/*! 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 41 A bead is formed from a sphere o... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 41

A bead is formed from a sphere of radius 5 by drilling through a diameter of the sphere with a drill bit of radius 3 a. Find the volume of the bead. b. Find the volume of the removed portion of the sphere.

Short Answer

Expert verified
Volume of the bead: ~240.85 cubic units; Removed volume: ~282.74 cubic units.

Step by step solution

01

Understand the Problem

We start with a sphere of radius 5 units. A cylindrical hole with radius 3 units is drilled through the sphere, extending across the full diameter. Our goals are to find the volume of the remaining bead and the volume of the cylindrical hole that has been removed.
02

Find the Volume of the Whole Sphere

The formula for the volume of a sphere is given by \(V = \frac{4}{3} \pi r^3\), where \(r\) is the radius of the sphere. Here, \(r = 5\). Thus, the volume of the whole sphere is: \[ V_{sphere} = \frac{4}{3} \pi (5)^3 = \frac{4}{3} \pi \times 125 = \frac{500}{3} \pi \approx 523.6 \text{ cubic units} \]
03

Find the Volume of the Cylinder

The volume of a cylinder is calculated using \(V = \pi r^2 h\), where \(r\) is the radius and \(h\) is the height. The cylinder radius is 3, and the height is the same as the sphere's diameter, which is 10. Therefore, the volume is: \[ V_{cylinder} = \pi (3)^2 \times 10 = 90 \pi \approx 282.74 \text{ cubic units} \]
04

Calculate the Volume of the Bead

The volume of the bead is the remaining volume after the cylindrical section is removed from the sphere. This is computed by subtracting the volume of the cylinder from the volume of the sphere: \[ V_{bead} = \frac{500}{3} \pi - 90 \pi \approx 523.6 - 282.74 \approx 240.85 \text{ cubic units} \]
05

Confirm the Volume of the Removed Cylindrical Portion

We verify that the volume of the removed portion is indeed just the volume of the cylindrical hole we calculated earlier, 90\( \pi\) cubic units, which we found to be approximately 282.74 cubic units.

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Key Concepts

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

Volume of Sphere
Understanding the volume of a sphere is crucial when dealing with geometry and calculus problems involving solids. A sphere is a perfectly symmetrical object in three-dimensional space. It is the set of all points that are equidistant from a given point, called the center. Calculating its volume involves using the radius, which is the distance from the center to any point on the surface.

The formula for the volume of a sphere is:
  • \( V = \frac{4}{3} \pi r^3 \)
Where \( r \) is the radius of the sphere.Let's break it down further:
  • The \( \pi \) represents the constant Pi, approximately 3.14159.
  • The radius is cubed \( (r^3) \), highlighting that it is a three-dimensional measure.
  • Finally, it is multiplied by \( \frac{4}{3} \) to account for the shape of the sphere compared to a cube.
To find the volume of any sphere, simply plug the given radius into this formula.
Volume of Cylinder
Cylinders are common shapes in geometry, characterized by their circular base and straight sides. Calculating the volume of a cylinder is straightforward if you know the radius of its base and its height.

The formula for the volume of a cylinder is:
  • \( V = \pi r^2 h \)
Where:
  • \( r \) is the radius of the base of the cylinder.
  • \( h \) is the height of the cylinder.
  • \( \pi \) again is the constant Pi.
The concept involves finding the area of the circular base, \( \pi r^2 \), and multiplying it by the height, \( h \), to extend that area into a volume. This multiplication by the height translates the area of the base into the three-dimensional space, giving us the volume.
Solid Geometry
Solid geometry deals with three-dimensional objects like spheres, cylinders, and other solid figures. Each object can be defined by its surfaces, edges, and vertices.

When addressing problems in solid geometry, it's essential to understand how different shapes interact in space. For instance, when a cylinder is removed from a sphere, as in the original problem, both shapes have to be understood individually before calculating the results of their interaction.
  • Identify the dimensions and shapes involved.
  • Apply the correct geometric formulas to calculate areas and volumes.
  • Consider how modifications, such as removing a section from a shape, affect its total volume.
Solid geometry problems often require combining basic knowledge of different shapes and their properties to solve complex problems, just like the bead problem where a cylinder is drilled out of a sphere.
Integral Calculus
Integral calculus can be used to find the volume of complex shapes or when the traditional volume formulas do not apply. It is particularly useful in calculating volumes of revolution or when a shape isn't uniform throughout. In the context of this problem, integral calculus helps reinforce how volume subtraction works when calculating the remaining volume of a complex shape. If one were to use integral calculus in such problems, it could involve:
  • Setting up integrals to calculate the volume of complex parts of the object.
  • Subtracting the integral results of the removed part from the whole.
  • Visualizing how integration adds up infinitesimal slices to form a total volume.
Although this problem is solvable through basic geometry, understanding integral calculus is valuable for tackling more complex problems involving irregular shapes or additional variables.

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Most popular questions from this chapter

The graph of \(y=x^{2}\) on \(0 \leq x \leq 2\) is revolved about the \(y\) -axis to form a tank that is then filled with salt water from the Dead Sea (weighing approximately 73 lb/ft \(^{3}\) ). How much work does it take to pump all of the water to the top of the tank?

The base of a solid is the region between the curve \(y=2 \sqrt{\sin x}\) and the interval \([0, \pi]\) on the \(x\) -axis. The cross-sections perpendicular to the \(x\) -axis are a. equilateral triangles with bases running from the \(x\) -axis to the curve as shown in the accompanying figure. b. squares with bases running from the \(x\) -axis to the curve. CANT COPY THE GRAPH

Stretching a rubber band A force of 2 N will stretch a rubber band \(2 \mathrm{cm}(0.02 \mathrm{m}) .\) Assuming that Hooke's Law applies, how far will a \(4-\mathrm{N}\) force stretch the rubber band? How much work does it take to stretch the rubber band this far?

Designing a wok You are designing a wok frying pan that will be shaped like a spherical bowl with handles. A bit of experimentation at home persuades you that you can get one that holds about \(3 \mathrm{L}\) if you make it \(9 \mathrm{cm}\) deep and give the sphere a radius of \(16 \mathrm{cm} .\) To be sure, you picture the wok as a solid of revolution, as shown here, and calculate its volume with an integral. To the nearest cubic centimeter, what volume do you really get? \(\left(1 \mathrm{L}=1000 \mathrm{cm}^{3}\right)\) CANT COPY THE GRAPH

A Bundt cake, well known for having a ringed shape, is formed by revolving around the \(y\) -axis the region bounded by the graph of \(y=\sin \left(x^{2}-1\right)\) and the \(x\) -axis over the interval \(1 \leq x \leq\) \(\sqrt{1+\pi} .\) Find the volume of the cake.
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