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A triple integral in cylindrical coordinates is given. Describe the region in space defined by the bounds of the integral. $$ \int_{0}^{\pi} \int_{0}^{1} \int_{0}^{2-r} r d z d r d \theta $$

Short Answer

Expert verified
The region is a 3D semicircular wedge with radius 1 and height \( z = 2-r \), from \( \theta = 0 \) to \( \pi \).

Step by step solution

01

Identify the Integration Order

The given integral is \( \int_{0}^{\pi} \int_{0}^{1} \int_{0}^{2-r} r \; dz \; dr \; d\theta \). The integration is performed first with respect to \( z \), then \( r \), and finally \( \theta \). This order indicates that \( z \), \( r \), and \( \theta \) are cylindrical coordinates defining a region in space.
02

Examine \( z \) Bounds

The innermost integral with respect to \( z \) has bounds from 0 to \( 2-r \). This means that for each fixed radius \( r \), the height above the \( xy \)-plane extends from 0 to \( 2-r \). This indicates a region where the top boundary slopes downwards as \( r \) increases.
03

Examine \( r \) Bounds

The bounds for \( r \) are from 0 to 1. Thus, the distance from the origin (or the radius) of the region in the \( xy \)-plane extends from 0 to 1. This restriction confines the region to a circular area with radius 1 in the \( xy \)-plane.
04

Examine \( \theta \) Bounds

The outermost integral with respect to \( \theta \) has bounds from 0 to \( \pi \). This covers half of the cylindrical coordinate plane, creating a semicircular region swiveling around the positive \( x \)-axis, from \( \theta = 0 \) to \( \pi \).
05

Describe the Overall Region

Combining all the bounds: the region is a 3D volume extending radially from the origin to 1, having a height extending up to \( z = 2-r \). The region forms a semicircular wedge, like the top half of a thick crescent slice, in the specified plane segment defined by \( \theta = 0 \) to \( \theta = \pi \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cylindrical Coordinates
When we talk about cylindrical coordinates, we are discussing a system that is especially useful for solving problems where there is some symmetry around an axis. This system replaces the traditional Cartesian coordinates
  • Instead of using \(x\), we use \(r\), which represents the radial distance from the origin in the \(xy\)-plane.
  • \(\theta\) is used instead of \(y\), denoting the angle in radians from the positive \(x\)-axis, moving counterclockwise in the \(xy\)-plane.
  • The \(z\) coordinate remains unchanged, representing the height above the \(xy\)-plane.
Cylindrical coordinates are particularly useful when dealing with circular or cylindrical shapes. They allow us to simplify calculations and better visualize the regions of integration, focusing on circular dimensions rather than boxy Cartesian boxes.
Using this system, problems with rotational symmetry become easier to describe and solve, as in this exercise where we consider a semicircular top wedge.
Integration Bounds
Integration bounds define the limits between which we operate mathematically, describing the region over which an integral operates. Here, let's break down the bounds used in the original exercise:
  • For \(z\), the bound \(0\) to \(2-r\) indicates how high above the \(xy\)-plane the region extends for any given \(r\). This variable upper limit creates a slope in the region.
  • The \(r\) bound from \(0\) to \(1\) confines the distance from the origin, defining the radial extent within the \(xy\)-plane.
  • Lastly, the \(\theta\) bound between \(0\) to \(\pi\) depicts a half-circle sweep, capturing 180 degrees of rotation.
Overall, these bounds work together to define a particular section of space. Correctly understanding these limits can help us visualize the complex 3-D shapes effectively represented by triple integrals.
Volume Integration
Volume integration in cylindrical coordinates helps us calculate the volume of a 3D shape by integrating over the defined bounds. Here, we use triple integrals:
  • The integration is in the order of \(z\), \(r\), and \(\theta\).
  • The integrand \(r\) serves as an adjustment factor, ensuring volumes are calculated correctly given the radial nature of cylindrical coordinates.
  • Accumulating contributions by integrating first along the height \(z\), then radius \(r\), and finally angle \(\theta\), creates a cumulative volume.
Each integral processes earlier results, using the integrand and bounds to account for variations influenced by contributions from the radial and angular dimensions. These integrations together map out the desired segment or volume, in this case, a half-cylinder region with a sloped cap.
Mathematical Region Description
Mathematical region descriptions provide a detailed account of the dimensional constraints and characteristics of a selected volume or shape. In this triple integral example:
  • The origin \(r=0\) extends outward to an edge defined by \(r=1\), giving it a quarter-circle area.
  • The regional height tapers down from \(z=2\) to \(z=0\) as \(r\) reaches its maximum boundary.
  • The angle sweeps from 0 to \(\pi\), covering exactly half of the potential cylindrical barrier.
Together, these features describe a semicircular wedge sitting on the plane, contained by boundaries that taper and curve based on cylindrical characteristics. Visualizing these components allows students to better grasp thought experiments based on radial and angular symmetry, crucial for mastering concepts like the segment described in this practice problem.

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Most popular questions from this chapter

Describe the curve, surface or region in space determined by the given bounds. Bounds in cylindrical coordinates: (a) \(1 \leq r \leq 2, \quad \theta=\pi / 2, \quad 0 \leq z \leq 1\) (b) \(r=2, \quad 0 \leq \theta \leq 2 \pi, \quad z=5\) Bounds in spherical coordinates: (c) \(0 \leq \rho \leq 2, \quad 0 \leq \theta \leq \pi, \quad \varphi=\pi / 4\) (d) \(\rho=2, \quad 0 \leq \theta \leq 2 \pi, \quad \varphi=\pi / 6\)

Find the average value of \(f\) over the region \(R .\) Notice how these functions and regions are related to the iterated integrals given in Exercises \(5-8\). \(f(x, y)=\sin x \cos y ; \quad R\) is bounded by \(x=0, x=\pi\) \(y=-\pi / 2\) and \(y=\pi / 2\).

Let \(z=f(x, y)\) and \(z=g(x, y)=2 f(x, y) .\) Why is the surface area of \(g\) over a region \(R\) not twice the surface area of \(f\) over \(R ?\)

Find the center of mass of the lamina described by the region \(R\) in the plane and its density function \(\delta(x, y)\) Note: these are the same lamina as in Exercises \(11-18\). \(R\) is the rectangle with corners (1,-3),(1,2),(7,2) and (7,-3)\(; \delta(x, y)=\left(x+y^{2}\right) \mathrm{gm} / \mathrm{cm}^{2}\)

Point masses are given along a line or in the plane. Find the center of \(\operatorname{mass} \bar{x}\) or \((\bar{x}, \bar{y}),\) as appropriate. (All masses are in grams and distances are in cm.) $$ \begin{array}{l} m_{1}=2 \text { at } x=-3 ; \quad m_{2}=2 \text { at } x=-1; \\ m_{3}=3 \text { at } x=0 ; \quad m_{4}=3 \text { at } x=7 \end{array} $$

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