Problem 6 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1–10, evaluate the iterated integrals. $$ \int_{0}^{5} \int_{0}^{3} \int_{z^{2}}^{9} x y z d x d z d y $$

Step-by-Step Solution

Answer

The value of the iterated integral is 1518.75.

Step 1: Understand the Order of Integration

The given iterated integral is \[\int_{0}^{5} \int_{0}^{3} \int_{z^{2}}^{9} x y z \, d x \, d z \, d y\]This means we will first integrate with respect to \(x\), then \(z\), and finally \(y\). Our limits for \(x\) are from \(z^2\) to \(9\), for \(z\) from \(0\) to \(3\), and for \(y\) from \(0\) to \(5\).

Step 2: Integrate with Respect to x

Integrate \(x y z\) with respect to \(x\):\[\int_{z^{2}}^{9} x y z \, dx = yz \left[ \frac{x^2}{2} \right]_{z^{2}}^{9}\]Evaluate the definite integral:\[yz \left( \frac{9^2}{2} - \frac{(z^2)^2}{2} \right) = yz \left( \frac{81}{2} - \frac{z^4}{2} \right) = \frac{yz}{2}(81 - z^4)\]

Step 3: Integrate with Respect to z

Take the result from Step 2 and integrate with respect to \(z\):\[\int_{0}^{3} \frac{yz}{2}(81 - z^4) \, dz = \frac{y}{2} \int_{0}^{3} z(81 - z^4) \, dz\]Distribute \(z\):\[\frac{y}{2} \left( \int_{0}^{3} 81z \, dz - \int_{0}^{3} z^5 \, dz \right)\]Calculate each integral:\[\frac{y}{2} \left( \left[ \frac{81z^2}{2} \right]_0^3 - \left[ \frac{z^6}{6} \right]_0^3 \right)\]Evaluate the limits:\[y \left( \frac{81(9)}{4} - \frac{729}{12} \right) = y \left( \frac{729}{4} - 60.75 \right) = y \left( \frac{729}{4} - \frac{243}{4} \right) = y \cdot 121.5\]

Step 4: Integrate with Respect to y

Now integrate the resulting expression with respect to \(y\):\[\int_{0}^{5} 121.5y \, dy\]Calculate the integral:\[\left[ 121.5 \frac{y^2}{2} \right]_0^5 = 121.5 \left[ \frac{25}{2} - 0 \right] = 121.5 \times 12.5 = 1518.75\]

Step 5: Conclusion: Evaluate the Iterated Integral

The value of the iterated integral \[\int_{0}^{5} \int_{0}^{3} \int_{z^{2}}^{9} x y z \, d x \, d z \, d y\]is 1518.75.

Key Concepts

Iterated Integrals

Iterated integrals are a way of handling multivariable integrals by breaking them down into simpler, one-dimensional integrations. They involve integrating a function of several variables step-by-step over each variable. For instance, given the integral \( \int \int \int \) of a function \( f(x, y, z) \), we would sequentially integrate with respect to \( x \), then \( z \), and then \( y \). This step-by-step approach makes it more feasible to solve complex integrals by reducing them to a series of single-variable problems. In simple terms, you imagine peeling off one layer of integration at a time which aids in approaching the problem of multidimensional spaces succinctly.

Triple Integrals

Triple integrals extend the concept of integrating a function over a region in three-dimensional space. They are often used in calculations involving volumes, center of mass, and other physical properties governed by a three variable system. A triple integral is denoted as: \[ \int \int \int f(x, y, z) \, dx \, dy \, dz \]In this scenario, you are integrating a function \( f(x, y, z) \) over a specified region inside a three-dimensional space, taking each of the variables \( x, y, \) and \( z \) into account. The computation typically begins with one variable while holding the others constant, and proceeds iteratively through each variable. They might seem daunting, but once you get used to iterated integrals, working with triple integrals becomes more predictable and mechanical.

Order of Integration

Understanding the order of integration is crucial when solving iterated integrals. The order often follows the sequence in which these integrals appear in notation, but it can sometimes be changed to simplify calculations. Changing the order of integration can make the limits of integration easier to handle or result in simpler intermediate expressions.The original problem displayed \[ \int_{0}^{5} \int_{0}^{3} \int_{z^{2}}^{9} x y z \, dx \, dz \, dy \]reminds us to start by integrating with respect to \(x\), followed by \(z\), and finally \(y\). It's important to adjust the limits accurately with each change in order, as this maintains the integrity of the problem while streamlining calculations.

Definite Integration

Definite integration is the process of finding the integral of a function over a specified range. In contrast to indefinite integration, where the result is a function plus a constant, definite integrals provide a numerical value. This result represents the area under the curve of the function between two boundary values. When performing definite integrals, you evaluate the integrated function at the upper limit and subtract its value at the lower limit. In the context of the problem, after calculating through different layers of integration, you arrive at a definite value that represents the cumulative effect of the integration over each axis, culminating in a specific total, which is 1518.75 in this case. This underscores how definite integration seeks to pin down an exact value within the bounds that transform theoretical calculations into practical numeric results.