/*! 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 47 Consider the region bounded by \... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Consider the region bounded by \(y=e^{x}\) the \(x\) -axis, and the lines \(x=0\) and \(x=1 .\) Find the volume of the following solids. The solid obtained by rotating the region about the horizontal line \(y=-3\).

Short Answer

Expert verified
The volume of the solid is \( \pi (e^2 + 6e - 7) \).

Step by step solution

01

Understand the Region

The given region is bounded by the curve \( y = e^x \), the \( x \)-axis, and the vertical lines \( x = 0 \) to \( x = 1 \). We need to visualize this region on the coordinate plane before performing any operation.
02

Identify the Method of Rotation

Since the region is rotated about the horizontal line \( y = -3 \), we use the method of cylindrical shells to find the volume. The axis of rotation is not at the boundary of the region, so the washer method is unsuitable.
03

Express the Radius and Height

For the cylindrical shell method, the radius of each shell is the distance from \( y = -3 \) to \( y = e^x \), which is \( (e^x + 3) \), and the height of the shell is determined by the function \( y = e^x \). The shell height is thus \( e^x \).
04

Setup the Integral for Volume

The volume \( V \) is calculated using the cylindrical shell formula:\[ V = 2\pi \int_{a}^{b} (radius)(height) \, dx \]In this case, \( a = 0 \) and \( b = 1 \). Substitute \( radius = (e^x + 3) \) and \( height = e^x \):\[ V = 2\pi \int_{0}^{1} (e^x + 3)e^x \, dx \]
05

Simplify and Integrate

Simplify the integrand:\[ (e^x + 3) e^x = e^{2x} + 3e^x \]The volume integral becomes:\[ V = 2\pi \int_{0}^{1} (e^{2x} + 3e^x) \, dx \]Compute the integral:\[ V = 2\pi \left[ \left( \frac{1}{2}e^{2x} \right) + 3e^x \right]_0^1 \]
06

Evaluate the Definite Integral

Evaluate the integral from 0 to 1:\[ V = 2\pi \left[ \left( \frac{1}{2}e^{2(1)} + 3e^{1} \right) - \left( \frac{1}{2}e^{2(0)} + 3e^{0} \right) \right] \]\[ = 2\pi \left[ \left( \frac{1}{2}e^2 + 3e \right) - \left( \frac{1}{2} + 3 \right) \right] \]\[ = 2\pi \left[ \frac{1}{2}e^2 + 3e - \frac{7}{2} \right] \]
07

Calculate the Final Volume

Simplify and calculate the final answer:\[ V = 2\pi \left( \frac{1}{2}e^2 + 3e - 3.5 \right) \]Distribute the \( 2\pi \):\[ V = \pi (e^2 + 6e - 7) \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Volume of Solids
In the world of geometry, finding the volume of solids, especially those with complex shapes, can be quite interesting! One special technique for finding volumes is through integration, specifically by using the method of cylindrical shells. This technique is best used when the solid is generated by rotating a region around an axis that is parallel to the boundaries of the region, but not directly at the edge.
This particular exercise involves a region defined by an exponential curve and bounded by vertical lines. When this region is rotated about a horizontal line that is below the region, it creates a donut-like shape. The key is to understand that each slice of the region turns into a 'shell' when rotated around an axis. These shells, which resemble 'hollow cylinders', can have their volumes added up to give the total volume of the solid.
The formula for calculating the volume using cylindrical shells is:
  • \[ V = 2\pi \int_{a}^{b} (\text{radius})(\text{height}) \, dx \]
This formula sums up the volumes of the cylindrical shells from position \(a\) to \(b\). By correctly identifying the radius, which is the distance to the axis of rotation, and the height, which is perpendicular to the axis, you're well-equipped to find the total volume of your solid.
Exponential Functions
Exponential functions are a cornerstone in mathematics, defined by their constant base raised to a variable exponent. An important characteristic of these functions is their rapid rate of growth, a property clearly displayed in the function \( y = e^x \) used in this exercise. The base \( e \) is a mathematical constant approximately equal to 2.71828 and is the foundation of natural logarithms.
The curve of the exponential function, \( y = e^x \), is upward-sloping, meaning it increases as the value of \( x \) increases. This feature plays a crucial role in the region’s shape to be used in the volume calculation. For example, between \( x = 0 \) and \( x = 1 \), the function creates a curve that is steeply sloped, which significantly affects the nature of the solids rotor to calculate their volumes.
Exponential functions are often used in various applications, such as modeling population growth, radioactive decay, and interest calculations in economics. Understanding how they behave and interact with other mathematical elements is key to mastering both theoretical and applied uses.
Definite Integrals
The definite integral is a fundamental concept in calculus that allows us to calculate the accumulated quantity, such as the area under a curve or the total volume of a solid. A definite integral has specific limits of integration, denoted by \(a\) and \(b\), which define the region over which the integration occurs.
In this exercise's context, the definite integral is used to sum up the volumes of infinitesimally thin cylindrical shells, thus calculating the total volume of the rotated solid. The integration bounds are determined by the boundaries of the region on the x-axis, which in this problem are \( x=0 \) and \( x=1 \).
The integral used in the method of cylindrical shells for this exercise is:
  • \[ \int_{0}^{1} (e^{2x} + 3e^x) \, dx \]
Solving this integral involves applying the basic rules of integration to each term, and then evaluating the result at the upper and lower bounds of the interval. This process finally gives the total volume when you apply the multiplication by \( 2\pi \), resulting from the method formula. This combination of solving for an integral and applying it, in reality, is what makes calculus so powerful!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In \(1980,\) before the unification of Germany in 1990 and the introduction of the Euro, West Germany made a loan of 20 billion Deutsche Marks to the Soviet Union, to be used for the construction of a natural gas pipeline connecting Siberia to Western Russia, and continuing to West Germany (Urengoi-Uschgorod- Berlin). Assume that the deal was as follows: In \(1985,\) upon completion of the pipeline, the Soviet Union would deliver natural gas to West Germany, at a constant rate, for all future times. Assuming a constant price of natural gas of 0.10 Deutsche Mark per cubic meter, and assuming West Germany expects \(10 \%\) annual interest on its investment (compounded continuously), at what rate does the Soviet Union have to deliver the gas, in billions of cubic meters per year? Keep in mind that delivery of gas could not begin until the pipeline was completed. Thus, West Germany received no return on its investment until after five years had passed. (Note: A more complex deal of this type was actually made between the two countries.)

Are the statements true or false? Give an explanation for your answer. The population density in a circular city of radius \(2 \mathrm{de}-\) pends on the distance \(r\) from the center by \(f(r)=\) \(10-3 r,\) so that the density is greatest at the center. Then the population of the inner city, \(0 \leq r \leq 1,\) is greater than the population of the suburbs, \(1 \leq r \leq 2\).

In May \(1991,\) Car and Driver described a Jaguar that sold for \(\$ 980,000 .\) At that price only 50 have been sold. It is estimated that 350 could have been sold if the price had been \(\$ 560,000 .\) Assuming that the demand curve is a straight line, and that \(\$ 560,000\) and 350 are the equilibrium price and quantity, find the consumer surplus at the equilibrium price.

give an example of: Supply and demand curves where producer surplus is smaller than consumer surplus.

Give an example of: A function \(f(x)\) whose graph passes through the points (0,0) and (1,1) and whose arc length between \(x=0\) and \(x=1\) is greater than \(\sqrt{2}\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.