/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 Evaluate the iterated integral. ... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate the iterated integral. \(\int_{-1}^{1} \int_{-1}^{1} \int_{-1}^{1} x^{2} y^{2} z^{2} d x d y d z\)

Short Answer

Expert verified
The value of the iterated integral is \(\frac{8}{27}\).

Step by step solution

01

Integrate with respect to x

Firstly, let's start by integrating the innermost integral with respect to \(x\). Due to the symmetry of the integrand, \(\int_{-1}^{1} x^{2} dx = 2\int_{0}^{1} x^{2} dx\). Apply the power rule for integral, the integral \(\int x^{2} dx\) becomes \(\frac{x^3}{3}\). Hence, the integral becomes \(2[\frac{1^{3}}{3}-\frac{0^{3}}{3}]=\frac{2}{3}\). So the triple integral simplifies to \(\frac{2}{3}\int_{-1}^{1} \int_{-1}^{1} y^{2} z^{2} dy dz\).
02

Integrate with respect to y

The next step is to integrate with respect to \(y\). Similarly as before, due to the symmetry of the integrand, \(\int_{-1}^{1} y^{2} dy = 2\int_{0}^{1} y^{2} dy\). Applying the power rule for integral again, the integral \(\int y^{2} dy\) becomes \(\frac{y^3}{3}\). Hence, the integral becomes \(2[\frac{1^{3}}{3}-\frac{0^{3}}{3}]=\frac{2}{3}\). Now the integral becomes \(\frac{2}{3}*\frac{2}{3}\int_{-1}^{1} z^{2} dz = \frac{4}{9}\int_{-1}^{1} z^{2} dz\).
03

Integrate with respect to z

Lastly, we need to integrate with respect to \(z\). Similar as the previous two steps, due to the symmetry \(\int_{-1}^{1} z^{2} dz = 2\int_{0}^{1} z^{2} dz\). Applying the power rule for integral finally, the integral \(\int z^{2} dz\) becomes \(\frac{z^3}{3}\). Hence, the integral becomes \(2[\frac{1^{3}}{3}-\frac{0^{3}}{3}]=\frac{2}{3}\). Now we have the final value for the triple integral: \(\frac{4}{9} * \frac{2}{3} = \frac{8}{27}\).

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Key Concepts

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

Triple Integration
Triple integration is a method of integrating over a three-dimensional region. In this context, it is used for iterated integrals that break down a complex integral into a sequence of simpler ones. Consider a triple integral in the form \[\int_{a}^{b}\int_{c}^{d}\int_{e}^{f} f(x, y, z) \, dx \, dy \, dz\]This represents integration over a volume, where the process is performed one dimension at a time. It's important to recognize:
  • The order of integration can vary depending on the boundaries and the function itself. This exercise integrates over the variables \( x, y, \) and \( z \).
  • A triple iterative integral first evaluates with \( x \), then with \( y \), and finally with \( z \) in the given problem statement.
Iterated integrals hone into particular areas of the function systematically, making them easier to calculate step-by-step. For symmetrical domains, such as here, the process is drastically simplified, often exploiting symmetry to handle more specific limits.
Power Rule for Integration
The power rule is a fundamental concept used during integration, especially noticeable in triple integrals and similar problems. This rule helps in managing polynomials, such as \( x^n \), when integrating. The rule states:\[\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\]Here, \( C \) is the constant of integration, which isn’t needed in definite integrals. This simplification applies perfectly to polynomials inside definite integrals, allowing us to find solutions straightforwardly. When applying the power rule:
  • Identify the power of \( x \), \( y \), or \( z \).
  • Integrate by adding one to the exponent and then dividing by the new exponent.
  • For definite integrals, substitute the limits and calculate the difference, handling one integral at a time.
This concept aids in transforming a complex-looking integral into a series of basic arithmetic operations, making calculations much simpler.
Symmetric Integrals
Symmetric integrals become invaluable in evaluating integrals, like the one presented within this exercise, over symmetric limits. The boundaries for \( x, y, \) and \( z \) lie symmetrically between \(-1\) and \(1\), aiding significant simplification. Often, integrals with symmetric limits yield simplified result forms due to the balance they provide. In this particular case:
  • The integral is calculated from \(-1\) to \(1\), but symmetry allows us to evaluate from \(0\) to \(1\) and then double the result obtained, simplifying calculations.
  • This represents that we only concern ourselves with one segment of symmetry and then utilize symmetrical properties to expand it.
  • Such symmetry exploits zeros or repeated patterns within the function, reducing unnecessary recalculation.
Understanding and recognizing symmetry helps foresee simplifications before delving into calculations, thus reducing errors and saving time.

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Most popular questions from this chapter

Find the average value of the function over the given solid. The average value of a continuous function \(f(x, y, z)\) over a solid region \(Q\) is $$\frac{1}{V} \iint_{Q} \int f(x, y, z) d V$$ where \(V\) is the volume of the solid region \(Q\). \(f(x, y, z)=x+y+z\) over the tetrahedron in the first octant with vertices \((0,0,0),(2,0,0),(0,2,0)\) and \((0,0,2)\)

Find the average value of the function over the given solid. The average value of a continuous function \(f(x, y, z)\) over a solid region \(Q\) is $$\frac{1}{V} \iint_{Q} \int f(x, y, z) d V$$ where \(V\) is the volume of the solid region \(Q\). \(f(x, y, z)=z^{2}+4\) over the cube in the first octant bounded by the coordinate planes, and the planes \(x=1, y=1\), and \(z=1\)

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The base of a pile of sand at a cement plant is rectangular with approximate dimensions of 20 meters by 30 meters. If the base is placed on the \(x y\) -plane with one vertex at the origin, the coordinates on the surface of the pile are \((5,5,3), \quad(15,5,6), \quad(25,5,4), \quad(5,15,2), \quad(15,15,7)\), and \((25,15,3)\). Approximate the volume of sand in the pile.

Using the description of the solid region, set up the integral for (a) the mass, (b) the center of mass, and (c) the moment of inertia about the \(z\) -axis. The solid in the first octant bounded by the coordinate planes and \(x^{2}+y^{2}+z^{2}=25\) with density function \(\rho=k x y\)

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