/*! 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 19 A round hole of radius \(a\) is ... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 19

A round hole of radius \(a\) is drilled through the center of a solid sphere of radius \(b\) (assume that \(b>a\) ). Find the volume of the solid that remains.

Short Answer

Expert verified
\(V = \frac{4}{3} \pi b^3 - 2\pi a^2 \sqrt{b^2 - a^2}\)

Step by step solution

01

Understand the Problem

We are tasked with finding the remaining volume when a cylindrical hole of radius \(a\) is drilled through a sphere of radius \(b\). The hole is centered at the sphere.
02

Visualize the Geometry

Visualize the sphere with a cylindrical hole. The cylinder has a height spanning through the entire sphere centred along the vertical axis from top to bottom.
03

Volume of the Sphere

The volume \(V_s\) of a complete sphere with radius \(b\) is given by \(V_s = \frac{4}{3} \pi b^3\).
04

Volume of the Cylinder

The volume \(V_c\) of the cylindrical hole with radius \(a\) and height \(h\) is given by \(V_c = \pi a^2 h\). The height \(h\) can be determined using the Pythagorean theorem in 3D, knowing that the cross-section of the sphere cut across the diameter is a circle of radius \(b\), through which the cylinder passes. This gives the height as \(2\sqrt{b^2-a^2}\).
05

Calculate the Volume of the Cylinder

Substitute the height \(h = 2\sqrt{b^2 - a^2}\) into the volume formula for the cylinder: \(V_c = \pi a^2 (2\sqrt{b^2 - a^2}) = 2\pi a^2 \sqrt{b^2 - a^2}\).
06

Find the Remaining Volume

Subtract the volume of the cylindrical hole from the volume of the sphere to obtain the remaining volume. This gives \(V = V_s - V_c = \frac{4}{3} \pi b^3 - 2\pi a^2 \sqrt{b^2 - a^2}\).

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Key Concepts

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

Solid Geometry
Solid geometry is a branch of mathematics that focuses on studying three-dimensional objects and the relationships between them. Key solid shapes include spheres, cylinders, cubes, and pyramids. These shapes have unique properties, such as surface area and volume. Solid geometry allows us to understand and calculate these characteristics which are crucial in fields such as architecture and engineering.

In this exercise, we're interested in a sphere and cylinder. A sphere is a perfectly symmetrical shape with all points on the surface equidistant from its center. A cylinder, on the other hand, consists of two parallel circular bases connected by a curved surface. Understanding these forms helps solve problems involving interactions between different shapes, such as the one in this exercise where a hole is drilled through a sphere.
Volume Calculation
Calculating volume is about determining how much space a three-dimensional object occupies. Each solid shape has its own formula for calculating volume. For example:
  • The volume of a sphere: \( V_s = \frac{4}{3} \pi b^3 \), where \( b \) is the radius of the sphere.
  • The volume of a cylinder: \( V_c = \pi a^2 h \), where \( a \) is the radius of the base and \( h \) is the height.

In our problem, we first calculate the volume of the entire sphere. Then, we find the volume of the cylindrical hole. This requires using geometry concepts to find the height of the cylinder, especially when part of it is inside another shape. Once we have both volumes, subtracting the volume of the cylinder from the sphere gives us the volume of the solid remaining after drilling.
Sphere and Cylinder Intersection
The intersection of a sphere and a cylinder is an interesting geometric problem. In our case, a cylinder is drilled through the center of a sphere. The challenge is to determine how much of the sphere's material remains after removing the cylinder.

To find this remaining volume, one must first understand the geometry of the intersection. The cylinder goes through the sphere along the vertical axis. Its height spans the inside portion of the sphere, which requires calculating using the intersection points between the two shapes. By applying the Pythagorean theorem in this 3D context, we find the height of the cylinder to be \(2\sqrt{b^2 - a^2}\), leveraging the fact that it goes through a sphere's largest cross-sectional diameter.

This approach shows how mathematical tools and geometric reasoning solve complex space problems by understanding shape interactions.

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Most popular questions from this chapter

Sketch the graph of the four-cusped hypocycloid \(x=a \sin ^{3} t, y=a \cos ^{3} t, 0 \leq t \leq 2 \pi,\) and find its length.

Find the volume of the solid generated by revolving about the \(y\) -axis the region bounded by the line \(y=4 x\) and the parabola \(y=4 x^{2}\).

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According to Coulomb's Law, two like electrical charges repel each other with a force that is inversely proportional to the square of the distance between them. If the force of repulsion is 10 dynes \(\left(1\right.\) dyne \(=10^{-5}\) newton \()\) when they are 2 centimeters apart, find the work done in bringing the charges from 5 centimeters apart to 1 centimeter apart.

A volume \(v\) of gas is confined in a cylinder, one end of which is closed by a movable piston. If \(A\) is the area in square inches of the face of the piston and \(x\) is the distance in inches from the cylinder head to the piston, then \(v=A x\). The pressure of the confined gas is a continuous function \(p\) of the volume, and \(p(v)=p(A x)\) will be denoted by \(f(x)\). Show that the work done by the piston in compressing the gas from a volume \(v_{1}=A x_{1}\) to a volume \(v_{2}=A x_{2}\) is $$ W=A \int_{x_{2}}^{x_{1}} f(x) d x $$ Hint: The total force on the face of the piston is \(p(v) \cdot A=\) \(p(A x) \cdot A=A \cdot f(x)\)
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