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Chapter 7: Problem 14

Finding the Volume of a Solid In Exercises \(13 - 16\) , find the volumes of the solids generated by revolving the region bounded by the graphs of the equations about the given lines. $$y = 2 x ^ { 2 } , \quad y = 0 , \quad x = 2$$ $$\begin{array} { l l } { \text { (a) the } y \text { -axis } } & { \text { (b) the } x \text { -axis } } \\ { \text { (c) the line } y = 8 } & { \text { (d) the line } x = 2 } \end{array}$$

Short Answer

Expert verified
The volumes of the solids obtained by rotating the region about the y-axis, x-axis, line \(y=8\), and line \(x=2\) are \(16\pi\), \(\frac{64}{5}\pi\), \(-24\pi\), and \(\frac{16}{3}\pi\) respectively.

Step by step solution

01

Solve for Rotation around Y-axis

The volume V of a solid that is obtained by rotating a region about the y-axis can be computed using the formula for the volume of a cylindrical shell: \[V=\int_{a}^{b}2\pi xf(x)dx\]. Here, \(f(x) = 2x^2, a = 0, b = 2\). Substituting these into the equation, we get \[V=\int_{0}^{2}2\pi x(2x^{2})dx = \int_{0}^{2}4\pi x^{3}dx\] which results in \(V = \pi \left [ x^4 \right ]_{0}^{2} = 16\pi\).
02

Solve for Rotation around X-axis

The volume V of a solid that is obtained by rotating a region about the x-axis can be computed using the formula for the volume of a disk: \[V=\int_{a}^{b}\pi [f(x)]^2dx\]. Here, \(f(x) = 2x^2, a = 0, b = 2\). Substituting these into the equation, we get \[V = \int_{0}^{2}\pi [2x^{2}]^{2}dx = \int_{0}^{2}\pi 4x^{4} dx\] which results in \(V = \pi \left [ \frac{4}{5} x^{5} \right ]_{0}^{2} = \frac{64}{5}\pi\).
03

Solve for Rotation around Line Y=8

For rotation around the line \(y=8\), the equation becomes \(V = \int_a^b 2 \pi (8 - 2x^2) x dx = \int_0^2 2 \pi (8x - 2x^3) dx\). Simplifying this we get \[V = 2 \pi \left [\int_0^2 x dx - \int_0^2 2 x^3 dx \right ], V = 2 \pi \left [ x^2 \right ]_0^2 - \pi \left [ x^4 \right ]_0^2 = 2 \pi (4 - 16) = -24 \pi\].The negative result indicates that the volume is oriented downwards.
04

Solve for Rotation around the Line X=2

For rotation around line \(x=2\) we have the expression \[V = \int_a^b 2 \pi (2 - x)(2x^2) dx = \int_0^2 2 \pi (4x^2 - 2x^3) dx\]. Simplifying this we get \[V = 2 \pi \left [\int_0^2 4x^2 dx - \int_0^2 2 x^3 dx\right], V = 2 \pi \left [ \frac{4}{3} x^3 \right ]_0^2 - \pi \left [ \frac{1}{2} x^4\right ]_0^2 = 2 \pi (16/3 - 8) = 16/3 \pi\].

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Key Concepts

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

Cylindrical Shells Method
The Cylindrical Shells Method is a technique used to find the volume of a solid of revolution. This method is particularly useful when the solid is generated by revolving a region around an axis that is parallel to the axis of the given function. In essence, imagine slicing the solid into thin cylindrical shells instead of disks.

For example, when rotating a function about the y-axis, each shell orients vertically. The volume of a shell is calculated using the formula: \[ V = \int_{a}^{b} 2 \pi x f(x) \, dx \]Here,
  • \( x \) represents the radius of the shell, which is the distance from the y-axis.
  • \( f(x) \) is the height of the shell.
  • \( dx \) is the thickness of the shell.
This method is advantageous when the function involves complex shapes or irregular bounds, providing a straightforward calculation.
Disk Method
The Disk Method is another popular approach in calculating the volume of solids of revolution. This method comes in handy when the region is revolved around the x-axis or a parallel line. When you look at a solid of revolution, you can visualize it as a series of disks stacked along the axis.

