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Chapter 6: Problem 10

Let \(R\) be the region bounded by the following curves. Use the shell method to find the volume of the solid generated when \(R\) is revolved about the \(y\) -axis. \(y=1-x^{2}, x=0,\) and \(y=0,\) in the first quadrant

Short Answer

Expert verified
Answer: The volume of the solid is \(\frac{\pi}{2}\).

Step by step solution

01

Identify the intersection points of the curves

To define the limits of integration, find the intersection points of y = 1 - x^2 and y = 0. To do this, set their equations equal: 1 - x^2 = 0 x^2 = 1 Thus, x = 卤1. The region R lies in the first quadrant, so x=1 will be an intersection and the other limit will be x=0 provided by the exercise.
02

Determine the radius and height of each shell

For the shell method, consider a thin infinitesimal vertical strip of length (y = 1 - x^2) between x = 0 and x = 1. The strip will have a height (1 - x^2) and a radius x (the distance from the y-axis). When revolving this strip around the y-axis, it forms a cylindrical shell with volume dV = 2蟺 * radius * height * thickness (length_dy). In this example: Radius: x Height: 1 - x^2 Thickness (length_dy): dx
03

Set up the integral for the volume

Using the shell method formula, the volume of the solid generated by revolving the region R around the y-axis is given by the integral: \(V = \int_{x=a}^{x=b} 2\pi \cdot x \cdot (1 - x^2) \;dx\), where a and b are the limits of integration (intersection points).
04

Evaluate the integral

Using the intersection points as limits of integration, the integral becomes: \(V = \int_{x=0}^{x=1} 2\pi \cdot x \cdot (1 - x^2) \;dx\) To integrate, expand the integrand: \(V = 2\pi \int_{x=0}^{x=1} (x - x^3) \;dx\) Now, integrate with respect to x: \(V = 2\pi \left [ \frac{x^2}{2} - \frac{x^4}{4} \right ]_{x=0}^{x=1}\)
05

Find the volume of the solid

Evaluate the integral at the limits of integration to find the volume of the solid: \(V = 2\pi \left( \frac{1^2}{2} - \frac{1^4}{4} - (\frac{0^2}{2} - \frac{0^4}{4}) \right) = 2\pi \left(\frac{1}{2} - \frac{1}{4}\right) = 2\pi \cdot \frac{1}{4} = \frac{\pi}{2}\) So, the volume of the solid generated by revolving region R around the y-axis is \(\frac{\pi}{2}\).

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Key Concepts

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

Volume of Revolution
Consider a flat region bounded by curves on a plane. When you revolve this region around a line (typically one of the axes), you create a three-dimensional figure called a solid of revolution. To find the volume of revolution, you鈥檇 calculate the space it occupies.

Mathematically, calculus comes into play. By slicing the solid into infinitely thin disks or shells, and adding up the volume of each, you can approximate the volume of the entire shape. Drawing intuition from real life, imagine taking a piece of paper and rolling it up; the resulting cylinder is similar to the cylinders we calculate in the method of cylindrical shells.
Method of Cylindrical Shells
The method of cylindrical shells is particularly powerful when the solid of revolution has a symmetry around an axis other than the one being revolved around. For instance, take a rectangle and revolve it around an outside line parallel to one of its sides. The result? A hollow cylinder.

To find its volume, think of slicing this shape vertically and focus on one thin slice. This slice represents a cylindrical shell. By calculating the volume of this shell and integrating over the range, you can find the total volume. You鈥檒l need the radius (the distance from the rotation axis to the shell) and height (the length of the shell) for the cylindrical volume formula, and integrate.
Integrals in Calculus
One of the central operations in calculus, integrals are about accumulation and are used to calculate areas, volumes, and other quantities that represent totals. The process of integration鈥攚hich involves integral symbols and the concept of the antiderivative鈥攖urns a function into a sum of all its infinitesimal parts.

In the context of calculating volume, integrals take the formula for a small piece of volume and sum it across the entire dimension you're interested in. For our case, this means integrating the volume of each shell from the first to the last, effectively adding them all up to get the total volume.
Solid of Revolution
A solid of revolution is the three-dimensional shape you get when you rotate a two-dimensional shape around a line. The resulting solid has a circular symmetry鈥攖hat鈥檚 why we call it 鈥榬evolution鈥欌�攍ike how a potter's wheel shapes clay.

