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Chapter 8: Problem 49

The given volume has a horizontal base. Let \(h\) be the height above the base of a slice with thickness \(\Delta h .\) Which of (I)-(IV) approximates the volume of this slice? I. \(\pi \sqrt{16-h^{2}} \Delta h\) II. \(\frac{\pi}{25}(20-h)^{2} \Delta h\) III. \(25 \pi(20-h)^{2} \Delta h\) IV. \(\pi\left(16-h^{2}\right) \Delta h\) V. \(\frac{1}{25}(20-h)^{2} \Delta h\) VI. \(25(20-h)^{2} \Delta h\) A hemisphere of radius \(r=4\).

Short Answer

Expert verified
Option IV, \( \pi (16-h^2) \Delta h \), approximates the volume of the slice.

Step by step solution

01

Understand the Problem

We are finding the volume of a thin slice of a 3D shape, specifically a hemisphere, with radius \( r = 4 \). The slice is parallel to the base and has a thickness of \( \Delta h \). We need to choose the correct formula to approximate this volume.
02

Identify the Shape of the Slice

The problem involves a hemisphere. A slice at height \( h \) with thickness \( \Delta h \) is a disk (horizontal cross-section). The radius of this disk depends on the distance from the center of the hemisphere.
03

Determine the Radius of the Disk

For a hemisphere of radius \( r \), the equation of a circle in 2D is \( x^2 + y^2 = r^2 \). For a slice at height \( h \), the radius of the disk is \( \sqrt{r^2 - h^2} \).
04

Calculate the Area of the Disk

The area of the disk, \( A \), at height \( h \) is given by the formula \( A = \pi r^2 \). Substitute the radius of the disk: \( A = \pi (\sqrt{r^2 - h^2})^2 = \pi (r^2 - h^2) \).
05

Volume of the Slice

The approximate volume \( V \) of the slice is the area of that disk multiplied by the thickness \( \Delta h \). So, \( V = \pi (r^2 - h^2) \Delta h \).
06

Select the Correct Option

Using \( r = 4 \), substitute into the volume formula: \( V = \pi (16 - h^2) \Delta h \). This matches option IV.

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Key Concepts

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

Hemisphere
A hemisphere is essentially half of a sphere. When imagining a sliced orange, each half represents a hemisphere. In geometry, hemispheres are three-dimensional shapes with properties derived from the sphere they belong to.
For the problem at hand, we consider a hemisphere of radius \( r = 4 \).Some important characteristics of a hemisphere include:
  • The base is a flat, circular shape from which the entire form arises.
  • The curved surface extends from the base to the top in a semi-circular manner.
Understanding these properties is crucial when dividing the hemisphere into smaller sections, as each slice at a particular height has a shape dictated by the geometry of the hemisphere.
Disk Area
In this context, when a slice is taken horizontally from a hemisphere, the shape of each slice is a disk. This disk is a two-dimensional, circular shape, similar to a flat plate.To determine the area of the disk slice from the hemisphere, you use the formula:\[ A = \pi r^2 \]However, because the radius of our slice changes with height \( h \) within the hemisphere, it becomes:\[ A = \pi (r^2 - h^2) \]Where:
  • \( r \) is the constant radius of the hemisphere, and
  • \( h \) is the height from the base, influencing the current slice's radius.
This formula helps in calculating the cross-sectional area of the slice which plays a significant role in finding the slice鈥檚 volume.
Thickness
Thickness in this problem refers to how thick the slice of the hemisphere is. It's represented by the symbol \( \Delta h \).This slice thickness is crucial as it impacts the overall volume of the slice.
  • By multiplying the area of the disk (cross-section) by the thickness \( \Delta h \), we find the approximate volume of each slice.
  • This small thickness allows us to treat each slice as having a roughly constant height, enabling us to use the disk area to approximate the volume accurately.
Thus, having a good grasp of what thickness denotes can significantly help in understanding how volume accumulates in 3-dimensional shapes.
Cross-Section
A cross-section is created when you cut through a 3-dimensional object horizontally. In our hemisphere, this results in a circular disk at a particular height \( h \). Cross-sections are incredibly useful for visualizing volume calculations and understanding 3D shapes.Key points about cross-sections in this scenario include:
  • The cross-section's shape depends on where the slice is cut on the hemisphere.
  • For a hemisphere, each cross-section parallel to the base is a disk whose radius changes with height.
Visualizing these cross-sections helps one to see how the hemisphere's volume is constructed from multiple small disks, each corresponding to a cross-section, reinforcing how integrating these disks renders the complete volume of the hemisphere.

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Most popular questions from this chapter

Explain what is wrong with the statement. The circumference of a circle with parametric equations \(x=\cos (2 \pi t), y=\sin (2 \pi t)\) is $$\int_{0}^{2} \sqrt{(-2 \pi \sin (2 \pi t))^{2}+(2 \pi \cos (2 \pi t))^{2}} d t$$

Concern the region bounded by \(y=x^{2}\) \(y=1,\) and the \(y\) -axis, for \(x \geq 0 .\) Find the volume of the solid. The solid whose base is the region and whose crosssections perpendicular to the \(y\) -axis are equilateral triangles.

Write an integral that represents the arc length of the portion of the graph of \(f(x)=-x(x-4)\) that lies above the \(x\) -axis. Do not evaluate the integral.

Concern the region bounded by \(y=x^{2}\) \(y=1,\) and the \(y\) -axis, for \(x \geq 0 .\) Find the volume of the solid. The solid obtained by rotating the region around the \(y\) axis.

(a) Write an integral which represents the circumference of a circle of radius \(r\). (b) Evaluate the integral, and show that you get the answer you expect.
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