Chapter 14: Problem 2
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.
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.
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.