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Magazine Chapter 14: Problem 13
From the limits of integration deserihe each region and find its volume. The inner integral has the inner limits. $$ \int_{e=0}^{2 \pi} \int_{z=0}^{1} \int_{r=0}^{2-z} r d r d z d \theta $$
Short Answer
Expert verified
The volume of the region is \( \frac{7\pi}{3} \).
Step by step solution
01
Identify the order of integration
The given triple integral is \( \int_{\theta=0}^{2 \pi} \int_{z=0}^{1} \int_{r=0}^{2-z} r \, dr \, dz \, d\theta \). It shows that the integration occurs first with respect to \( r \), then \( z \), and finally \( \theta \). These integrations correspond to cylindrical coordinates.
02
Describe the region of integration
The limits \( r=0 \) to \( r=2-z \) describe a region in the \( r \)-direction that is dependent on \( z \), starting at 0 and extending to \( 2-z \). The limits \( z=0 \) to \( z=1 \) indicate it is confined between these planes along the \( z \)-axis. Finally, \( \theta=0 \) to \( \theta=2\pi \) describes a full rotation around the \( z \)-axis.
03
Interpret the geometric region
Since \( r = 2 - z \) describes a slanted plane intersecting the \( z \)-axis and the \( r \)-axis in the cylindrical coordinate system, the region bounded by these limits is a cone truncated at \( z = 1 \). It opens upwards with the vertex at \( z = 2 \), \( r = 0 \).
04
Evaluate the inner integral (with respect to \( r \))
Compute \( \int_{r=0}^{2-z} r \, dr \). The antiderivative of \( r \) is \( \frac{1}{2}r^2 \). Evaluating it, \[ \left[ \frac{1}{2}r^2 \right]_0^{2-z} = \frac{1}{2}(2-z)^2 \].
05
Evaluate the middle integral (with respect to \( z \))
Next, solve \( \int_{z=0}^{1} \frac{1}{2}(2-z)^2 \, dz \). First, expand \( (2-z)^2 = 4 - 4z + z^2 \). The integrand becomes \( \frac{1}{2}(4 - 4z + z^2) \). Compute the integral, \[ \int_{z=0}^{1} \left(2 - 2z + \frac{1}{2}z^2\right) \, dz. \] Find the antiderivative: \( 2z - z^2 + \frac{1}{6}z^3 \). Evaluate from 0 to 1: \[ \left( 2(1) - 1^2 + \frac{1}{6}(1)^3 \right) - \left( 2(0) - 0^2 + \frac{1}{6}(0)^3 \right) = 2 - 1 + \frac{1}{6} = \frac{7}{6}. \]
06
Evaluate the outer integral (with respect to \( \theta \))
Finally, evaluate the integral \( \int_{\theta=0}^{2\pi} \frac{7}{6} \, d\theta \). Since \( \frac{7}{6} \) is constant with respect to \( \theta \), the integral becomes \( \frac{7}{6} \times [\theta]_0^{2\pi} = \frac{7}{6} \times (2\pi) = \frac{14\pi}{6} = \frac{7\pi}{3} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Cylindrical coordinates
Cylindrical coordinates are a three-dimensional coordinate system that extends polar coordinates into three dimensions by adding a height dimension, often denoted as \( z \). They are particularly beneficial for problems involving symmetry around a central axis, often simplifying integrals and calculations.
- The three components in cylindrical coordinates are \( r \), the radial distance from the origin to the point in the plane, \( \theta \), the angle from the positive x-axis, and \( z \), the height.
- The transformations between Cartesian and cylindrical coordinates are: \( x = r \cos(\theta) \), \( y = r \sin(\theta) \), and \( z = z \).
- For volumetric problems, these coordinates streamline integration as they naturally incorporate radial symmetry around the \( z \)-axis.
Region of integration
The region of integration is the domain over which the function is integrated. For our triple integral, the limits of integration dictate this region in cylindrical coordinates.
- The limits \( r=0 \) to \( r=2-z \) determine a region where \( r \) changes with \( z \), illustrating a dependency that forms part of the boundary of the integration region.
- Next, \( z=0 \) to \( z=1 \) describes height restrictions, creating a vertical boundary from the plane \( z=0 \) to \( z=1 \).
- Finally, the limits \( \theta=0 \) to \( \theta=2\pi \) imply a full rotation around the \( z \)-axis, covering the entire radial angle plane.
Antiderivative
An antiderivative is a function that is derived from another function by reversing differentiation. In integration, finding an antiderivative is a key step in evaluating definite integrals.
- For example, when integrating \( r \) with respect to \( r \), the antiderivative is \( \frac{1}{2}r^2 \) which we then evaluate across our specified limits.
- In the given problem, we found the antiderivative of the function inside the integral with respect to \( r \) successfully moving from \( r = 0 \) to \( r=2-z \).
- Subsequently, the integrals with respect to \( z \) and \( \theta \) were computed using similar processes of finding the antiderivatives and evaluating them over their respective limits.
Geometric interpretation
The geometric interpretation provides a visual understanding of the region being solved with the integral. It translates mathematical expressions into a recognizable geometric shape or volume.
- For this problem, the equation \( r = 2 - z \) outlines a slanted boundary that extends the region horizontally as \( z \) decreases, and vertically as \( z \) increases.
- This translates into a truncated cone: a cone segment chopped off at \( z=1 \) yet encompassing a full 360-degree rotation around the \( z \)-axis.
- The cone nature is clarified through limits such as \( r=0 \), hinting at the cone's vertex, and \( \theta \) spanning \( 0 \) to \( 2\pi \), illustrating the wrap-around.
