/*! 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 28 Determine whether the statement ... [FREE SOLUTION] | 91Ó°ÊÓ

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Determine whether the statement is true or false. Explain your answer. [In these exercises, assume that a solid \(S\) of volume \(V\) is bounded by two parallel planes perpendicular to the \(x\) -axis at \(x=a\) and \(x=b\) and that for each \(x\) in \([a, b], A(x)\) denotes the cross-sectional area of \(S\) perpendicular to the \(x\) -axis.] If each cross section of \(S\) is a disk or a washer, then \(S\) is a solid of revolution.

Short Answer

Expert verified
True; each circular cross section implies \(S\) is a solid of revolution.

Step by step solution

01

Understanding Cross Sections

The problem states that the solid \(S\) is bounded by two planes perpendicular to the \(x\)-axis at \(x=a\) and \(x=b\), and for each point \(x\), there is a cross-sectional area\(A(x)\) that is perpendicular to the \(x\)-axis.
02

Interpreting Disk or Washer Cross Sections

If the cross sections of \(S\) are disks or washers, it implies that they are circular in shape. Disks are full circles, while washers are annular rings, possibly with a hole in the middle.
03

Determining a Solid of Revolution

A solid of revolution is a solid formed by rotating a region around an axis. If cross sections perpendicular to the axis of rotation are circular, this suggests that the region has been revolved around an axis to form such cross sections.
04

Conclusion Based on Cross Sections

Since each cross section of \(S\) is a disk or a washer, this suggests that the solid \(S\) was formed by revolving a region around the \(x\)-axis. Thus, \(S\) is a solid of revolution as circular cross sections are typical for such solids.

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Key Concepts

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

Solid of Revolution
Solids of revolution are fascinating geometric structures. These solids are created when a 2-dimensional region is revolved around an axis, forming a 3D object. Imagine taking a flat shape like a rectangle or circle and spinning it around a line, much like a potter spins clay on a wheel. This rotational action generates a solid figure, often with a symmetrical, curved form.
To visualize, picture rotating a rectangle about one of its edges. The result would be a cylinder, a perfect example of a solid of revolution. A solid of revolution can include more complex shapes too. By rotating various curves or lines, intricate and beautiful solids can emerge.
When thinking about solids of revolution, recall that the axis of rotation is crucial. Typically, this axis can be one of the coordinate axes, like the x-axis or y-axis. By examining the shape of initial regions and understanding their rotation around an axis, the properties of the solid can be determined.
Cross-Sectional Area
The concept of cross-sectional area is central in understanding the volume of a solid. The cross-section refers to the shape you would see if you "sliced" through the solid along a plane. For solids of revolution, each slice is perpendicular to the axis of rotation.
When you look at a cross-section, it reveals the profile of the solid at that particular location. These can be of various shapes, but for solids of revolution, they often appear circular. In this case, each cross-sectional area is represented by a function, commonly denoted as \(A(x)\), where \(x\) is the position along the axis.
Understanding the cross-sectional area is important because it helps calculate the volume of the solid. By integrating these areas along the axis of rotation, you can sum up all "slices" to find the total volume. This integration effectively adds together all the infinitesimally thin disks or washers that form the solid.
Disk and Washer Method
The disk and washer method are two powerful techniques for finding the volume of a solid of revolution. They're both integral-based methods that utilize the concept of cross-sectional area.
  • Disk Method: This is used when the solid's cross-sections are disks. To apply this method, imagine slicing the solid perpendicular to the axis of rotation, forming a series of circular disks stacked along the axis. The formula for the volume of each disk involves its radius and thickness, calculated as \(V = \int_a^b \pi [f(x)]^2 \, dx\).
  • Washer Method: The washer method extends the disk method by accounting for a hollow center, much like an annular ring. If a solid has a hole in the middle, it means each cross-section resembles a washer. The volume is calculated by subtracting the volume of the inner hole from the outer disk, expressed as \(V = \int_a^b \pi ([R(x)]^2 - [r(x)]^2) \, dx\), where \(R(x)\) and \(r(x)\) represent the outer and inner radii, respectively.
These methods are essential tools in calculus for transforming the idea of a solid's volume from theoretical calculations into practical applications.

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Most popular questions from this chapter

Suppose that a hollow tube rotates with a constant angular velocity of \(\omega\) rad/s about a horizontal axis at one end of the tube, as shown in the accompanying figure. Assume that an object is free to slide without friction in the tube while the tube is rotating. Let \(r\) be the distance from the object to the pivot point at time \(t \geq 0,\) and assume that the object is at rest and \(r=0\) when \(t=0 .\) It can be shown that if the tube is horizontal at time \(t=0\) and rotating as shown in the figure, then $$r=\frac{g}{2 \omega^{2}}[\sinh (\omega t)-\sin (\omega t)]$$ during the period that the object is in the tube. Assume that \(t\) is in seconds and \(r\) is in meters, and use \(g=9.8 \mathrm{m} / \mathrm{s}^{2}\) and \(\omega=2 \mathrm{rad} / \mathrm{s}\) (a) Graph \(r\) versus \(t\) for \(0 \leq t \leq 1\) (b) Assuming that the tube has a length of \(1 \mathrm{m},\) approximately how long does it take for the object to reach the end of the tube? (c) Use the result of part (b) to approximate \(d r / d t\) at the instant that the object reaches the end of the tube. (GRAPH CAN'T COPY).

Show that (a) \(\frac{d}{d x}\left[\operatorname{sech}^{-1}|x|\right]=-\frac{1}{x \sqrt{1-x^{2}}}\) (b) \(\frac{d}{d x}\left[\operatorname{csch}^{-1}|x|\right]=-\frac{1}{x \sqrt{1+x^{2}}}\)

True-False Determine whether the statement is true or false. Explain your answer. [In Exercise \(34,\) assume that the (rotated) square lies in the \(x y\) -plane to the right of the \(y\) -axis.] The centroid of an equilateral triangle is the intersection of the medians of the triangle.

Find \(d y / d x\). $$y=\sinh ^{-1}(1 / x)$$

Find \(d y / d x\). $$y=\left(\operatorname{coth}^{-1} x\right)^{2}$$

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