/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 100 Compute the volume of the follow... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 100

Compute the volume of the following solids. The wedge sliced from the cylinder \(x^{2}+y^{2}=1\) by the planes \(z=1-x\) and \(z=x-1\)

Short Answer

Expert verified
Question: Compute the volume of the wedge created by the intersection of a cylinder defined by the equation \(x^2 + y^2 = 1\), and two planes: \(z = 1 - x\) and \(z = x - 1\). Answer: The volume of the wedge can be calculated by evaluating the triple integral, set up in cylindrical coordinates: \(V = \left|\int_{0}^{1} [2r^2\sin(2\pi - 2\arccos{(1/r)}) - 2r^2\sin(0)] \,dr\right|\).

Step by step solution

01

Set up the problem in cylindrical coordinates

Since we are working with a cylinder, it's best to use cylindrical coordinates. Replace \(x^2 + y^2\) by \(r^2\), where \(r\) is the radial distance from the \(z\)-axis. Our equation becomes: \(r^2 = 1\) The two planes can also be written in cylindrical coordinates. First, solve the plane equations for \(z\): \(z = 1 - x\) \(z = x - 1\) Now, replace \(x\) by \(r\cos\theta\): \(z = 1 - r\cos\theta\) \(z = r\cos\theta - 1\)
02

Find the boundary of the intersection

We need to figure out the boundaries of the intersection between the cylinder and two planes. To do this, first find the intersection of the planes by setting the \(z\)-coordinates equal to each other: \(1 - r\cos\theta = r\cos\theta - 1\) Now solve for \(r\cos\theta\): \(r\cos\theta = 1\)
03

Set up the triple integral

Our goal is to compute the volume by integrating over the region with the appropriate kernel. In cylindrical coordinates, we have: \(dV = r\,dr\,d\theta\,dz\) The bounds for \(r\) are \(0 \leq r \leq 1\), for \(\theta\), we need to determine the bounds where the intersection of the planes occurs with the cylinder: \(0 \leq \theta \leq 2\pi - 2\arccos{(1/r)}\) And for \(z\), \(1-r\cos\theta \leq z \leq r\cos\theta - 1\): \(\int_{0}^{1}\int_{0}^{2\pi - 2\arccos{(1/r)}}\int_{1-r\cos\theta}^{r\cos\theta - 1} r\,dz\,d\theta\,dr\)
04

Perform the integration

Let's first integrate with respect to \(z\): \(\int_{0}^{1}\int_{0}^{2\pi - 2\arccos{(1/r)}} r[(r\cos\theta - 1) - (1 - r\cos\theta)]\,d\theta\,dr\) This simplifies to: \(\int_{0}^{1}\int_{0}^{2\pi - 2\arccos{(1/r)}} 2r^2\cos\theta\,d\theta\,dr\) Next, integrate with respect to \(\theta\): \(\int_{0}^{1} [2r^2\sin\theta]_{0}^{2\pi - 2\arccos{(1/r)}} \,dr\) Finally, integrate with respect to \(r\), remembering to add the absolute value because the volume is always positive: \(V = \left|\int_{0}^{1} [2r^2\sin(2\pi - 2\arccos{(1/r)}) - 2r^2\sin(0)] \,dr\right|\) Evaluate this integral to get the volume of the wedge.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Cylindrical Coordinates
When tackling problems related to cylinders, switching to cylindrical coordinates can significantly simplify the math.
In cylindrical coordinates, a point in space is represented by three values:
  • \(r\) - the radial distance from the origin (akin to the radius in polar coordinates).
  • \(\theta\) - the angle of rotation from the positive x-axis.
  • \(z\) - the height above the xy-plane, which is a natural extension from the Cartesian z-coordinate.
For the specific problem at hand, we translate the cylinder equation \(x^2 + y^2 = 1\) to \(r^2 = 1\).
This equation states that the radial distance is constant at 1, defining a unit cylinder.
Triple Integral
Computing the volume of a solid shape often requires the application of a triple integral.
This involves integrating a function over a three-dimensional region. In this case, the function to integrate is simply the differential volume \(dV = r \, dr \, d\theta \, dz\).
  • The order of integration can be carefully selected to simplify the process. Here, the integration occurs first in \(z\), then \(\theta\), and finally in \(r\).
  • Each integral has specific bounds, derived from the geometry formed by the cylinder and the intersecting planes.
This structured integration efficiently accumulates the total volume of the defined solid within the given boundaries.
Solid Geometry
Understanding the geometry of combined solids is crucial.
In this case, the wedge is defined by intersecting a unit cylinder with two planes.
By examining the equations \(z = 1 - x\) and \(z = x - 1\) in cylindrical coordinates:
  • These describe planes slicing through the cylinder, producing a wedge-shaped section.
  • The intersection conditions determine limits of integration and essential properties like symmetry.
Evaluating the shape in cylindrical terms allows insights into how these geometric elements interact, crucial for effectively setting up the integral bounds.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A cylindrical soda can has a radius of \(4 \mathrm{cm}\) and a height of \(12 \mathrm{cm} .\) When the can is full of soda, the center of mass of the contents of the can is \(6 \mathrm{cm}\) above the base on the axis of the can (halfway along the axis of the can). As the can is drained, the center of mass descends for a while. However, when the can is empty (filled only with air), the center of mass is once again \(6 \mathrm{cm}\) above the base on the axis of the can. Find the depth of soda in the can for which the center of mass is at its lowest point. Neglect the mass of the can and assume the density of the soda is \(1 \mathrm{g} / \mathrm{cm}^{3}\) and the density of air is \(0.001 \mathrm{g} / \mathrm{cm}^{3}\)

Find equations for the bounding surfaces, set up a volume integral, and evaluate the integral to obtain a volume formula for each region. Assume that \(a, b, c, r, R,\) and h are positive constants. Find the volume of the cap of a sphere of radius \(R\) with height \(h\)

Use a change of variables to evaluate the following integrals. \(\iiint_{D} d V ; D\) is bounded by the planes \(y-2 x=0, y-2 x=1\) \(z-3 y=0, z-3 y=1, z-4 x=0,\) and \(z-4 x=3\)

Consider the following two-and three-dimensional regions. Specify the surfaces and curves that bound the region, choose a convenient coordinate system, and compute the center of mass assuming constant density. All parameters are positive real numbers. A tetrahedron is bounded by the coordinate planes and the plane \(x / a+y / a+z / a=1 .\) What are the coordinates of the center of mass?

Evaluate the following integrals using the method of your choice. A sketch is helpful. $$\iint_{R} \sqrt{x^{2}+y^{2}} d A ; R=\left\\{(x, y): 1 \leq x^{2}+y^{2} \leq 4\right\\}$$
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.