Problem 15 - Calculus Early Transcendentals: Pearson New International Edition

Question

Sketch the solid S. Then write an iterated integral for $$ \iiint_{S} f(x, y, z) d V $$ $$ \begin{aligned} S=\\{(&x, y, z): 0 \leq x \leq 3 z \\ &0 \leq y \leq 4-x-2 z, 0 \leq z \leq 2\\} \end{aligned} $$

Step-by-Step Solution

Answer

Visualize solid S, then set the iterated integral: \(\int_{0}^{2} \int_{0}^{3z} \int_{0}^{4-x-2z} f(x,y,z) \, dy \, dx \, dz\).

Step 1: Explore the Bounds for x, y, and z

To understand the region of integration, analyze the inequalities.- For x: The bounds are from 0 to \(3z\), meaning x extends from zero to the plane \(x = 3z\).- For y: The bounds are from 0 to \(4 - x - 2z\), suggesting y is limited by the plane \(y = 4 - x - 2z\).- For z: The bounds are from 0 to 2, which forms two parallel planes at \(z = 0\) and \(z = 2\).

Step 2: Visualize the Solid S

Imagine the region where each variable x, y, z is within its specified bounds. - At \(z = 0\), the region is determined by \(x\) and \(y\), forming a triangle where \(0 \leq x \leq 0\) and \(y = 4 - x\).- As \(z\) increases towards 2, x extends up to the plane \(x = 3z\) and further restricts the range for \(y\).- The solid is bound between two planes at \(z = 0\) and \(z = 2\).

Step 3: Set Up the Iterated Integral

Using the information from the bounds, the iterated integral is arranged as follows:\[\int_{0}^{2} \int_{0}^{3z} \int_{0}^{4-x-2z} f(x,y,z) \, dy \, dx \, dz\]This integral performs the integration over y first, then x, and finally z, covering the entire region S as defined.

Key Concepts

Iterated Integrals

Iterated integrals are a fundamental concept in multiple integrals, especially useful for evaluating functions over a multi-dimensional region. They involve successive integrations over different variables.
In this context, we proceed with integration starting from the innermost to the outermost integral. This layered approach simplifies solving the complex multi-dimensional integrals.
For the given exercise, the integral is structured as:

• First, integrate with respect to the y-variable while keeping x and z as constants.
• Next, integrate the result with respect to the x-variable, considering z as constant.
• Finally, integrate the resultant expression with respect to the z-variable, covering the described solid region.

Iterated integrals help to gradually calculate the volume, area, or any other required quantity in a specified region, by handling one dimension at a time.

Volume Calculation

Calculating volume using integrals involves finding the region where the function extends in three-dimensional space. The iterated integral provides a way to efficiently compute this volume.
Here, the function under consideration, denoted as \( f(x, y, z) \), is integrated over the given solid region. The integration bounds for each variable narrow down the space to exactly the desired solid configuration.
This process computes the summation of infinitesimal volumes, contributing to the total volume of the solid region.

• Each integration layer corresponds to a dimension: y for width, x for depth, and z for height.
• The function \( f(x, y, z) \) adjusts the density or weight of each differential volume element \( dV \).

This method ensures that you can find not just the volume but also evaluate other properties like mass, if the density function \( f \) is known.

Solid Region

A solid region refers to the three-dimensional space over which the function is integrated. Visualizing this area is crucial for setting up the correct integration bounds.
In the given problem, the region S is described by inequalities that define the limits for x, y, and z. This forms a geometric shape bounded by:

• The planes at \( z = 0 \) and \( z = 2 \), which cap the solid from below and above.
• The plane \( x = 3z \), marking the extension limit for x relative to z.
• The plane \( y = 4 - x - 2z \), shaping the limit for y based on x and z.

The formed solid can be constructed step by step by imagining each intersection and face, typically helping to sketch or visualize the shape before proceeding with setup of iterated integrals.

Integration Bounds

Integration bounds define the start and end points of each integral and are crucial in multiple integrals to ensure the accuracy of results. They specify the scope within which the function is integrated in each variable.
The problem provides explicit bounds:

• For x: From 0 to \(3z\), indicating a variable boundary dependent on z.
• For y: From 0 to \(4-x-2z\), making the y-boundary a function of both x and z.
• For z: From 0 to 2, a constant boundary across all x and y values.

Correctly interpreting these limits allows the construction of a precise mathematical model for the iterated integral. Each integral’s limits are dependent on the variables integrated before it, requiring careful attention.