Problem 7 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

The evaluated iterated integral is \( \frac{8}{9} \).

Step 1: Understand the Problem

The problem asks us to evaluate an iterated triple integral \( \int_{0}^{2} \int_{1}^{z} \int_{0}^{\sqrt{x / z}} 2 x y z \, d y \, d x \, d z \). This means we need to integrate the function \( 2xyz \) first with respect to \( y \), then \( x \), and finally \( z \).

Step 2: Integrate with respect to y

Evaluate the innermost integral by integrating \( 2xyz \) with respect to \( y \) while treating \( x \) and \( z \) as constants. The integral becomes: \[ \int_{0}^{\sqrt{x/z}} 2xyz \, dy = 2xz \int_{0}^{\sqrt{x/z}} y \, dy = 2xz \left[ \frac{y^2}{2} \right]_{0}^{\sqrt{x/z}} = xz \left[ \left( \frac{x}{z} \right) \right] = \frac{x^2}{z} \]

Step 3: Integrate with respect to x

Now integrate the result \( \frac{x^2}{z} \) with respect to \( x \) within the limits from \( 1 \) to \( z \): \[ \int_{1}^{z} \frac{x^2}{z} \, dx = \frac{1}{z} \int_{1}^{z} x^2 \, dx = \frac{1}{z} \left[ \frac{x^3}{3} \right]_{1}^{z} = \frac{1}{z} \left( \frac{z^3}{3} - \frac{1}{3} \right) = \frac{z^2}{3} - \frac{1}{3z} \]

Step 4: Integrate with respect to z

Finally, integrate the result \( \frac{z^2}{3} - \frac{1}{3z} \) with respect to \( z \) from \( 0 \) to \( 2 \): \[ \int_{0}^{2} \left( \frac{z^2}{3} - \frac{1}{3z} \right) \, dz = \frac{1}{3} \int_{0}^{2} z^2 \, dz - \frac{1}{3} \int_{0}^{2} \frac{1}{z} \, dz \]Evaluating each part separately:For \( \frac{1}{3} \int_{0}^{2} z^2 \, dz \), we get:\[ \frac{1}{3} \left[ \frac{z^3}{3} \right]_{0}^{2} = \frac{1}{3} \left( \frac{8}{3} - 0 \right) = \frac{8}{9} \]For \( -\frac{1}{3} \int_{0}^{2} \frac{1}{z} \, dz \), we use the property that \( \int \frac{1}{z} \, dz = \ln |z| \):\[ -\frac{1}{3} \left( \ln |2| - \ln |0| \right) = -\frac{1}{3} \ln 2 \]Since \( \ln|0| \) is undefined, we consider limit as z approaches 0, which simplifies to:\[ -\frac{1}{3} \ln 2 \]Thus, the entire integral evaluates to:\[ \frac{8}{9} - \frac{0}{3} \ln 2 = \frac{8}{9} \]

Step 5: Final Evaluation and Verification

The result we obtained is \( \frac{8}{9} \). Double-checking each step makes sure all calculations have been carried out correctly according to the standard integration techniques.

Key Concepts

Triple Integral

A triple integral is a way to integrate a function in three-dimensional space. It's a bit like stacking sheets of paper on top of one another. Each sheet represents a slice through 3D space. With triple integrals, we calculate the volume under the surface described by the function. In our exercise, we are computing the triple integral to find the volume beneath the function \( 2xyz \) over a specific region. This allows us to apply calculus to solve complex three-dimensional problems.

Integration Techniques

When evaluating a triple integral, you have to use specific integration techniques. We evaluate each integral one at a time, starting from the innermost to the outermost. Here, we integrate with respect to \( y \) first, then \( x \), and finally \( z \). This process is called iterated integration. We rely on the fundamental theorem of calculus, which allows us to find the antiderivative of the function and evaluate it within given limits. Alongside, we can utilize substitution and partial fraction decomposition, as needed.

Mathematical Problem Solving

Solving mathematical problems using triple integrals involves understanding the problem layout and the logic behind integration. We broke down the computation into manageable steps:

• Identify the function: We pinpoint the function we want to integrate, \( 2xyz \), and the order of integration.
• Establish limits of integration: Each integral has specific upper and lower bounds, guiding how far the integration should go in each dimension.
• Proceed stepwise: Integrate iteratively from innermost to outermost function, computing each step meticulously.

This meticulous approach helps in finding accurate solutions to complex mathematical problems.

Calculus Education

Calculus education often includes learning through practice and application of concepts like triple integrals. By working through exercises like these, students solidify their grasp on abstract mathematical ideas and their application to real-world problems. They develop problem-solving strategies and enhance their analytical skills by repeatedly working through methodical step-by-step solutions. Mastery over calculus ensures a deeper understanding of the mechanics of the world around us, making it a cornerstone in fields such as physics, engineering, and economics.