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Magazine Chapter 4: Problem 19
In the following exercises, find the volume of the solid \(E\) whose boundaries are given in rectangular coordinates. \(E\) is bounded by the circular cone \(z=\sqrt{x^{2}+y^{2}}\) and \(z=1\).
Short Answer
Expert verified
The volume of the solid \(E\) is \(\frac{\pi}{3}\) cubic units.
Step by step solution
01
Understand the Solid's Boundaries
The solid is bounded above by the plane where \(z = 1\) and below by the circular cone where \(z = \sqrt{x^2 + y^2}\). This means the solid lies between the cone and the plane, essentially creating a truncated cone or a frustum of a cone.
02
Set Up the Triple Integral
We will calculate the volume of the solid \(E\) using triple integration in cylindrical coordinates. In cylindrical coordinates, the equations are \(z = r\) and \(z = 1\), where \(r = \sqrt{x^2 + y^2}\).
03
Determine the Limits of Integration
For \(z\), the limits are \(r\) to \(1\). For \(r\), since \(z = r\) and \(z = 1\), \(r\) ranges from \(0\) to \(1\). For \(\theta\), it ranges from 0 to \(2\pi\) because the solid revolves fully around the z-axis.
04
Write and Solve the Triple Integral
The volume \(V\) is given by the integral:\[V = \int_{0}^{2\pi} \int_{0}^{1} \int_{r}^{1} r \, dz \, dr \, d\theta\]Integrate with respect to \(z\) first:\[= \int_{0}^{2\pi} \int_{0}^{1} [z]_{r}^{1} r \, dr \, d\theta = \int_{0}^{2\pi} \int_{0}^{1} (1 - r) r \, dr \, d\theta\]Now integrate with respect to \(r\):\[= \int_{0}^{2\pi} \left[\frac{r^2}{2} - \frac{r^3}{3}\right]_{0}^{1} \, d\theta = \int_{0}^{2\pi} \left(\frac{1}{2} - \frac{1}{3}\right) d\theta\]Compute the integral for \(\theta\):\[= \frac{1}{6} \int_{0}^{2\pi} d\theta = \frac{1}{6} [\theta]_{0}^{2\pi} = \frac{1}{6} (2\pi) = \frac{\pi}{3}\]
05
Conclusion
The volume of the solid \(E\) is \(\frac{\pi}{3}\) cubic units.
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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 way to extend polar coordinates into three dimensions. They are particularly useful for symmetric solids around a central axis, such as cones or cylinders. In this system, any point in space is identified by three values:
- The radial distance \( r \), which measures the horizontal distance from the chosen axis (usually the z-axis).
- The angle \( \theta \), which is similar to the angle in polar coordinates and measures how far the point is around from the reference direction (usually the x-axis). \( \theta \) ranges from \( 0 \) to \( 2\pi \).
- The height \( z \), similar to the z-coordinate in Cartesian coordinates, which measures how far up or down the point is from the xy-plane.
Triple Integration
Triple integration is a method used to find volumes in three-dimensional space. When you perform a triple integral, you may integrate over three variables that describe each dimension, such as \( r \), \( \theta \), and \( z \) in cylindrical coordinates. In solving the problem, the triple integral is set up to calculate the volume of the solid: \[ V = \int_{0}^{2\pi} \int_{0}^{1} \int_{r}^{1} r \, dz \, dr \, d\theta \]Here's how it works step by step:
- First, integrate with respect to \( z \), keeping \( r \) and \( \theta \) constant. This determines the height contribution to the volume.
- Next, integrate with respect to \( r \). This step adds the radial dimension to the volume calculation.
- Finally, integrate with respect to \( \theta \). This covers the angular symmetry of the solid, completing the full volume calculation.
Circular Cone
A circular cone is a three-dimensional geometric shape with a circular base and a pointed top called the apex. A right circular cone is symmetric around a central axis passing through its apex and base center. In the given exercise, the circular cone is described by the equation \( z = \sqrt{x^2 + y^2} \).This equation implies that every point on the cone has a height \( z \) that is equal to its radial distance from the axis (or \( r \) in cylindrical coordinates). Key features of a circular cone:
- The base circle lies in a plane parallel to the xy-plane.
- The apex is directly above or below the center of the base.
- The equation \( z = r \) illustrates that the cone's radius reduces to zero as it approaches the apex.
Frustum of a Cone
A frustum of a cone is formed when a cone is cut by a plane parallel to its base, removing the top portion. What's left is the frustum, which has circular top and bottom surfaces and a lateral surface connecting them. The exercise involves calculating the volume of a frustum of a cone bounded by the planes \( z = \sqrt{x^2 + y^2} \) and \( z = 1 \).Key characteristics of a frustum of a cone:
- It maintains the circular symmetry of the original cone.
- The top and bottom surfaces are parallel to each other, as in this exercise, bounded by the circular top at \( z = 1 \) and open at the base of the cone \( z = r \).
- You calculate the volume by integrating between these two boundaries.
