Problem 13 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 11-20, sketch the solid \(S\). Then write an iterated integral for $$ \iiint_{S} f(x, y, z) d V $$ $$ S=\left\\{(x, y, z): 0 \leq x \leq \frac{1}{2} y, 0 \leq y \leq 4,0 \leq z \leq 2\right\\} $$

Step-by-Step Solution

Answer

The iterated integral is \( \int_0^4 \int_0^{\frac{1}{2}y} \int_0^2 f(x, y, z) \, dz \, dx \, dy \).

Step 1: Understand the Region

The solid \( S \) is given by the inequality constraints: \( 0 \leq x \leq \frac{1}{2} y \), \( 0 \leq y \leq 4 \), and \( 0 \leq z \leq 2 \). This implies that \( x \) is dependent on \( y \), which means that for each value of \( y \), \( x \) has a range from 0 to \( \frac{1}{2} y \). The \( y \) and \( z \) ranges are independent with \( y \) ranging from 0 to 4, and \( z \) ranging from 0 to 2.

Step 2: Visualize the Region

Sketch the region in the \( xy \)-plane described by \( 0 \leq x \leq \frac{1}{2} y \) and \( 0 \leq y \leq 4 \). This results in a right triangle with vertices at (0,0), (2,4), and (0,4). Then extend this triangular region vertically from \( z = 0 \) to \( z = 2 \) to form a triangular prism.

Step 3: Set Up the Iterated Integral

To write the iterated integral, first decide the order of integration. A convenient order is \( dz \, dy \, dx \), due to the simple bounds of \( z \). Integrate first with respect to \( z \, (z: 0 \text{ to } 2) \), then \( y \, (y: 0 \text{ to } 4) \), and finally \( x \, (x: 0 \text{ to } \frac{1}{2}y) \).

Step 4: Express the Iterated Integral

The iterated integral can be expressed as:\[\int_0^4 \int_0^{\frac{1}{2}y} \int_0^2 f(x, y, z) \, dz \, dx \, dy\]This integral first accumulates values along the \( z \) axis, then the \( x \) axis, and finally spans the entire solid through the \( y \) axis.

Key Concepts

Iterated Integrals

In multivariable calculus, iterated integrals are an essential tool used to evaluate integrals over a multi-dimensional region. They are especially critical when working with regions defined by boundaries in more than one variable. The idea is to break down the integration process into simpler, one-variable integrals that can be tackled in sequence. For instance, in a three-dimensional space, an iterated integral may take the form

• Nested integrals that cover different dimensions.
• Order of integration chosen based on the function's boundaries.

In the given problem, you derive and organize the integral boundaries by simplifying these nested operations. We start by integrating with respect to one variable, employ its limits, and then proceed to the next. Each integration step processes accumulative data from the previous dimensions, eventually covering the entire volume or area defined by the limits. Understanding these steps makes iterated integrals much less daunting.

Triple Integrals

Triple integrals extend the idea of double integrals to the third dimension, which is particularly useful for finding volumes under surfaces or inside solid regions. The objective is to integrate a function over a three-dimensional region, effectively generalizing the one-dimensional integral operation to encompass two more dimensions.
Triple integrals allow you to:

• Calculate volume by integrating over three variables: typically denoted as \( x \), \( y \), and \( z \).
• Sum infinitesimal elements of volume, leading to applications in physics and engineering, like finding mass or charge distributions.

In practical terms, each dimension is integrated iteratively, where each subsequent integral accounts for accumulated values from the previous integrals. For the problem provided, the triple integral helps evaluate the entire function \( f(x, y, z) \) across the solid region \( S \), capturing variations in all three variables. The formulation \[ \int_0^4 \int_0^{\frac{1}{2}y} \int_0^2 f(x, y, z) \, dz \, dx \, dy \] highlights how each integration stage builds onto the last, ultimately providing a comprehensive calculation across the defined region.

Solid Regions

Solid regions in multivariable calculus refer to the three-dimensional spaces over which integrals are performed. These regions are typically bounded by surfaces or by coordinate planes and can take on various shapes, such as cubes, spheres, or more complex geometric configurations.
In our context:

• The solid region \( S \) is defined by constraints on \( x \), \( y \), and \( z \).
• Each dimensional boundary affects the others, like how \( x \)'s maximum is tied to \( y \) with the equation \( x \leq \frac{1}{2} y \).

Visualizing solid regions is key. For this exercise, imagine the region as a triangular prism. It starts as a triangle on the \( xy \)-plane and extends into space along the \( z \)-axis. Drawing these areas step by step is helpful in understanding the overall structure. Once visualized, you are better equipped to apply iterated triple integrals, rooted firmly in grasping these solid regions comprehensively. Having a mental model aids significantly in setting up the correct integral expressions effectively.