/*! 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 42 A solid is generated by revolvin... [FREE SOLUTION] | 91Ó°ÊÓ

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A solid is generated by revolving the region bounded by \(y=\sqrt{9-x^{2}}\) and \(y=0\) about the \(y\) -axis. A hole, centered along the axis of revolution, is drilled through this solid so that one-third of the volume is removed. Find the diameter of the hole.

Short Answer

Expert verified
The diameter of the hole that removes one third of the original volume of the solid is approximately \(2 \sqrt[3]{9}\).

Step by step solution

01

Compute the volume V of the solid of revolution

The solid is generated by revolving the region bounded by \(y=\sqrt{9-x^{2}}\) (a semi-circle of radius 3) around the \(y\)-axis. The volume of the solid V is thus \(V = \frac{4}{3}\pi (3)^3 = 36\pi\).
02

Determine the volume of the hole

The hole removed from the solid is supposed to contain one third of its initial volume. This is \(\frac{1}{3}V = \frac{1}{3} * 36\pi = 12\pi\).
03

Compute the radius of the hole

The volume of the hole is given by the formula for the volume of a sphere, \(V = \frac{4}{3}\pi r^3\). If we set \(12\pi = \frac{4}{3}\pi r^3\), we can easily solve for \(r\), obtaining \(r = \sqrt[3]{9}\).
04

Determine the hole's diameter

The diameter of the hole (D) is simply double the radius. Therefore, \(D = 2r = 2 \sqrt[3]{9}\).

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Key Concepts

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

Volume calculation
Understanding volume calculation is essential when dealing with solids of revolution. In essence, volume calculation for these solids typically involves mathematical integration.
Using symmetry and geometry, we apply calculus principles, particularly the disk or washer method, to determine amount of space a 3D shape occupies.
In the original exercise, the solid was derived from revolving a semicircle defined by the equation \( y = \sqrt{9-x^{2}} \). The semicircle has a radius of 3, and when revolved around the \( y \)-axis, it forms a sphere.
To find the volume \( V \) of this sphere, use the formula for the sphere's volume:
  • \( V = \frac{4}{3}\pi r^3 \)
  • Where \( r \) corresponds to the radius, which is 3.
  • Ultimately yielding \( 36\pi \), illustrating that the solid's volume is tightly knit to its geometrical boundaries.

Ensuring such calculations align with the geometry's properties is crucial in determining accurate volumes.
Revolving solid
A revolving solid, or solid of revolution, is essentially a three-dimensional shape created by rotating a two-dimensional shape around an axis.
This concept plays a pivotal role in visualizing shapes like cones, cylinders, and spheres.
The original exercise discussed revolved a semicircle around the \( y \)-axis. This process effectively transforms the semicircle into a 3D object—a sphere.
Key characteristics when dealing with revolving solids include:
  • Choosing the correct axis ensures the symmetry and geometry remain consistent.
  • Understanding that the revolution transforms basic shapes into more complex solids.
Revolution helps identify how the 2D shape fills in space as it spins, creating substantial objects integral in engineering and design.
Geometry of holes
The geometry of holes involves cutting out sections or voids within a solid to alter its mass, structure, or function.
In our exercise, a hole was made by removing a cylindrical shape from the solid sphere. This hole's volume was designated to be one-third of the original solid's volume.
Understanding this concept is crucial in fields like architecture and engineering, where precise alterations need to be calculated for optimized structure.
To find the diameter of a hole:
  • Calculate its volume using the solid's original volume. Here, one-third of \( 36\pi \) resulted in \( 12\pi \).
  • Apply the volume formula, solve for radius \( r \) of the hole, and then find the diameter by \( 2r \).

In practice, understanding the geometry of holes allows designers to optimize material usage without compromising the solid structure’s integrity.

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Most popular questions from this chapter

The region bounded by \(y=\sqrt{x}, y=0, x=0\), and \(x=4\) is revolved about the \(x\) -axis. (a) Find the value of \(x\) in the interval \([0,4]\) that divides the solid into two parts of equal volume. (b) Find the values of \(x\) in the interval \([0,4]\) that divide the solid into three parts of equal volume.

\mathrm{\\{} T h i n k ~ A b o u t ~ I t ~ C o n s i d e r ~ t h e ~ e q u a t i o n ~ \(\frac{x^{2}}{9}+\frac{y^{2}}{4}=1\). (a) Use a graphing utility to graph the equation. (b) Set up the definite integral for finding the first quadrant arc length of the graph in part (a). (c) Compare the interval of integration in part (b) and the domain of the integrand. Is it possible to evaluate the definite integral? Is it possible to use Simpson's Rule to evaluate the definite integral? Explain. (You will learn how to evaluate this type of integral in Section 8.8.)

A cylindrical gasoline tank is placed so that the axis of the cylinder is horizontal. Find the fluid force on a circular end of the tank if the tank is half full, assuming that the diameter is 3 feet and the gasoline weighs 42 pounds per cubic foot.

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If the area of the region bounded by the graphs of \(f\) and \(g\) is 1, then the area of the region bounded by the graphs of \(h(x)=f(x)+C\) and \(k(x)=g(x)+C\) is also \(1 .\)

Find the center of mass of the point masses lying on the \(x\) -axis. $$ \begin{aligned} &m_{1}=7, m_{2}=4, m_{3}=3, m_{4}=8 \\ &x_{1}=-3, x_{2}=-2, x_{3}=5, x_{4}=6 \end{aligned} $$

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