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Chapter 7: Problem 41

\(39-41=\) Use cylindrical shells to find the volume of the solid. A right circular cone with height \(h\) and base radius \(r\)

Short Answer

Expert verified
The volume of the cone is \( \frac{1}{3}\pi r^2 h \).

Step by step solution

01

Identify the Problem

We are tasked to find the volume of a right circular cone with height \( h \) and base radius \( r \) using cylindrical shells.
02

Set Up the Integral for Cylindrical Shells

The volume of the solid can be calculated using the formula for the volume of a solid of revolution with cylindrical shells:\[V = \int_a^b 2\pi x \, f(x) \, dx\]For a cone, we consider rotating the line \( y = \frac{h}{r}x \) from \( x = 0 \) to \( x = r \) around the y-axis, as the y-axis serves as the axis of revolution.
03

Express Function in Terms of x

Given the equation of the line forming the slant height of the cone is \( y = \frac{h}{r}x \), \( f(x) \) is therefore the same \( y \) value which represents height of the shell at a given \( x \). Thus, the integrand function in terms of \( x \) is: \(2\pi x \left( \frac{h}{r}x \right) = 2\pi \frac{h}{r} x^2\)
04

Establish Limits of Integration

The limits of integration represent the range of \( x \) over which the shells are integrated. As the base of the cone extends from the y-axis to the edge at x = r, the limits of integration are from 0 to \( r \).
05

Evaluate the Integral

Integrate the function from 0 to \( r \): \[V = \int_0^r 2\pi \frac{h}{r} x^2 \, dx\]This integrates to\[V = 2\pi \frac{h}{r} \left[ \frac{x^3}{3} \right]_0^r\]Calculating further, we have\[V = 2\pi \frac{h}{r} \left( \frac{r^3}{3} \right) = 2\pi \frac{h}{r} \cdot \frac{r^3}{3} = \frac{2\pi hr^2}{3}\]
06

Final Formula for Volume

The volume \( V \) of the cone using cylindrical shells is derived as \[V = \frac{1}{3}\pi r^2 h\]This matches the general volume formula for a cone as expected.

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Key Concepts

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

Volume of a Cone
Understanding the volume of a cone is important for various calculations in mathematics and engineering. A cone is a three-dimensional geometric figure with a circular base and a pointed top, or apex. Its volume can be thought of as the amount of space it occupies.
To find the volume, we use the formula:
  • The formula for the volume of a cone is: \[ V = \frac{1}{3} \pi r^2 h \]
  • Here, \( r \) is the radius of the base and \( h \) is the height measured from the base to the tip.
This formula makes intuitive sense, as it states that the volume of a cone is one-third the volume of a cylinder with the same base radius and height. This is because a cone is essentially a pointed homogeneous distribution of space that narrows to an apex. When calculating using the cylindrical shell method, this outcome is confirmed as we deconstruct the cone into many infinitesimally small cylindrical shells and integrate their volumes to calculate the total volume.
Integration Techniques
Integration is a powerful mathematical tool used to solve a wide range of problems, especially when calculating areas under curves and volumes of solids. When dealing with a problem that involves integration, such as finding the volume of a cone, you often need to apply different techniques.
In this specific problem, we use the method of cylindrical shells. Here鈥檚 how it steps through:
  • Identify the shape and understand the problem, which in this case is the cone.
  • Set up an integral to describe the volume using a shell approach. The formula is: \[ V = \int_a^b 2\pi x \, f(x) \, dx \]
  • Express the function \( f(x) \) in terms of \( x \), which describes the height of each shell. For our cone, \( f(x) = \frac{h}{r}x \).
  • Determine the correct limits of integration. For the cone, it's from \( x = 0 \) to \( x = r \).
  • Finally, evaluate the integral to find the total volume.
Mastering these steps allows you to tackle a wide array of problems using integration techniques. The integration process for the volume of our cone shows how breaking a problem into small, manageable steps allows for a comprehensive solution.
Solid of Revolution
The concept of a solid of revolution is fundamental in understanding how to use integration to find volumes of complex shapes. When a two-dimensional shape is revolved around an axis, the volume of the resulting three-dimensional solid can often be calculated.
For our cone exercise:
  • The line forming the slant height of the cone is revolved around the y-axis. This creates the cone shape from the original flat shape.
  • To calculate the volume of this solid of revolution, cylindrical shells are used. These are thin, hollow cylinders stacked from one end of the solid to the other.
  • The volume of each shell is proportional to its radius and height. Summing the volumes of these shells using integration gives the overall volume.
Using cylindrical shells for solids of revolution involves visualizing how a flat line can revolve to fill space volumetrically. This technique allows you to consider the contribution of every infinitesimal segment across the rotation, ensuring an accurate volume measurement.

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Most popular questions from this chapter

\(33-38=\) The region bounded by the given curves is rotated about the specified axis. Find the volume of the resulting solid by any method. $$x=(y-3)^{2}, x=4 ; \quad \text { about } y=1$$

Sketch the region bounded by the curves, and visually estimate the location of the centroid. Then find the exact coordinates of the centroid. \(y=\sin x, \quad y=0, \quad 0 \leqslant x \leqslant \pi\)

Some of the pioneers of calculus, such as Kepler and Newton, were inspired by the problem of finding the volumes of wine barrels. (In fact Kepler published a book Stereometria doliorum in 1715 devoted to methods for finding the volumes of barrels.) They often approximated the shape of the sides by parabolas. (a) A barrel with height \(h\) and maximum radius \(R\) is constructed by rotating about the \(x\) -axis the parabola \(y=R-c x^{2},-h / 2 \leqslant x \leqslant h / 2,\) where \(c\) is a positive constant. Show that the radius of each end of the barrel is \(r=R-d,\) where \(d=c h^{2} / 4\) (b) Show that the volume enclosed by the barrel is $$ V=\frac{1}{3} \pi h\left(2 R^{2}+r^{2}-\frac{2}{5} d^{2}\right) $$

The table shows values of a force function \(f(x),\) where \(x\) is measured in meters and \(f(x)\) in newtons. Use the Midpoint Rule to estimate the work done by the force in moving an object from \(x=4\) to \(x=20\) . $$\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline x & {4} & {6} & {8} & {10} & {12} & {14} & {16} & {18} & {20} \\ \hline f(x) & {5} & {5.8} & {7.0} & {8.8} & {9.6} & {8.2} & {6.7} & {5.2} & {4.1} \\ \hline\end{array}$$

\(29-32=\) Sketch the direction field of the differential equation. Then use it to sketch a solution curve that passes through the given point. $$y^{\prime}=y+x y, \quad(0,1)$$
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