To use the Disk Method, we focus on the following formula:\[ V = \int_{a}^{b} \pi [f(x)]^{2} \, dx \]In this formula:
  • \( \pi [f(x)]^{2} \) represents the area of a single disk.
  • \( dx \) gives us the thickness of each disk.
Calculating each disk's volume and summing them up from some starting point \( a \) to an endpoint \( b \), you get the entire volume. This approach is useful when the solid has a simpler geometry, like those bounded on all sides by given functions.
Axis of Rotation
In integral calculus, the axis of rotation is a critical line about which a two-dimensional region is revolved to create a three-dimensional solid. This axis can be horizontal or vertical, affecting the computation method used.

Choosing an axis of rotation influences whether you will use the disk or cylindrical shells method:
  • The y-axis or lines parallel to it typically involve the Cylindrical Shells Method.
  • The x-axis or lines parallel to it are usually associated with the Disk Method.
To determine the volume correctly, it's essential to have a clear understanding of how shifting the axis of rotation changes the dimensions and orientations of the slices, whether they are disks, washers (a variation of disks), or shells.
Integral Calculus
Integral calculus is a branch of mathematics dealing with integrals and their applications. It primarily helps in finding areas under curves, but more broadly, it's powerful in computing volumes of complex geometrical shapes like solids of revolution.

Integral calculus involves two main types: definite and indefinite integrals. For volumes of revolution:
  • Definite integrals play a crucial role, providing a precise calculation between given bounds \( a \) and \( b \).
  • The integral sums up infinitely small slices or elements鈥攕uch as disks or shells鈥攖o determine the total volume.
With practice, integrating under various circumstances becomes intuitive, leading to mastery in solving real-world problems that involve complex figures and volumes.

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Most popular questions from this chapter

Hydraulic Press In Exercises \(45-48,\) use the integration capabilities of a graphing utility to approximate the work done by a press in a manufacturing process. A model for the variable force \(F\) (in pounds) and the distance \(x\) (in feet) the press moves is given. \(F(x)=1000[1.8-\ln (x+1)] \quad 0 \leq x \leq 5\)

Think About It Consider the equation \(\frac{x^{2}}{9}+\frac{y^{2}}{4}=1\) (a) Use a graphing utility to graph the equation. (b) Write the definite integral for finding the first-quadrant arc length of the graph in part (a). (c) Compare the interval of integration in part \((\mathrm{b})\) and the domain of the integrand. Is it possible to evaluate the definite integral? Explain. (You will learn how to evaluate this type of integral in Section \(8.8 . )\)

Lateral Surface Area of a Cone A right circular cone is generated by revolving the region bounded by \(y=h x / r\) , \(y=h,\) and \(x=0\) about the \(y\) -axis. Verify that the lateral surface area of the cone is \(S=\pi r \sqrt{r^{2}+h^{2}}\)

Modeling Data A pond is approximately circular, with a diameter of 400 feet. Starting at the center, the depth of the water is measured every 25 feet and recorded in the table (see figure). $$\begin{array}{|c|c|c|c|}\hline x & {0} & {25} & {50} \\ \hline \text { Depth } & {20} & {19} & {19} \\ \hline\end{array}$$ $$\begin{array}{|c|c|c|c|}\hline x & {75} & {100} & {125} \\ \hline D e p t h & {17} & {15} & {14} \\ \hline\end{array}$$ $$\begin{array}{|c|c|c|c|}\hline x & {150} & {175} & {200} \\ \hline \text { Depth } & {10} & {6} & {0} \\ \hline\end{array}$$ $$\begin{array}{l}{\text { (a) Use the regression capabilities of a graphing utility to find }} \\ {\text { a quadratic model for the depths recorded in the table. Use }} \\ {\text { the graphing utility to plot the depths and graph the model. }}\end{array}$$ $$\begin{array}{l}{\text { (b) Use the integration capabilities of a graphing utility and }} \\ {\text { the model in part (a) to approximate the volume of water }} \\ {\text { in the pond. }}\end{array}$$ $$\begin{array}{l}{\text { (c) Use the result of part (b) to approximate the number of }} \\ {\text { gallons of water in the pond. (Hint: } 1 \text { cubic foot of water }} \\ {\text { is approximately } 7.48 \text { gallons.) }}\end{array}$$

In Exercises 33-36, (a) use a graphing utility to graph the region bounded by the graphs of the equations, and (b) use the integration capabilities of the graphing utility to approximate the volume of the solid generated by revolving the region about the y-axis. $$y=\sqrt[3]{(x-2)^{2}(x-6)^{2}}, \quad y=0, \quad x=2, \quad x=6$$
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