Fortunately, we can create mathematical descriptions of these shapes and use tools from calculus to find their properties, such as volume. Whether you rotate a function around the x-axis or y-axis will determine the approach you take: disks, washers, or shells. Each 'slice' of the solid can be thought about as a small, manageable piece of the whole鈥攑erfect for calculus to handle with precision.

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Most popular questions from this chapter

Consider the functions \(f(x)=a \sin 2 x\) and \(g(x)=(\sin x) / a,\) where \(a>0\) is a real number. a. Graph the two functions on the interval \([0, \pi / 2],\) for \(a=\frac{1}{2}, 1\) and 2. b. Show that the curves have an intersection point \(x^{*}\) (other than \(x=0)\) on \([0, \pi / 2]\) that satisfies \(\cos x^{*}=1 /\left(2 a^{2}\right),\) provided \(a>1 / \sqrt{2}\) c. Find the area of the region between the two curves on \(\left[0, x^{*}\right]\) when \(a=1\) d. Show that as \(a \rightarrow 1 / \sqrt{2}^{+}\). the area of the region between the two curves on \(\left[0, x^{*}\right]\) approaches zero.

Determine whether the following statements are true and give an explanation or counterexample. a. \(\frac{d}{d x}(\sinh \ln 3)=\frac{\cosh \ln 3}{3}.\) b. \(\frac{d}{d x}(\sinh x)=\cosh x\) and \(\frac{d}{d x}(\cosh x)=-\sinh x.\) c. Differentiating the velocity equation for an ocean wave \(v=\sqrt{\frac{g \lambda}{2 \pi} \tanh \left(\frac{2 \pi d}{\lambda}\right)}\) results in the acceleration of the wave. d. \(\ln (1+\sqrt{2})=-\ln (-1+\sqrt{2}).\) e. \(\int_{0}^{1} \frac{d x}{4-x^{2}}=\frac{1}{2}\left(\operatorname{coth}^{-1} \frac{1}{2}-\cot ^{-1} 0\right).\)

A strong west wind blows across a circular running track. Abe and Bess start running at the south end of the track, and at the same time, Abe starts running clockwise and Bess starts running counterclockwise. Abe runs with a speed (in units of \(\mathrm{mi} / \mathrm{hr}\) ) given by \(u(\varphi)=3-2 \cos \varphi\) and \(\mathrm{Bess}\) runs with a speed given by \(v(\theta)=3+2 \cos \theta,\) where \(\varphi\) and \(\theta\) are the central angles of the runners. a. Graph the speed functions \(u\) and \(v,\) and explain why they describe the runners' speeds (in light of the wind). b. Compute the average value of \(u\) and \(v\) with respect to the central angle. c. Challenge: If the track has a radius of \(\frac{1}{10} \mathrm{mi}\), how long does it take each runner to complete one lap and who wins the race?

Some species have growth rates that oscillate with an (approximately) constant period \(P\). Consider the growth rate function $$N^{\prime}(t)=r+A \sin \frac{2 \pi t}{P}$$ where \(A\) and \(r\) are constants with units of individuals/yr, and \(t\) is measured in years. A species becomes extinct if its population ever reaches 0 after \(t=0\) a. Suppose \(P=10, A=20,\) and \(r=0 .\) If the initial population is \(N(0)=10,\) does the population ever become extinct? Explain. b. Suppose \(P=10, A=20,\) and \(r=0 .\) If the initial population is \(N(0)=100,\) does the population ever become extinct? Explain. c. Suppose \(P=10, A=50,\) and \(r=5 .\) If the initial population is \(N(0)=10,\) does the population ever become extinct? Explain. d. Suppose \(P=10, A=50,\) and \(r=-5 .\) Find the initial population \(N(0)\) needed to ensure that the population never becomes extinct.

Find the volume of the solid generated in the following situations. The region \(R\) in the first quadrant bounded by the graphs of \(y=x\) and \(y=1+\frac{x}{2}\) is revolved about the line \(y=3\).
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