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In the following exercises, use two circular permutations of the variables \(x, y,\) and \(z\) to write new integrals whose values equal the value of the original integral. A circular permutation of \(x, y,\) and \(z\) is the arrangement of the numbers in one of the following orders: \(y, z,\) and \(x\) or \(z, x,\) and \(y\) \(\int_{0}^{1} \int_{-y}^{y} \int_{0}^{1-x^{4}-y^{4}} \ln x d z d x d y\)

Short Answer

Expert verified
The two new integrals are \( \int_{0}^{1} \int_{-x}^{x} \int_{0}^{1-y^{4}-x^{4}} \ln y \, d x \, d y \, d z \) and \( \int_{0}^{1} \int_{-z}^{z} \int_{0}^{1-x^{4}-z^{4}} \ln z \, d y \, d z \, d x \).

Step by step solution

01

Identify the Original Integral

The given integral to work with is:\[\int_{0}^{1} \int_{-y}^{y} \int_{0}^{1-x^{4}-y^{4}} \ln x \, d z \, d x \, d y\]
02

Apply First Circular Permutation

For the first circular permutation, we swap the variables \(x\), \(y\), and \(z\) into the order \(y, z, x\). This means that:\[x \rightarrow y, \; y \rightarrow z, \; z \rightarrow x \]So, the integral becomes:\[\int_{0}^{1} \int_{-x}^{x} \int_{0}^{1-y^{4}-x^{4}} \ln y \, d x \, d y \, d z\]
03

Apply Second Circular Permutation

For the second circular permutation, we rearrange them into \(z, x, y\). So:\[x \rightarrow z, \; y \rightarrow x, \; z \rightarrow y\]Transforming the integral to:\[\int_{0}^{1} \int_{-z}^{z} \int_{0}^{1-x^{4}-z^{4}} \ln z \, d y \, d z \, d x\]
04

Confirm Equivalence of New Integrals

The values of both newly derived integrals from the circular permutations, \[\int_{0}^{1} \int_{-x}^{x} \int_{0}^{1-y^{4}-x^{4}} \ln y \, d x \, d y \, d z\]and\[\int_{0}^{1} \int_{-z}^{z} \int_{0}^{1-x^{4}-z^{4}} \ln z \, d y \, d z \, d x\]match the value of the original integral due to the symmetry of the problem.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Circular Permutations
Circular permutations involve rearranging the order of variables in a specific manner. Unlike linear permutations, where each element can be placed anywhere, circular permutations rotate elements in a sequence.
This means that the positions of the variables change in a way that they follow each other in a continuous loop.
It's like rotating a key ring around so that the keys fall into a different order.

In mathematics, when dealing with functions or integrals that involve multiple variables, circular permutations can simplify calculations or reveal properties.
This is particularly useful in integrals where symmetry plays a crucial role, as it can help find equivalent expressions or simplify the process of integration.
  • Consider three variables, say, \(x\), \(y\), and \(z\).
  • You can permute these in a circular way as \((y, z, x)\) or \((z, x, y)\).
  • It ensures that all variables partake in different positions across the expression.
    By applying circular permutations to an integral, you can produce equivalent forms of that integral without changing its value, due to the inherent symmetry of the function.
    This is why circular permutations are critical for recognizing and utilizing symmetries in problems involving multiple integrals.
Change of Variables
A change of variables is a method used in calculus to simplify the evaluation of integrals and solve problems. By changing variables, complicated integrals can be transformed into simpler ones, making them easier to solve.
This technique involves substituting old variables with new expressions in terms of different variables.

There are a few key advantages to making a change of variables:
  • Simplifies integration by altering variables to a simpler function form.
  • Shifts the limits of integration to make calculations easier.
  • Exploits symmetrical properties of the functions.
    In the context of multiple integrals, changing variables allows us to switch the roles of limits and integrands.
    This doesn't change the value of the integral, as each new integral essentially describes the same volume or region.
    For our case, rotating the variables was a strategic move to uncover symmetries hidden by the original setup.
    Thus, a change of variables is not merely a mathematical trick but an essential tool for tackling integrals in a structured and logical manner.
Symmetry in Integrals
Symmetry in integrals is a vital concept, offering significant simplifications when evaluating complex expressions.
When an integrand, the function being integrated, has symmetrical properties, it allows us to exploit these to find solutions more easily.
Symmetries often occur in integrals of functions involving multiple variables, especially if they resemble well-known geometric or algebraic properties.

In the context of the provided exercise, symmetry helps us anticipate and confirm that different permutations of the integral will yield the same value.
This is due to the symmetrical nature of the limits and integrands.
  • The original integral is transformed using the circular permutations of variables.
  • All permutations remain equivalent in terms of value.
  • This equivalency proves the function's symmetry concerning those variables.
    Such symmetry not only validates the results of permutation but also ensures calculations are correct without the need for further verification in many cases.
    Utilizing symmetry effectively can mean less work and fewer errors, which is why it is a powerful aspect of working with integrals.

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Most popular questions from this chapter

The midpoint rule for the triple integral \(\iiint_{B}f(x, y, z) d V\) over the rectangular solid box \(B\) is a generalization of the midpoint rule for double integrals. The region \(B\) is divided into subboxes of equal sizes and the integral is approximated by the triple Riemann sum \(\sum_{i=1}^{l} \sum_{j=1}^{m} \sum_{k=1}^{n} f\left(\overline{x}_{i}, \overline{y}_{j}, \overline{z_{k}}\right) \Delta V,\) where \(\left(\overline{x}_{i}, \overline{y}_{j}, \overline{z}_{k}\right)\) is the center of the box \(B_{i j k}\) and \(\Delta V\) is the volume of each subbox. Apply the midpoint rule to approximate \(\iiint_{B}x^{2} d V\)over the solid \(B=\\{(x, y, z) | 0 \leq x \leq 1,0 \leq y \leq 1,0 \leq z \leq 1\\}\) by using a partition of eight cubes of equal size. Round your answer to three decimal places.

For the following problems, find the specified area or volume. The volume an ice cream cone that is given by the solid above \(z=\sqrt{\left(x^{2}+y^{2}\right)}\) and below \(z^{2}+x^{2}+y^{2}=z\)

Evaluate the following integrals. \(\iiint_{R} 3 y d V\) where \(R=\left\\{(x, y, z) | 0 \leq x \leq 1,0 \leq y \leq x, 0 \leq z \leq \sqrt{9-y^{2}}\right\\}\)

The transformation \(T_{\theta} : \mathrm{R}^{2} \rightarrow \mathrm{R}^{2}, T_{\theta}(u, v)=(x, y)\) where \(x=u \cos \theta-v \sin \theta, \quad y=u \sin \theta+v \cos \theta,\) is called a rotation of angle \(\theta .\) Show that the inverse transformation of \(T_{\theta}\) satisfies \(T_{\theta}^{-1}=T_{-\theta},\) where \(T_{-\theta}\) is the rotation of angle \(-\theta\)

Lame ovals have been consistently used by designers and architects. For instance, Gerald Robinson, a Canadian architect, has designed a parking garage in a shopping center in Peterborough, Ontario, in the shape of a superellipse of the equation \(\left(\frac{x}{a}\right)^{n}+\left(\frac{y}{b}\right)^{n}=1\) with \(\frac{a}{b}=\frac{9}{7}\) and \(n=e .\) Use a CAS to find an approximation of the area of the parking garage in the case \(a=900\) yards, \(b=700\) yards, and \(n=2.72\) yards.

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