dx\] Solving this last integral results in \[\frac{1}{6}\]

Setting up the second volume integral

For part (b), the equation \(x=y=z\) is along the main diagonal of the cube, and these equal values range from 0 to 1. As this is an infinitesimally thin line in the three-dimensional space, the volume along this line would always be 0. Thus, \(Q(C) = 0\) .

Conclusion

Thus, for part (a), we calculated \(Q(C) = \frac{1}{6}\), and for part (b), we found that \(Q(C) = 0\) since we are calculating the volume of a line in a 3-dimensional space.

Key concepts

Volume Integration

Volume integration is an essential method in mathematics used to calculate the volume occupied by a three-dimensional region. It involves the use of triple integrals, which allow us to sum up infinitely many infinitesimally small volumes over a particular region. In the context of our exercise, we find this region within a cube in the first octant defined by three simple inequalities.

To grasp this, imagine dividing the space into tiny cubes or boxes, each with a volume represented by the product of small changes in each direction, denoted by dx, dy, and dz. The triple integral \( \iiint_C dx\,dy\,dz \) signifies the summation of the volumes of these infinitesimal boxes over the region C. For part (a), we've set up the integral with the specific limits that represent the inequalities defining region C. As we integrate, we're essentially 'adding up' all the tiny volumes that fit within the region C.

When it comes to part (b), the volume calculation is conceptually different. We're asked to calculate the volume of a subset within the cube that is infinitely thin—an infinitesimally thin line along the main diagonal. Since a line has no 'thickness,' it occupies zero volume in three-dimensional space, hence the result of the integral is 0.

Mathematical Statistics

While mathematical statistics may not seem directly related to volume integration at first glance, it's important to understand that many statistical problems, particularly those involving probability distributions, use volume integration for calculations in higher dimensions.

For instance, when dealing with multivariate probability distributions, the probability that a random vector falls within a certain region could be computed through a triple integral over that region. Just as we calculated the volume of a subset of a cube in our exercise, one could calculate the 'volume' under a probability density function over a particular region to find the probability of the random vector falling within that region.

In more advanced applications, triple integrals are used in statistical mechanics and to find moments of inertia in physics. The concepts behind volume integration thus serve as a bridge between pure mathematical operations and practical applications in statistics and other scientific fields.

Multivariable Calculus

Multivariable calculus extends the principles of calculus to functions of several variables. In essence, we are not just looking at one rate of change, as in regular calculus, but at rates of change across multiple dimensions. This is why in our exercise, we have a triple integral: it signifies integration across three independent variables—x, y, and z—which define our three-dimensional space.

The cube problem we've tackled is a classic example of how multivariable calculus can be used to understand and solve real-world problems that involve several variables and dimensions. Multivariable calculus provides the tools to find maximum and minimum values of functions, optimize systems, and analyze the behavior of complex systems over various dimensions, which is invaluable in fields such as engineering, economics, and the physical sciences.

Moreover, multivariable calculus is fundamental for understanding concepts like gradient, divergence, and curl, which are central to vector calculus and physics, and for solving differential equations that describe numerous physical phenomena.