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Magazine Chapter 13: Problem 13
Sketch the solid \(S .\) Then write an iterated integral for \(\iiint_{S} f(x, y, z) d V\). $$ S=\left\\{(x, y, z): 0 \leq x \leq \frac{1}{2} y, 0 \leq y \leq 4,0 \leq z \leq 2\right\\} $$
Short Answer
Expert verified
The iterated integral is \(\int_{0}^{2} \int_{0}^{4} \int_{0}^{\frac{1}{2}y} f(x, y, z) \, dx \, dy \, dz\).
Step by step solution
01
Understand the region's boundaries
Region \(S\) is defined in the space with specific boundaries for \(x\), \(y\), and \(z\). We observe \(x\) is constrained by \(0 \leq x \leq \frac{1}{2}y\), \(y\) by \(0 \leq y \leq 4\), and \(z\) by \(0 \leq z \leq 2\). These constraints define a solid region in the first octant of the three-dimensional space.
02
Define limits for each variable
Recognizing that \(x\) depends on \(y\), we order the integration first with \(x\), then \(y\), and finally \(z\). The bounds for \(x\) are from \(0\) to \(\frac{1}{2}y\); for \(y\) from \(0\) to \(4\); and for \(z\) from \(0\) to \(2\).
03
Set up the iterated integral
Based on the order of integration determined in step 2, the iterated integral for \(\iiint_{S} f(x, y, z) \, dV\) becomes: \[\int_{0}^{2} \int_{0}^{4} \int_{0}^{\frac{1}{2}y} f(x, y, z) \, dx \, dy \, dz\] This represents integration with respect to \(x\), then \(y\), then \(z\).
04
Sketch the solid region
Start by sketching the region on the \(xy\)-plane. Since \(x\leq \frac{1}{2}y\), this gives a triangular region with points \((0,0)\), \((4,0)\), and \((4,2)\). Extend this region in the \(z\)-direction from \(0\) to \(2\), forming a prism extending vertically along the \(z\)-axis.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Solid Region
When we talk about a solid region like the solid \( S \) in the given exercise, we're focusing on a specific part of three-dimensional space bounded by the set conditions. This solid is described by three inequalities:
This might seem complex, but envision each inequality as describing a wall or edge that confines \( S \). For example, \( x \leq \frac{1}{2}y \) forms a slanting plane that angles off the \( x \)-axis and limits \( x \) based on \( y \).'s value.
The result is a three-dimensional shape, specifically a prism with triangular cross-sections.”],
- For \( x \), \( 0 \leq x \leq \frac{1}{2}y \).
- For \( y \), \( 0 \leq y \leq 4 \).
- For \( z \), \( 0 \leq z \leq 2 \).
This might seem complex, but envision each inequality as describing a wall or edge that confines \( S \). For example, \( x \leq \frac{1}{2}y \) forms a slanting plane that angles off the \( x \)-axis and limits \( x \) based on \( y \).'s value.
The result is a three-dimensional shape, specifically a prism with triangular cross-sections.”],
Three-Dimensional Space
In this context, three-dimensional space refers to the environment where the variables \( x \), \( y \), and \( z \) interact to define a volume or solid. This space is divided into eight regions called octants, much like quadrants on a two-dimensional plane.
The first octant is the region where all coordinates \( x \), \( y \), and \( z \) are non-negative. Our region \( S \) fits here because all extracted data meet these conditions.
Visualizing three-dimensional space involves imagining how these axes create a grid or scaffold where solids can exist. By picturing slices along these axes, we can break down and analyze the volume more simply. Each step of integration effectively calculates the volume of a slice before adding them together to find the entire solid.”},{
The first octant is the region where all coordinates \( x \), \( y \), and \( z \) are non-negative. Our region \( S \) fits here because all extracted data meet these conditions.
Visualizing three-dimensional space involves imagining how these axes create a grid or scaffold where solids can exist. By picturing slices along these axes, we can break down and analyze the volume more simply. Each step of integration effectively calculates the volume of a slice before adding them together to find the entire solid.”},{
Integration Limits
Establishing integration limits is a cornerstone of solving integrals over solid regions. These limits define the boundaries over which integration occurs and dictate the order and shape of the solid.
In the given task, each integral's limits are constructed based on how the variable interactions vary with the region's dimensions.
In the given task, each integral's limits are constructed based on how the variable interactions vary with the region's dimensions.
- The innermost integral, with respect to \( x \), varies from \( 0 \) to \( \frac{1}{2}y \). This considers how \( x \) changes for a constant \( y \).
- The middle integral, with respect to \( y \), stretches from \( 0 \) to \( 4 \). It encompasses all possible vertical heights the cross-section can take.
- The outermost integral brings \( z \) into play, spanning from \( 0 \) to \( 2 \). This accounts for the entire prism's elevation in three-dimensional space.
